- Read the constant rate from the vertical axis. - Identify the length of the time interval. - Multiply the rate by the elapsed time.

1. The transfer rate is constant at \(200\,\text{kg/min}\). 2. The accumulated mass is the rectangular area under the rate graph: \((200\,\text{kg/min})(2.5\,\text{min})=500\,\text{kg}\).

\(500\,\text{kg}\)

- Read the constant rate from the vertical axis. - Find the altitude change shown on the horizontal axis. - Multiply the rate by the altitude interval.

1. The magnitude of the decrease rate is constant at \(5\,\text{K/km}\). 2. Over an altitude change of \(4\,\text{km}\), the temperature decrease is \(5\,\frac{\text{K}}{\text{km}}\cdot4\,\text{km}=20\,\text{K}\).

The temperature decreases by \(20\,\text{K}\).

- Identify what quantity the rate measures before integrating. - Think about how rate units combine with the differential's time unit. - The integral represents accumulation over the stated interval, not an instantaneous rate.

1. Integrating a production rate over time accumulates the number of packages produced during that time interval. 2. The rate unit packages per hour multiplied by the time unit hours gives packages.

It represents the number of packages produced from hour \(2\) through hour \(5\), measured in packages.

- The signed area under a net-rate graph gives the change in the accumulated amount. - Treat the two constant-rate time intervals separately. - Apply the net change to the initial queue size.

1. From \(t=0\) to \(t=3\), the queue changes by \((2)(3)=6\) jobs. 2. From \(t=3\) to \(t=5\), it changes by \((-1)(2)=-2\) jobs. 3. The net change is \(4\) jobs, so the queue has \(8+4=12\) jobs at \(t=5\).

\(12\) jobs

- A rate tells you the change during one unit of time. - Multiply the signed rate by the time interval. - Decide whether a negative result means an increase or a decrease.

1. The change is rate multiplied by duration: \((-5)\cdot2=-10\) sheets. 2. The negative sign shows that the paper supply fell by \(10\) sheets.

\(-10\) sheets. There are \(10\) fewer sheets than at the start.

- Probability is represented by area, not by the curve's height alone. - Find the area of the rectangle above the requested interval.

1. The region from \(0\) to \(1\) is a rectangle of width \(1\) and height \(\frac12\). 2. Its area, and therefore the probability, is \(1\cdot\frac12=\frac12\).

\(\frac12\), or \(50\%\).

- Integrating a growth rate gives accumulated growth. - Find an antiderivative of the rate. - Use the initial population to determine the constant. - Check the units of the integral.

1. An antiderivative of \(r(t)\) is \(R(t)=t^3+20t^2\). 2. \(\int_0^4r(t)\,\text{d}t=[t^3+20t^2]_0^4=384\). 3. The integral is the increase in the number of bacteria during the first \(4\) hours. 4. Since \(N'(t)=r(t)\), \(N(t)=t^3+20t^2+C\). Using \(N(0)=1000\) gives \(C=1000\). 5. Thus, \(N(t)=t^3+20t^2+1000\).

a) \(384\) bacteria; this is the increase during the first \(4\) hours. b) \(N(t)=t^3+20t^2+1000\)

- Integrate the material-transfer rate over the full run. - Use \(0\) and \(12\) as the limits of integration. - Convert grams to kilograms at the end.

1. The total mass in grams is \(\int_0^{12}(120t-10t^2)\,\mathrm{d}t\). 2. An antiderivative is \(60t^2-\frac{10}{3}t^3\). 3. Therefore, \(\left[60t^2-\frac{10}{3}t^3\right]_0^{12}=8640-5760=2880\,\text{g}\). 4. Since \(1000\,\text{g}=1\,\text{kg}\), \(2880\,\text{g}=2.88\,\text{kg}\).

\(2.88\,\text{kg}\)

- Treat accumulated volume as area under the flow-rate graph. - Split the interval where the shape of the graph changes. - Use the graph's scale to read the needed heights exactly.

1. From \(t=1\) to \(t=2\), the region under the graph is a trapezoid with heights \(0.2\) and \(0.4\) liters per second and width \(1\) second, so its area is \(0.3\,\text{L}\). 2. From \(t=2\) to \(t=3\), the rate is constant at \(0.4\,\text{L/s}\), giving a rectangular area of \(0.4\,\text{L}\). 3. The delivered volume is \(0.3+0.4=0.7\,\text{L}\).

\(0.7\,\text{L}\)

- Break the area under the graph into familiar geometric shapes. - Accumulated data has units of rate multiplied by time. - For an instantaneous rate, read the graph's vertical value at the requested time rather than finding an area.

1. From \(0\) to \(1\), the area is a triangle with base \(1\,\text{s}\) and height \(0.4\,\text{MB/s}\), so it represents \(0.2\,\text{MB}\). 2. From \(1\) to \(3\), the area is a trapezoid with parallel sides \(0.4\) and \(0.2\,\text{MB/s}\) and width \(2\,\text{s}\), so it represents \(0.6\,\text{MB}\). 3. The total data received is \(0.2+0.6=0.8\,\text{MB}\). 4. Reading the graph at \(t=2\) gives \(r(2)=0.3\,\text{MB/s}\). Integrating a rate over time produces megabytes, while an instantaneous rate remains in megabytes per second.

a) \(0.8\,\text{MB}\) b) \(0.3\,\text{MB/s}\); part a is an accumulated amount, while part b is an instantaneous rate.

- Start with the initial reservoir volume and add the net accumulated inflow. - Use a different integration variable inside the accumulation integral if you define \(V(t)\). - Check that integrating cubic meters per day over days produces cubic meters.

1. The amount in the reservoir equals the initial amount plus accumulated inflow: \(V(t)=5000+\int_0^t(-0.25u^4+4u^3)\,\text{d}u\). 2. Evaluating the integral gives \(V(t)=5000-\frac{t^5}{20}+t^4\). 3. Therefore, \(V(12)=5000-\frac{12^5}{20}+12^4=13{,}294.4\,\text{m}^3\).

\(V(t)=5000-\frac{t^5}{20}+t^4\), and \(V(12)=13{,}294.4\,\text{m}^3\).

- Use Hooke's law with the observed force and stretch to determine the spring constant. - Convert every stretch distance to meters before integrating. - Work is the definite integral of force with respect to displacement.

1. Convert \(6\,\text{cm}\) to \(0.06\,\text{m}\). Hooke's law gives \(k=\frac{15}{0.06}=250\,\text{N/m}\). 2. The work from equilibrium to \(0.10\,\text{m}\) is \(\int_0^{0.1}250s\,\text{d}s=1.25\,\text{J}\). 3. The additional work from \(0.10\) to \(0.20\,\text{m}\) is \(\int_{0.1}^{0.2}250s\,\text{d}s=3.75\,\text{J}\).

a) \(250\,\text{N/m}\) b) \(1.25\,\text{J}\) c) \(3.75\,\text{J}\)

- Write work as the integral of force with respect to displacement. - Use the power rule for integration. - When substituting \(3a\), cube the entire expression. - Simplify the resulting terms.

1. Work is the definite integral of force: \(W=\int_a^{3a}cs^2\,\text{d}s\). 2. Evaluate the integral: \(W=\left[\frac{c}{3}s^3\right]_a^{3a}=\frac{c}{3}(27a^3-a^3)=\frac{26}{3}ca^3\).

\(W=\frac{26}{3}ca^3\)

- Integrate the production rate to obtain the total-production function. - Use the initial completed quantity to determine the constant. - Substitute the requested time into the function.

1. Integrating the production rate gives \(N(t)=-0.2t^3+3t^2+4t+C\). 2. Since \(N(0)=50\), \(C=50\). Thus, \(N(t)=-0.2t^3+3t^2+4t+50\). 3. \(N(5)=-0.2(5)^3+3(5)^2+4(5)+50=120\) components.

a) \(N(t)=-0.2t^3+3t^2+4t+50\) b) \(120\) components

- Integrating power over time gives electrical energy. - For average power, divide total accumulated energy over the interval by the interval length. - Track the distinction between kilowatts and kilowatt-hours.

1. Energy is the integral of power, so \(\int_0^6(-t^2+12t)\,\text{d}t=144\,\text{kWh}\). 2. The average power is \(\frac{1}{12}\int_0^{12}(-t^2+12t)\,\text{d}t=24\,\text{kW}\).

1) \(144\,\text{kWh}\) 2) \(24\,\text{kW}\)

- Integrate the polynomial term by term. - Apply the Fundamental Theorem of Calculus at the two stated times. - An integral of a rate over time has units of accumulated volume. - On the graph, identify the boundaries of the region represented by the integral without assuming it is already shaded.

1. Integrating term by term gives \(F(t)=-\frac16t^3+2t^2+2t+C\). 2. The accumulated volume is \(\int_1^7f(t)\,\text{d}t=F(7)-F(1)=51\,\text{m}^3\). 3. The integral is the total volume entering the basin from hour \(1\) through hour \(7\). On the graph, it is the region under \(f\) and above the \(t\)-axis between \(t=1\) and \(t=7\).

a) \(F(t)=-\frac16t^3+2t^2+2t+C\) b) \(51\,\text{m}^3\) c) The accumulated inflow from \(t=1\) to \(t=7\), represented by the area under the rate graph over that interval.

- Multiply the horizontal and vertical units to determine the integral's units. - Break the region into simple geometric shapes. - Add the areas of the two triangles and the rectangle.

1. Mathematically, the integral is the area under the graph of \(f\) from \(t=0\) to \(t=10\). In context, it is the total number of visitors who enter during the first \(10\) minutes. 2. Divide the region into two triangles and one rectangle: \(\frac{1}{2}\cdot4\cdot20=40\), \(4\cdot20=80\), and \(\frac{1}{2}\cdot2\cdot20=20\). Therefore, \(\int_0^{10}f(t)\,\text{d}t=140\) visitors.

a) It is the area under the rate graph and represents the total number of visitors entering during the first \(10\) minutes. b) \(140\) visitors

- Interpret area under a power-time graph in physical units. - Determine the energy represented by one full grid square. - Combine partial squares rather than using only completely filled squares.

1. Electrical energy is the area under a power-versus-time graph. 2. Each grid square is \(1\,\text{h}\times10\,\text{kW}=10\,\text{kWh}\). 3. The region from \(t=4\) to \(t=8\) contains about \(25\) grid-square equivalents, giving an estimate of about \(250\,\text{kWh}\).

About \(250\,\text{kWh}\); estimates reasonably close to this value are supported by the grid.

- Break the full area under the graph into simple geometric pieces. - For the accumulation function, first determine the equation of the first line segment from its displayed endpoints. - Integrate only over the interval where that first segment applies.

1. The total area is a triangle of area \(10\), a rectangle of area \(20\), and a second triangle of area \(10\). Thus \(40\) bicycles are returned. 2. On \([0,1]\), the graph is the line through \((0,0)\) and \((1,20)\), so \(r(s)=20s\). 3. Therefore, \(B(t)=\int_0^t20s\,\text{d}s=10t^2\) for \(0\le t\le1\).

a) \(40\) bicycles b) \(B(t)=10t^2\) for \(0\le t\le1\)

- Check the physical nonnegativity condition separately from the total-volume condition. - Integrate each proposed rate over the full interval. - A model must satisfy both conditions; satisfying only the integral condition is not enough.

1. A valid inflow model must be nonnegative throughout the interval and have integral \(1\,\text{L}\) over \([0,2]\). 2. For \(f\), the minimum value on \([0,2]\) is \(f(2)=0.25>0\), so the rate is nonnegative. But \(\int_0^2(-0.5t+1.25)\,\text{d}t=1.5\,\text{L}\), so the stated total is wrong. 3. For \(g\), \(g(0)=-1<0\), so it fails the nonnegative-rate condition. Although \(\int_0^2(1.5t-1)\,\text{d}t=1\,\text{L}\), the model still fails the full claim.

Neither model satisfies both conditions. The function \(f\) is nonnegative but delivers \(1.5\,\text{L}\), while \(g\) has integral \(1\,\text{L}\) but is negative on part of the interval.

- Volume is represented by signed area under the flow-rate graph. - Break each requested region into rectangles or trapezoids. - For the interval from \(3\) to \(4\), first read the rate at \(t=4\) from the decreasing segment. - Add adjacent interval volumes for the larger interval.

1. The area over \([0,1]\) is a rectangle: \((1)(1)=1\,\text{L}\). 2. The area over \([1,3]\) is a trapezoid: \(\frac{1+4}{2}(2)=5\,\text{L}\). 3. The line decreases from \((3,4)\) to \((5,0)\), so \(q(4)=2\). The trapezoid over \([3,4]\) has area \(\frac{4+2}{2}(1)=3\,\text{L}\). 4. Adding the adjacent volumes from \(1\) to \(3\) and from \(3\) to \(4\) gives \(5+3=8\,\text{L}\).

a) \(1\,\text{L}\) b) \(5\,\text{L}\) c) \(3\,\text{L}\) d) \(8\,\text{L}\)

- Stored mass increases where the displayed rate is positive and decreases where it is negative. - Look for the time where the sign changes from positive to negative. - Compare the two halves of the graph by rotational symmetry, paying attention to signed area.

1. The rate is positive before \(t=4\) and negative after \(t=4\), so the hopper mass increases until \(t=4\) and decreases afterward. Therefore, the mass is greatest at \(t=4\,\text{min}\). 2. The graph has 180-degree rotational symmetry about \((4,0)\). Thus the negative signed area on \([4,8]\) has the same magnitude as the positive signed area on \([0,4]\). 3. The two signed areas cancel, so the net change over \([0,8]\) is \(0\,\text{kg}\).

a) \(t=4\,\text{min}\) b) \(0\,\text{kg}\)

- Integrate the inflow rate over the full six-minute interval. - Keep the trigonometric antiderivative exact until the endpoint evaluation is complete. - Check that liters per minute multiplied by minutes gives liters.

1. The accumulated inflow is \(\int_0^6\left(10\sin\left(\frac{\pi}{6}t\right)+15\right)\,\text{d}t\). 2. An antiderivative is \(-\frac{60}{\pi}\cos\left(\frac{\pi}{6}t\right)+15t\). 3. Evaluating from \(0\) to \(6\) gives \(90+\frac{120}{\pi}\,\text{L}\approx128.20\,\text{L}\). 4. The integral represents the total volume entering the tank over the first \(6\) minutes.

\(90+\frac{120}{\pi}\,\text{L}\approx128.20\,\text{L}\); it is the accumulated inflow during the first \(6\) minutes.

- Estimate the accumulated inflow as area under the rate graph. - Determine the volume represented by one grid square before counting. - Multiply the estimated volume by the treatment cost per cubic meter only after finding the area.

1. The incoming volume is the area under the rate graph from \(t=4\) to \(t=12\). 2. Each grid square represents \((1\,\text{h})(1\,\text{m}^3/\text{h})=1\,\text{m}^3\). Counting whole and partial squares gives about \(57\,\text{m}^3\). 3. The treatment cost is therefore about \((57)(\$1.50)=\$85.50\), which rounds to approximately \(\$86\).

Approximately \(\$86\)

- Use the supplied antiderivative at the two endpoints. - The units of a concentration-time integral are concentration multiplied by time. - Interpret the definite integral as cumulative exposure rather than as the concentration at one instant.

1. By the Fundamental Theorem of Calculus, \(\int_0^4f(t)\,\text{d}t=F(4)-F(0)\). 2. This gives \(16-48e^{-2}\,\text{mg}\cdot\text{h}/\text{L}\approx9.50\,\text{mg}\cdot\text{h}/\text{L}\). 3. The value is the area under the concentration-time curve during the first \(4\) hours, which represents cumulative drug exposure over that period.

\(16-48e^{-2}\,\text{mg}\cdot\text{h}/\text{L}\approx9.50\,\text{mg}\cdot\text{h}/\text{L}\); this is the cumulative drug exposure during the first \(4\) hours.

- Identify the geometric shape between the line and the horizontal axis. - Read the growth rates at the beginning and end of the interval. - Use the area formula for a trapezoid.

1. The accumulated increase in diameter is the area under the rate graph from \(t=0\) to \(t=3\). 2. The region is a trapezoid with parallel sides \(2\,\text{mm/week}\) and \(8\,\text{mm/week}\) and width \(3\,\text{weeks}\). 3. Its area is \(\frac{2+8}{2}(3)=15\,\text{mm}\).

The diameter increases by \(15\,\text{mm}\).

- Interpret the area under the rate graph as accumulated mass. - Identify the base and height of the triangular region. - Use the area formula for a triangle.

1. The accumulated mass is the area under the rate graph from \(t=0\) to \(t=2\). 2. This region is a right triangle with base \(2\,\text{min}\) and height \(400\,\text{kg/min}\). 3. The mass is \(\frac{1}{2}(2)(400)=400\,\text{kg}\).

\(400\,\text{kg}\)

- Compare the heights of the rate curves to find when one pipe is flowing faster. - The constant-rate pipe produces a rectangular area under the graph. - To compare totals, compare the full areas over the same ten-minute interval, not just the peak heights. - Estimate the curved area by grouping the grid squares or using several trapezoids.

1. The changing-rate curve lies above the constant-rate line between the intersections, at about \(t=1.9\) and \(t=8.1\) minutes. 2. The constant rate is \(5\,\text{L/min}\). Its area over \(10\) minutes is \(5\cdot10=50\,\text{L}\). 3. The area under the changing-rate curve is approximately \(63\,\text{L}\), compared with \(50\,\text{L}\) for the constant-rate pipe. The first pipe therefore carries about \(13\,\text{L}\) more water.

a) Approximately \(1.9<t<8.1\) minutes b) \(50\,\text{L}\) c) The changing-rate pipe: its area is about \(63\,\text{L}\), compared with the constant-rate pipe's \(50\,\text{L}\). That is about \(13\,\text{L}\) more.

- Interpret processed quantity as area under the rate graph. - The full interval and the first hour each form triangular regions. - Split the third interval at the graph's peak and use trapezoid areas.

1. The full region is a triangle with base \(3\) hours and height \(16\) boards per hour, so \(24\) boards are processed. 2. From \(0\) to \(1\), the triangular area is \(\frac{1}{2}(1)(8)=4\) boards. 3. From \(1.5\) to \(2\), the trapezoid area is \(\frac{12+16}{2}(0.5)=7\) boards. From \(2\) to \(2.5\), it is \(\frac{16+8}{2}(0.5)=6\) boards, for a total of \(13\) boards.

1) \(24\) boards 2) \(4\) boards 3) \(13\) boards are processed from \(t=1.5\) to \(t=2.5\).

- Determine the volume represented by one grid rectangle. - Use symmetry about the graph's center to relate intervals on opposite sides. - Compare areas under the curve rather than only individual rate values.

1. Each grid rectangle represents \((0.2\,\text{min})(1\,\text{L/min})=0.2\,\text{L}\). Counting whole and partial rectangles from \(0\) to \(0.5\) gives about \(1.6\,\text{L}\). 2. The graph is symmetric about \(t=1\), so the area on \([0.5,1]\) equals the area on \([1,1.5]\). The curve lies higher on \([1,1.5]\) than on \([1.5,2]\), so more volume is added on \([0.5,1]\) than on \([1.5,2]\).

a) Approximately \(1.6\,\text{L}\) b) More volume is added on \([0.5,1]\).

- Volume is area under the flow-rate graph. - Break the requested interval into a trapezoid and a rectangle. - For the full interval, use a triangle and a rectangle.

1. a) From \(1\) to \(2\), the area is a trapezoid: \(\frac{2+4}{2}(1)=3\,\text{L}\). From \(2\) to \(4\), the area is a rectangle: \((2)(4)=8\,\text{L}\). Therefore, the volume is \(3+8=11\,\text{L}\). 2. b) The area from \(0\) to \(2\) is a triangle with area \(\frac12(2)(4)=4\,\text{L}\), and the area from \(2\) to \(6\) is a rectangle with area \((4)(4)=16\,\text{L}\). The total is \(20\,\text{L}\).

a) \(11\,\text{L}\) b) The total volume is \(4+16=20\,\text{L}\).

- Accumulated mass is area under the deposition-rate graph. - Use the grid to identify the first triangle’s dimensions. - For part b), first determine the graph’s height at \(t=3\). - Then use the area formula for a triangle.

1. a) The area from \(0\) to \(1\) is a triangle with base \(1\) and height \(4\): \(\frac12(1)(4)=2\,\text{g}\). 2. b) The line from \((1,4)\) to \((5,0)\) has equation \(r(t)=-t+5\), so \(r(3)=2\,\text{g/min}\). The region from \(3\) to \(5\) is a triangle with base \(2\) and height \(2\): \(\frac12(2)(2)=2\,\text{g}\).

a) \(2\,\text{g}\) b) \(2\,\text{g}\)

- Equal basin volumes occur when the signed accumulated change between the two times is zero. - Look for a symmetry of the rate graph around its central zero. - Compare a positive area before the central zero with a corresponding negative area after it.

1. To return to the same water volume, the accumulated change \(\int_{0.5}^{t}g(s)\,\text{d}s\) must equal \(0\). 2. The graph is point-symmetric about \((2,0)\). Therefore, the positive signed area from \(t=0.5\) to \(t=2\) is canceled by the negative signed area over the reflected interval from \(t=2\) to \(t=3.5\). 3. Hence \(\int_{0.5}^{3.5}g(s)\,\text{d}s=0\), so the basin has the same volume again at \(t=3.5\) hours.

\(t=3.5\) hours

- Separate the ideas of instantaneous rate, net change, and accumulated amount. - Which given quantity is already a definite integral over the full interval? - Use the initial amount only when recovering the final amount.

1. Since \(A'(t)=r(t)\), the instantaneous rate at \(t=6\) is \(r(6)=4\). 2. By the Fundamental Theorem of Calculus, \(A(6)-A(0)=\int_0^6 r(t)\,\mathrm{d}t=-18\). 3. Therefore, \(A(6)=100-18=82\).

a) \(4\) units per unit time b) \(-18\) units c) \(82\) units

- Which values of \(X\) satisfy the event, and which are possible in this model? - What simple shape gives the area below this straight graph over the chosen interval?

1. From \(0\) to \(1\), the graph encloses a triangle with base \(1\) and height \(\frac{1}{2}\). 2. Its area is \(\frac{1}{2}\cdot1\cdot\frac{1}{2}=\frac{1}{4}\), so \(P(X\leq1)=\frac{1}{4}\).

\(P(X\leq1)=\frac{1}{4}=25\%\)

- The function decreases to zero at the right endpoint. - Express the triangular area in terms of \(k\). - Use the given total-area condition.

1. The graph on \([1,3]\) is a triangle with base \(2\) and height \(2k\). 2. Its area is \(\frac12\cdot2\cdot2k=2k\). 3. Setting \(2k=1\) gives \(k=\frac12\).

\(k=\frac12\)

- Integrate the rate to find the change, which can include negative contributions. - Keep the change separate from the original amount. - Check where the rate is zero to decide whether its sign changes.

1. Net change is \(\int_0^3(4-2t)\,\mathrm{d}t=[4t-t^2]_0^3=12-9=3\) tokens. 2. The final supply is \(10+3=13\) tokens. 3. Since \(r(2)=0\) and \(r(t)<0\) for \(2<t\leq3\), the rate is not always positive.

Net change: \(+3\) tokens. Final supply: \(13\) tokens. No; the rate becomes negative after \(2\) minutes.

- The stated total comes from integrating the hourly sales rate. - Keep the unknown factor outside the integral. - Divide the rate units by the units of time to get the units of \(k\).

1. The total number sold is \(\int_0^2kt\,\mathrm{d}t=k\left[\frac{t^2}{2}\right]_0^2=2k\). 2. Therefore \(2k=12\), giving \(k=6\). 3. Since \(r(t)\) is tickets per hour and \(t\) is hours, \(k\) has units of tickets per hour squared.

\(k=6\,\text{tickets/h}^2\)

- Where is the density taller, on the left or the right of the interval? - The mean formula weights each possible value by its density. - Compare your integral result with the midpoint after evaluating it.

1. The density is larger near \(x=3\), so values toward the right have more weight and the mean should exceed \(1.5\). 2. Using the given mean formula, \(E(X)=\int_0^3x\cdot\frac{2x}{9}\,\text{d}x=\frac{2}{9}[\frac{x^3}{3}]_0^3=2\). 3. The mean \(2\) is greater than the midpoint \(1.5\), as predicted.

The mean is greater than \(1.5\): \(E(X)=2\). Larger values receive more weight.

- Treat the upper endpoint as a letter while finding an antiderivative. - How are the areas from \(0\) to two different endpoints related to the area between them? - Check that the final probability lies between zero and one.

1. Integrating up to the variable endpoint gives \(F(a)=\int_0^a\frac{2-x}{2}\,\text{d}x=[x-\frac{x^2}{4}]_0^a=a-\frac{a^2}{4}\). 2. The probability of the interval is \(F(1.5)-F(1)\). 3. Since \(F(1.5)=\frac{15}{16}\) and \(F(1)=\frac{3}{4}\), the probability is \(\frac{3}{16}\).

a) \(F(a)=a-\frac{a^2}{4}\) for \(0\le a\le2\) b) \(P(1\le X\le1.5)=\frac{3}{16}\)

- The second moment weights larger values of \(x\) more heavily than the ordinary mean does. - How does the constant density simplify the first integral? - Check that the expression under the square root is nonnegative.

1. The second moment is \(E(X^2)=\int_0^2\frac{x^2}{2}\,\text{d}x=[\frac{x^3}{6}]_0^2=\frac{4}{3}\). 2. The variance is \(E(X^2)-[E(X)]^2=\frac{4}{3}-1=\frac{1}{3}\). 3. Therefore the standard deviation is \(\sigma=\sqrt{\frac{1}{3}}=\frac{\sqrt{3}}{3}\).

\(E(X^2)=\frac{4}{3}\) and \(\sigma=\frac{\sqrt{3}}{3}\)

- Read the density height on each half of the interval. - Find the contribution of each half separately in the given mean integral. - The midpoint treats both halves equally; does the density do that?

1. Reading the graph gives \(f(x)=\frac{1}{4}\) for \(0\le x<1\) and \(f(x)=\frac{3}{4}\) for \(1\le x\le2\). 2. The mean is \(E(X)=\int_0^1\frac{x}{4}\,\text{d}x+\int_1^2\frac{3x}{4}\,\text{d}x=\frac{1}{8}+\frac{9}{8}=\frac{5}{4}\). 3. The density is higher in the second half, so larger values count more than smaller values. The mean lies to the right of the midpoint.

\(E(X)=\frac{5}{4}\), which exceeds the midpoint \(1\) because the second half has greater probability density.

- Integrate the data-transfer rate over the requested time interval. - Use the initial stored amount to determine the constant in the accumulation function. - Read the variable limits as a moving \(5\)-minute interval. - Track the units from megabytes per minute to megabytes.

1. An antiderivative is \(F(t)=t^{3/2}+20t\). 2. \(\int_{10}^{40}f(t)\,\mathrm{d}t=[t^{3/2}+20t]_{10}^{40}=(600+70\sqrt{10})\,\text{MB}\approx821.36\,\text{MB}\). This is the amount of backup data received between minute \(10\) and minute \(40\). 3. Since \(D(0)=500\), \(D(t)=t^{3/2}+20t+500=t\sqrt{t}+20t+500\). 4. The equation in part c asks for a starting time \(t\) such that exactly \(150\,\text{MB}\) is received during the next \(5\) minutes.

a) \((600+70\sqrt{10})\,\text{MB}\approx821.36\,\text{MB}\), the amount received from minute \(10\) through minute \(40\) b) \(D(t)=t\sqrt{t}+20t+500\) c) It asks when a \(5\)-minute interval begins during which exactly \(150\,\text{MB}\) is received.

- Integrate the exponential rate. - Distinguish accumulated increase from total population. - Use the initial population to determine the constant. - Interpret the variable limits as a \(2\)-hour interval.

1. An antiderivative is \(W(t)=2000e^{0.1t}\). 2. The increase is \(\int_0^{10}200e^{0.1t}\,\text{d}t=2000(e-1)\approx3436.56\), or about \(3437\) bacteria. 3. A total-population function has the form \(N(t)=2000e^{0.1t}+C\). Since \(N(0)=5000\), \(C=3000\). 4. Thus, \(N(t)=2000e^{0.1t}+3000\). 5. The equation asks for the starting time of a \(2\)-hour interval during which the population grows by exactly \(2000\) bacteria.

a) \(2000(e-1)\approx3436.56\) bacteria, or about \(3437\) bacteria b) \(N(t)=2000e^{0.1t}+3000\) c) “At what time does a \(2\)-hour period begin during which the culture grows by exactly \(2000\) bacteria?”

- Integrate the rate over the full shift to find the mass captured per square meter. - Find the combined area of all filter elements. - Convert square centimeters to square meters and milligrams to grams.

1. The mass captured per square meter is \(\int_{-4}^{4}500\left(1-\frac{t^2}{16}\right)\,\mathrm{d}t =\left[500t-\frac{500}{48}t^3\right]_{-4}^{4} =\frac{8000}{3}\,\frac{\text{mg}}{\text{m}^2}\). 2. The total filter area is \(120\cdot2500\,\text{cm}^2 =300{,}000\,\text{cm}^2 =30\,\text{m}^2\). 3. Therefore, the total mass is \(30\,\text{m}^2\cdot\frac{8000}{3}\,\frac{\text{mg}}{\text{m}^2} =80{,}000\,\text{mg} =80\,\text{g}\).

The system captures \(80\,\text{g}\) of particulate matter.

- Start with initial volume plus the accumulated inflow. - Keep the exponential term exact when evaluating at \(10\) and \(20\) minutes. - After \(20\) minutes, replace the original rate model by a constant-rate accumulation beginning from \(V(20)\).

1. Accumulated inflow from time \(0\) to \(t\) is \(\int_0^t(12-12e^{-0.4u})\,\text{d}u=12t+30e^{-0.4t}-30\). 2. Adding the initial \(40\,\text{L}\) gives \(V(t)=12t+30e^{-0.4t}+10\). 3. At \(t=10\), \(V(10)=130+30e^{-4}\approx130.55\,\text{L}\). 4. At \(t=20\), \(V(20)=250+30e^{-8}\). For \(t\ge20\), the volume is \(250+30e^{-8}+12(t-20)\). 5. Setting this equal to \(500\) gives \(t=\frac{245}{6}-\frac{5}{2}e^{-8}\approx40.83\,\text{min}\).

a) \(V(t)=12t+30e^{-0.4t}+10\) b) \(130+30e^{-4}\,\text{L}\approx130.55\,\text{L}\) c) \(t=\frac{245}{6}-\frac{5}{2}e^{-8}\,\text{min}\approx40.83\,\text{min}\)

- A population increases when its rate of change is positive. - Add accumulated change to the initial population. - Keep track of the “thousands of bacteria” unit. - Interpret each side of the equation in part c as a population.

1. The population increases where \(f(t)>0\). The zeros are \(t=2\) and \(t=6\), and the parabola opens downward, so \(f(t)>0\) on \((2,6)\). 2. In thousands of bacteria, \(B(6)=10+\int_0^6(-0.1t^2+0.8t-1.2)\,\text{d}t\). 3. The integral equals \(\left[-\frac{t^3}{30}+0.4t^2-1.2t\right]_0^6=0\). Therefore, \(B(6)=10\), or \(10{,}000\) bacteria. 4. The equation in part c asks for all times \(k\) when the population is \(9.5\) thousand, or \(9500\), bacteria.

a) \((2,6)\) hours b) \(10{,}000\) bacteria c) “At what times \(k\) does the culture contain \(9500\) bacteria?”

- A maximum of an accumulated quantity occurs when its rate changes from positive to negative. - Integrate the net inventory rate to find the net change. - A zero accumulated change means equal starting and ending inventory amounts.

1. The inventory is greatest when the rate changes from positive to negative. Solve \(4e^{-0.5t}-2=0\): \(t=2\ln2\approx1.39\,\text{h}\). 2. The net change is \(\int_0^4(4e^{-0.5t}-2)\,\mathrm{d}t=[-8e^{-0.5t}-2t]_0^4=-8e^{-2}\approx-1.08\) thousand components. 3. A solution would mean that the inventory at time \(x\) equals the inventory at \(t=2\). No such solution exists for \(2<x\le5\), because \(h(t)<0\) throughout that interval.

a) \(t=2\ln2\approx1.39\,\text{h}\) b) \(-8e^{-2}\approx-1.08\) thousand components, a decrease c) It would mean the inventory at time \(x\) equals the inventory at \(t=2\); no such \(x\) exists in \((2,5]\).

- Apply the chain rule to the exponential function. - Check the behavior of the antiderivative at both infinite limits. - Match the integral in part b) to the supplied formula for a mean. - What does reflecting the density about the vertical axis suggest?

1. By the chain rule, \(\frac{\text{d}}{\text{d}z}(-e^{-z^2/2})=ze^{-z^2/2}\), so \(g\) is an antiderivative of \(h\). 2. Thus \(\int_{-\infty}^{\infty}z\phi(z)\,\text{d}z=\frac{1}{\sqrt{2\pi}}[-e^{-z^2/2}]_{-\infty}^{\infty}=0\), because the exponential tends to zero at both ends. 3. The integral has the mean formula's form with \(f=\phi\) and \(X=Z\), so \(E(Z)=0\). 4. Since \(\phi\) is even, \(z\phi(z)\) is odd. The negative and positive contributions cancel.

a) \(g'(z)=ze^{-z^2/2}=h(z)\) b) \(0\) c) The integral equals \(E(Z)=0\); the contributions on opposite sides of zero cancel by symmetry.

- What integral condition determines the missing constant in a density? - In the mean formula, why is the density multiplied by time? - The last probability covers only the upper part of the possible lifetimes.

1. Since \(t+1>0\) on \([0,4]\), the density is nonnegative when \(k>0\). 2. Normalizing gives \(1=\int_0^4k(t+1)\,\text{d}t=k[\frac{t^2}{2}+t]_0^4=12k\), so \(k=\frac{1}{12}\). 3. The mean is \(E(T)=\frac{1}{12}\int_0^4(t^2+t)\,\text{d}t=\frac{1}{12}[\frac{t^3}{3}+\frac{t^2}{2}]_0^4=\frac{22}{9}\,\text{years}\). 4. The probability of lasting longer than two years is \(P(T>2)=\frac{1}{12}\int_2^4(t+1)\,\text{d}t=\frac{2}{3}\).

a) \(k=\frac{1}{12}\) b) \(E(T)=\frac{22}{9}\,\text{years}\) c) \(P(T>2)=\frac{2}{3}\)

- Evaluate \(r(t)\) directly only for the instantaneous-rate question. - Use a definite integral of \(r(t)\) for each accumulated-volume question. - Match each requested time interval to the limits of integration. - Check that liters per hour and liters are not interchanged.

1. The instantaneous rate at \(t=0.5\) is \(r(0.5)=6(0.5-0.25)=1.5\,\text{L/h}\). 2. An antiderivative of \(6(t-t^2)\) is \(R(t)=3t^2-2t^3\). 3. The volume during the first \(0.5\) hour is \(R(0.5)-R(0)=0.5\,\text{L}\). 4. The volume from \(t=0.2\) to \(t=0.8\) is \(R(0.8)-R(0.2)=0.792\,\text{L}\). 5. The volume from \(t=0.75\) to \(t=1\) is \(R(1)-R(0.75)=0.15625\,\text{L}\).

a) \(1.5\,\text{L/h}\) b) \(0.5\,\text{L}\) c) \(0.792\,\text{L}\) d) \(0.15625\,\text{L}\)

- Use the full-cycle volume condition to determine the rate parameter first. - After finding \(k\), integrate the rate over each requested subinterval. - Check that each subinterval volume is smaller than the full-cycle volume.

1. The full-cycle condition gives \(\int_0^4k(4-t)\,\text{d}t=1\), so \(8k=1\) and \(k=\frac18\). 2. With this value of \(k\), an accumulation function is \(R(t)=\frac18\left(4t-\frac{t^2}{2}\right)\). 3. The volume from \(t=2\) to \(t=4\) is \(R(4)-R(2)=1-\frac68=0.25\,\text{L}\). 4. The volume from \(t=1\) to \(t=3\) is \(R(3)-R(1)=0.9375-0.4375=0.5\,\text{L}\).

a) \(k=\frac18=0.125\) b) \(0.25\,\text{L}\) c) \(0.5\,\text{L}\)

- Integrate the rate to find the net inventory change. - Divide the accumulated change by the interval length to find an average rate. - Account for the scale factor of \(100\) in the rate units. - Use the sign of the rate to determine when inventory decreases and increases.

1. a) The net inventory change is \(100\int_0^4f(t)\,\text{d}t\). An antiderivative is \(-0.025t^4+0.2t^3-0.4t^2\), and evaluation from \(0\) to \(4\) gives \(0\). Thus, the net inventory change is \(0\) packages. 2. b) The average model rate on \([0,2]\) is \(\frac{1}{2}\int_0^2f(t)\,\text{d}t=-0.2\). Since the model is measured in hundreds of packages per hour, the average net inventory rate is \(-20\) packages per hour. 3. c) Factor \(f(t)=-0.1t(t-2)(t-4)\). The rate is negative on \((0,2)\) and positive on \((2,4)\), so inventory decreases until \(t=2\) and increases afterward. Therefore, the minimum occurs at \(t=2\,\text{h}\).

a) \(0\) packages b) \(-20\) packages per hour c) \(t=2\,\text{h}\)

- Integrate the emission rate to find accumulated mass. - Divide the total mass by \(10\) hours to find the average rate. - Evaluate the definite integral separately on \([0,5]\) and \([5,10]\).

1. An antiderivative is \(F(t)=-0.0025t^4+0.04t^3+0.5t\). 2. a) \(\int_0^{10}E(t)\,\text{d}t=F(10)-F(0)=-25+40+5=20\,\text{kg}\). 3. b) The average rate is \(\frac{1}{10}\int_0^{10}E(t)\,\text{d}t=2\,\text{kg/h}\). 4. c) During the first half, \(\int_0^5E(t)\,\text{d}t=\frac{95}{16}=5.9375\,\text{kg}\). During the second half, \(\int_5^{10}E(t)\,\text{d}t=\frac{225}{16}=14.0625\,\text{kg}\). Therefore, substantially more pollutant is released during the second half.

a) \(20\,\text{kg}\) b) \(2\,\text{kg/h}\) c) First \(5\) hours: \(5.9375\,\text{kg}\). Final \(5\) hours: \(14.0625\,\text{kg}\). The second half releases \(8.125\,\text{kg}\) more pollutant.

- Integrate the flow rate over the requested time interval to obtain volume. - For average flow rate, divide the accumulated volume over the interval by the interval length. - Keep the volume units and rate units distinct when interpreting the two results.

1. An antiderivative is \(G(t)=(-500t^2-5000t-25{,}000)e^{-0.2t}+200t\). 2. The total volume during the first \(12\) hours is \(G(12)-G(0)\approx13{,}157.28\,\text{m}^3\). 3. The average flow rate on \([6,18]\) is \(\frac{1}{12}\int_6^{18}g(t)\,\text{d}t=\frac{G(18)-G(6)}{12}\approx1401.54\,\text{m}^3/\text{h}\).

a) \(13{,}157.28\,\text{m}^3\) b) \(1401.54\,\text{m}^3/\text{h}\)

- Integrate the transfer rate to find the accumulated data. - Account for the inner derivative when integrating the sine function. - Add the accumulated data to the initial stored amount for part b).

1. An antiderivative of the data-transfer rate is \(G(t)=-\frac{160}{\pi}\cos\left(\frac{\pi}{4}t\right)+50t\). 2. a) \(\int_0^4g(t)\,\text{d}t=\frac{320}{\pi}+200\approx301.86\,\text{MB}\). 3. b) \(\int_0^6g(t)\,\text{d}t=300+\frac{160}{\pi}\approx350.93\,\text{MB}\). Adding the initial amount gives \(200+350.93\approx550.93\,\text{MB}\).

a) \(200+\frac{320}{\pi}\,\text{MB}\approx301.86\,\text{MB}\) b) \(500+\frac{160}{\pi}\,\text{MB}\approx550.93\,\text{MB}\)

- The sign of net power indicates whether stored energy is increasing or decreasing. - Integrate power over time to find the net energy change. - Use radian measure when evaluating the trigonometric functions.

1. At \(t=3\), \(v(3)=20\cos\left(\frac{3\pi}{4}\right)=-10\sqrt{2}\approx-14.14\,\text{kW}\). The negative sign means the battery bank is discharging at that time. 2. An antiderivative is \(\frac{80}{\pi}\sin\left(\frac{\pi}{4}t\right)\). 3. Therefore, \(\int_2^6v(t)\,\mathrm{d}t=\frac{80}{\pi}\left(\sin\left(\frac{3\pi}{2}\right)-\sin\left(\frac{\pi}{2}\right)\right)=-\frac{160}{\pi}\approx-50.93\,\text{kWh}\). 4. The stored energy decreases by approximately \(50.93\,\text{kWh}\) from \(t=2\) to \(t=6\).

a) \(v(3)=-10\sqrt{2}\approx-14.14\,\text{kW}\); the battery bank is discharging. b) \(\int_2^6v(t)\,\mathrm{d}t=-\frac{160}{\pi}\approx-50.93\,\text{kWh}\); the stored energy decreases by approximately \(50.93\,\text{kWh}\).

- Interpret a negative rate as a decrease. - Integrate the rate over the full time interval to find the net change. - Compare the cosine values at the two endpoints.

1. At \(t=5\), \(k(5)=-0.5\sin\left(\frac{\pi}{2}\right)=-0.5\). The concentration is decreasing at a rate of \(0.5\,\text{mol}/(\text{L}\cdot\text{min})\). 2. An antiderivative is \(\frac{5}{\pi}\cos\left(\frac{\pi}{10}t\right)\). Therefore, \(\int_0^{20}k(t)\,\mathrm{d}t =\frac{5}{\pi}\left(\cos(2\pi)-\cos(0)\right) =0\). The net change in concentration over the \(20\) minutes is zero, so the final concentration equals the initial concentration.

a) \(k(5)=-0.5\,\text{mol}/(\text{L}\cdot\text{min})\); the concentration is decreasing. b) \(\int_0^{20}k(t)\,\mathrm{d}t=0\); the final concentration equals the initial concentration.

- Use the zeros and sign of the rate function. - A definite integral of a rate gives net change. - Add accumulated change to the initial amount. - A positive-to-negative rate change produces a local maximum of the accumulated quantity.

1. The zeros of \(f(t)\) are \(t=2\) and \(t=10\). Since the parabola opens downward, \(f(t)<0\) on \([0,2)\) and \((10,12]\), and \(f(t)>0\) on \((2,10)\). 2. The integral from \(4\) to \(9\) is the net change in the grain amount, in metric tons, during that time interval. 3. \(M(10)=100+\int_0^{10}(-t^2+12t-20)\,\text{d}t\). 4. The integral is \(\left[-\frac{t^3}{3}+6t^2-20t\right]_0^{10}=\frac{200}{3}\), so \(M(10)=\frac{500}{3}\approx166.67\) metric tons. 5. At \(t=10\), the rate changes from positive to negative, so the amount changes from increasing to decreasing and has a local maximum.

a) Removed on \([0,2)\) and \((10,12]\); added on \((2,10)\) b) The net change in the grain amount from hour \(4\) to hour \(9\) c) \(\frac{500}{3}\) metric tons, or approximately \(166.67\) metric tons d) The rate changes from positive to negative at \(t=10\).

- Factor the rate to make a sign chart. - Add the accumulated rate to the initial volume. - The sign of the total integral determines whether the ending volume is above or below the initial volume.

1. Factor the rate: \(v(t)=0.2t(t-8)(t-10)\). The sign is positive on \((0,8)\) and negative on \((8,10)\). 2. \(V(8)=500+\int_0^80.2(t^3-18t^2+80t)\,\text{d}t\). 3. An antiderivative is \(0.2\left(\frac{t^4}{4}-6t^3+40t^2\right)\). Its value at \(8\) is \(102.4\), so \(V(8)=602.4\,\text{m}^3\). 4. The net change from \(0\) to \(10\) is \(100\,\text{m}^3\), so \(V(10)=600\,\text{m}^3\), greater than the initial volume.

a) \(v(t)=0.2t(t-8)(t-10)\), which has the stated signs b) \(602.4\,\text{m}^3\) c) Greater; \(V(10)=600\,\text{m}^3\)

- Use the maximum value of sine. - Energy is the integral of power over time. - Compare the interval length with the period. - Scale one-system, one-shift energy by the numbers of systems and shifts.

1. The sine term is greatest when its argument is \(\frac{\pi}{2}\). This gives \(t=6\), and \(P(6)=35\,\text{kW}\). 2. The interval length \(12\) is one full period, so the sinusoidal term has integral zero. Therefore, \(\int_0^{12}P(t)\,\text{d}t=20(12)=240\,\text{kWh}\). 3. For \(8\) systems and \(2\) shifts, the total is \(240\cdot8\cdot2=3840\,\text{kWh}\).

a) \(35\,\text{kW}\) at \(t=6\,\text{h}\) b) \(240\,\text{kWh}\) c) \(3840\,\text{kWh}\)

- An integral of a rate gives the net change. - Compare the numerator with the derivative of the denominator. - Interpret a negative accumulated change as a loss of free storage capacity.

1. Since the numerator is a constant multiple of the derivative of the denominator, an antiderivative is \(-30\ln(t^2+36)\). 2. Therefore, \(\int_0^8f(t)\,\mathrm{d}t=\left[-30\ln(t^2+36)\right]_0^8=-30\ln\left(\frac{25}{9}\right)\approx-30.65\,\text{GB}\). The free storage capacity decreases by approximately \(30.65\,\text{GB}\) during the first \(8\) hours. 3. The inequality states that the net change in free capacity is less than \(-40\,\text{GB}\), meaning the server has lost more than \(40\,\text{GB}\) of free capacity.

a) \(-30\ln\left(\frac{25}{9}\right)\approx-30.65\,\text{GB}\); free storage capacity decreases by approximately \(30.65\,\text{GB}\). b) “After how many hours has the server lost more than \(40\,\text{GB}\) of free storage capacity?”

- Integrate the data rate to find the accumulated amount. - Compare the numerator with the derivative of the denominator. - Use the requested time interval as the limits of integration.

1. Because the numerator is a constant multiple of the derivative of the denominator, an antiderivative is \(H(t)=50\ln\left((t+2)^2+16\right)\). 2. For the first \(6\) hours, \(\int_0^6h(t)\,\mathrm{d}t=50\ln(80)-50\ln(20)=50\ln4\approx69.31\,\text{GB}\). 3. From \(t=2\) to \(t=4\), \(\int_2^4h(t)\,\mathrm{d}t=50\ln(52)-50\ln(32)=50\ln\left(\frac{13}{8}\right)\approx24.28\,\text{GB}\). This is the amount of data sent between the second and fourth hours.

a) \(50\ln4\approx69.31\,\text{GB}\) b) \(50\ln\left(\frac{13}{8}\right)\approx24.28\,\text{GB}\); this is the data sent from \(t=2\) to \(t=4\).

- Express the given work as a definite integral containing the unknown spring constant. - Convert centimeters to meters before using Hooke's law or the work integral. - After finding \(k\), evaluate the force law at the requested stretch.

1. Converting to meters, the work condition is \(0.72=\int_{0.04}^{0.08}ks\,\text{d}s=\frac{k}{2}(0.08^2-0.04^2)\). 2. Since \(\frac{1}{2}(0.08^2-0.04^2)=0.0024\), \(k=\frac{0.72}{0.0024}=300\,\text{N/m}\). 3. At a stretch of \(0.15\,\text{m}\), \(F(0.15)=300(0.15)=45\,\text{N}\).

a) \(k=300\,\text{N/m}\) b) \(45\,\text{N}\)

- Use the force at \(s=0\) to determine one constant immediately. - The area under a force-versus-displacement graph represents work. - Set the definite integral equal to the given work. - Solve the resulting linear equation.

1. Since \(F(0)=d=10\), \(d=10\,\text{N}\). 2. Use the work condition: \(80=\int_0^4(ks+10)\,\text{d}s=\left[\frac{k}{2}s^2+10s\right]_0^4=8k+40\). 3. Solving \(8k+40=80\) gives \(k=5\,\text{N/m}\).

\(k=5\,\text{N/m}\) and \(d=10\,\text{N}\)

- Convert the surface radius and final center-to-spacecraft distance to meters. - Work is the integral of the position-dependent gravitational force. - The final center-to-spacecraft distance is twice the lunar radius. - Round only after substituting the supplied constants.

1. The distance from the Moon's center changes from \(r_1=1.738\times10^6\,\text{m}\) to \(r_2=3.476\times10^6\,\text{m}\). 2. The work is \(W=\int_{r_1}^{r_2}\frac{GMm}{r^2}\,\text{d}r=GMm\left(\frac{1}{r_1}-\frac{1}{r_2}\right)\). 3. Because \(r_2=2r_1\), this simplifies to \(W=\frac{GMm}{2r_M}\). 4. Using the supplied rounded constants gives \(W\approx5.643\times10^8\,\text{J}=564.3\,\text{MJ}\) to four significant figures.

\(W\approx5.643\times10^8\,\text{J}\), or \(564.3\,\text{MJ}\)

- Integrate the rate using a negative exponent. - Use the initial condition first. - Use the one-hour measurement to determine \(k\). - Substitute the constant back into the mass function.

1. Integrating the rate gives \(m(t)=-\frac{k}{t+1}+C\). 2. Since \(m(0)=0\), \(-k+C=0\), so \(C=k\). 3. Since \(m(1)=2\), \(-\frac{k}{2}+k=2\), so \(k=4\). 4. Therefore, \(m(t)=4-\frac{4}{t+1}\).

\(k=4\) and \(m(t)=4-\frac{4}{t+1}\)

- Multiply the units represented by a grid square's width and height. - The rate's greatest and least values indicate the fastest increase and decrease. - Use the sign of the rate to determine when inventory is increasing. - Integrate the rate to find the net inventory change.

1. Each grid square is \(1\,\text{h}\) wide and \(2\) pallets per hour high, so it represents \(2\) pallets. 2. Inventory increases fastest when \(f\) is greatest. The vertex occurs at \(t=5\), where \(f(5)=9\). Inventory decreases fastest when \(f\) is least. On the interval, the minimum value is \(f(0)=f(10)=-16\), so this occurs at \(t=0\) and \(t=10\). Because \(f(t)>0\) for \(2<t<8\), inventory increases throughout that interval and is greatest at \(t=8\). 3. Expanding gives \(f(t)=-t^2+10t-16\), with antiderivative \(F(t)=-\frac{1}{3}t^3+5t^2-16t\). Therefore, \(\int_2^8f(t)\,\text{d}t=36\). The warehouse gains a net total of \(36\) pallets from \(t=2\) to \(t=8\).

a) \(2\) pallets b) Increasing fastest at \(t=5\); decreasing fastest at \(t=0\) and \(t=10\); greatest inventory on \([2,8]\) at \(t=8\) c) \(36\) pallets, the net inventory gain from \(t=2\) to \(t=8\)

- Integrating an arrival rate gives accumulated arrivals. - Use the accumulation function for the threshold total. - To analyze the line, compare the arrival rate with the fixed processing rate.

1. \(I_0(x)\) is the total number of visitors who arrive from opening until time \(x\). 2. An antiderivative of \(f\) is \(-10t^3+120t^2\), so \(I_0(2)=400\) visitors. 3. Solving \(-10x^3+120x^2=2160\) on \([0,8]\) gives \(x=6\). 4. The arrival rate equals the processing capacity when \(-30t^2+240t=360\), giving \(t=2\) and \(t=6\). The arrival rate exceeds capacity between these times and falls below capacity after \(t=6\), so the line can begin to shrink after \(6\) hours.

a) The total number of visitors arriving from opening through time \(x\) b) \(400\) visitors c) \(x=6\) hours d) After \(t=6\) hours

- Accumulated production and shipping are areas under their rate graphs. - Break the regions into triangles and rectangles. - Subtract accumulated shipping from accumulated production. - Include the initial inventory of \(20\) machines.

1. From \(t=0\) to \(t=4\), production is the triangular area \(\frac{1}{2}(4)(8)=16\) machines, and shipping is \(\frac{1}{2}(4)(2)=4\) machines. Thus, inventory is \(20+16-4=32\) machines. 2. From \(t=0\) to \(t=8\), production is \(16+(4)(8)=48\) machines, and shipping is \(\frac{1}{2}(8)(4)=16\) machines. Thus, inventory is \(20+48-16=52\) machines. 3. From \(t=0\) to \(t=12\), production is \(48+\frac{1}{2}(4)(8)=64\) machines, and shipping is \(\frac{1}{2}(12)(6)=36\) machines. Thus, inventory is \(20+64-36=48\) machines.

After \(4\) hours: \(32\) machines After \(8\) hours: \(52\) machines After \(12\) hours: \(48\) machines

- Relate the sign of power to whether stored energy is increasing or decreasing. - A minimum can occur where power changes from negative to positive. - Interpret signed area under a power graph as a change in energy. - Use both axis scales to determine the energy represented by one grid square.

1. a) The power is negative on \([0,5)\) and positive on \((5,8]\). Therefore, stored energy decreases until \(t=5\) and increases afterward, so its minimum occurs at \(t=5\,\text{h}\). 2. b) The change in stored energy is \(\int_0^5p(t)\,\text{d}t\). Each grid square represents \((1\,\text{h})(10\,\text{kW})=10\,\text{kWh}\). The region below the axis from \(0\) to \(5\) has an area of about \(10\) squares, so the energy changes by about \(-100\,\text{kWh}\). 3. Thus, the stored energy is approximately \(400-100=300\,\text{kWh}\).

a) \(t=5\,\text{h}\), because the power changes from negative to positive there. b) Approximately \(300\,\text{kWh}\)

- Material volume accumulated over time is represented by area under the rate graph. - Divide the region into triangles, rectangles, or trapezoids. - First find half of the total volume. - Locate the interval in which the cumulative volume reaches that value, then use the constant rate on that interval.

1. The total area consists of a triangle on \([0,1]\), a rectangle on \([1,2]\), and a triangle on \([2,4]\): \(\frac12(1)(40)+(1)(40)+\frac12(2)(40)=20+40+40=100\,\text{cm}^3\). 2. From \(1\) to \(2\), the area is \(40\,\text{cm}^3\). From \(2\) to \(3\), the trapezoid area is \(\frac{40+20}{2}(1)=30\,\text{cm}^3\). Thus, the volume is \(70\,\text{cm}^3\). 3. Half of the total volume is \(50\,\text{cm}^3\). By \(t=1\), \(20\,\text{cm}^3\) has been deposited, and by \(t=2\), \(60\,\text{cm}^3\) has been deposited, so \(m\in[1,2]\). 4. On this interval, the rate is \(40\,\text{cm}^3/\text{min}\). Solve \(20+40(m-1)=50\), which gives \(m=1.75\,\text{min}\).

a) \(100\,\text{cm}^3\) b) \(70\,\text{cm}^3\) c) \(m=1.75\,\text{min}\)

- Read the equations of the two line segments from their displayed endpoints. - Build the accumulation formula separately on each interval where the rate rule changes. - Carry the accumulated value at \(t=2\) into the next piece. - Once the rate becomes zero, the cumulative amount stays constant.

1. From the graph, \(r(t)=20t\) on \([0,2]\) and \(r(t)=-20t+80\) on \([2,4]\). 2. For \(t<0\), no arrivals have occurred, so \(A(t)=0\). 3. For \(0\le t<2\), \(A(t)=\int_0^t20s\,\text{d}s=10t^2\). 4. By \(t=2\), \(40\) people have arrived. For \(2\le t\le4\), \(A(t)=40+\int_2^t(-20s+80)\,\text{d}s=-10t^2+80t-80\). 5. At \(t=4\), \(A(4)=80\), and the rate is zero afterward, so \(A(t)=80\) for \(t>4\).

\(A(t)=\begin{cases}0 & t<0\\10t^2 & 0\le t<2\\-10t^2+80t-80 & 2\le t\le4\\80 & t>4\end{cases}\)

- Factor the density to test its sign. - Use an integral over the requested interval for the probability. - For the cumulative probability, treat the upper bound as a variable. - Interpret the mean using the formula given in the question.

1. On \([0,3]\), \(f(x)=\frac4{27}x^2(3-x)\ge0\), and \(\int_0^3f(x)\,\text{d}x=[\frac4{27}x^3-\frac1{27}x^4]_0^3=1\). 2. The probability is \(P(X\le1.5)=\int_0^{3/2}f(x)\,\text{d}x=\frac5{16}\). 3. For \(0\le x\le3\), \(F(x)=\int_0^xf(t)\,\text{d}t=\frac4{27}x^3-\frac1{27}x^4\). 4. The mean is \(E(X)=\int_0^3xf(x)\,\text{d}x=1.8\,\text{hours}\), or \(1\) hour \(48\) minutes.

a) \(f(x)\ge0\) and \(\int_0^3f(x)\,\text{d}x=1\) b) \(P(X\le1.5)=\frac5{16}\) c) \(F(x)=\frac4{27}x^3-\frac1{27}x^4\) for \(0\le x\le3\) d) \(E(X)=1.8\,\text{hours}=1\,\text{hour}\;48\,\text{minutes}\)

- Stored energy increases when net power is positive and decreases when net power is negative. - Use the sign changes of the cosine model to locate candidates for a maximum. - A definite integral of power over time is an energy change. - Add the accumulated change to the initial stored energy.

1. The power is positive on \((0,2)\) and negative on \((2,6)\), so stored energy changes from increasing to decreasing at \(t=2\). The later positive accumulation from \(6\) to \(8\) does not recover the earlier loss, so the greatest stored energy occurs at \(t=2\). 2. An antiderivative is \(P(t)=3\sin\left(\frac{\pi}{4}t\right)\), so \(\int_2^6p(t)\,\text{d}t=-6\,\text{kWh}\). This represents a net energy loss of \(6\,\text{kWh}\). 3. The accumulated change from \(0\) to \(6\) is \(-3\,\text{kWh}\), so the stored energy after \(6\) hours is \(15-3=12\,\text{kWh}\).

a) \(t=2\) hours b) \(-6\,\text{kWh}\), a net loss of \(6\,\text{kWh}\) c) \(12\,\text{kWh}\)

- Inventory equals initial inventory plus accumulated net rate. - Evaluate the accumulation function at the two requested times. - To rule out a hidden lower value between endpoints, determine where the net rate is positive and negative.

1. Since \(I'(t)=f(t)\), \(I(t)=\frac14t^4-\frac{11}{3}t^3+12t^2+C\). The initial condition \(I(0)=100\) gives \(C=100\). 2. Thus \(I(3)=\frac{517}{4}=129.25\) units and \(I(8)=\frac{44}{3}\approx14.67\) units. 3. Factoring \(f(t)=t(t-3)(t-8)\) shows that inventory increases on \((0,3)\) and decreases on \((3,8)\). Therefore, the minimum on \([0,8]\) is at an endpoint. Since \(I(0)=100\) and \(I(8)=\frac{44}{3}>0\), the warehouse never runs out.

a) \(I(t)=\frac14t^4-\frac{11}{3}t^3+12t^2+100\) b) \(I(3)=129.25\) units and \(I(8)=\frac{44}{3}\approx14.67\) units c) The inventory never reaches \(0\) on \([0,8]\).

- Net departures occur where the net entry rate is negative. - Add the accumulated net change to the initial attendance. - For an absolute maximum, use the rate sign changes and compare relevant endpoint/critical-time attendance values.

1. Factor \(f(t)=-3(t-2)(t-10)\). The rate is negative on \([0,2)\) and \((10,12]\), so net departures occur from noon to 2:00 p.m. and from 10:00 p.m. to midnight. 2. The accumulated change by \(t=6\) is \(\int_0^6f(t)\,\text{d}t=72\), so attendance is \(272\). 3. The rate is negative before \(t=2\), positive on \((2,10)\), and negative after \(t=10\), so attendance has a local maximum at \(t=10\). 4. Comparing \(A(0)=200\), \(A(10)=400\), and \(A(12)=344\) shows that the absolute maximum is \(400\) attendees at 10:00 p.m.

a) From noon to 2:00 p.m. and from 10:00 p.m. to midnight b) \(272\) attendees c) \(400\) attendees at 10:00 p.m.

- Separate the behavior of a rate from the behavior of the accumulated amount. - Compare total delivered volumes using complete definite integrals, not isolated rate values. - A high rate near the end does not determine the total accumulated amount over the whole interval.

1. The first statement is false. A positive but decreasing inflow rate still increases the cumulative amount; it only increases it more slowly. 2. Total supplied volume is a definite integral. The three totals are \(\int_0^{60}a(t)\,\text{d}t=\frac{896}{9}\,\text{m}^3\), \(\int_0^{60}b(t)\,\text{d}t=180\,\text{m}^3\), and \(\int_0^{60}c(t)\,\text{d}t=\frac{256}{3}\,\text{m}^3\). Therefore, pipe \(B\) supplies the greatest total volume. 3. The third statement is false because \(\frac{896}{9}>\frac{256}{3}\), so pipe \(A\) supplies more total water than pipe \(C\) despite the late-time rate comparison. 4. The fourth statement is true because the integral of \(b\) is exactly \(180\,\text{m}^3\).

1) False; a positive decreasing rate still increases cumulative volume. 2) True; pipe \(B\) supplies \(180\,\text{m}^3\), the greatest total. 3) False; pipe \(A\) supplies \(\frac{896}{9}\,\text{m}^3\), while pipe \(C\) supplies \(\frac{256}{3}\,\text{m}^3\). 4) True.

- A negative net flow rate makes stored volume decrease. - Candidate extrema occur where the rate changes sign or at an endpoint. - Use signed geometric areas for volume changes. - Add the full accumulated change to the initial volume.

1. The volume decreases when \(v(t)<0\), which occurs on \((3,8)\). 2. The volume increases until \(t=3\), decreases until \(t=8\), and then increases. The positive signed area from \(0\) to \(3\) is \(5\,\text{L}\), while the negative signed area from \(3\) to \(8\) is \(-7\,\text{L}\), so the absolute minimum occurs at \(t=8\,\text{min}\). 3. From \(0\) to \(5\), the signed geometric areas sum to \(4+1-1-2=2\,\text{L}\). 4. Over the full \(10\) minutes, the signed area is \(5-7+2=0\,\text{L}\). Adding this to the initial volume gives \(50\,\text{L}\).

a) \((3,8)\) minutes b) \(t=8\,\text{min}\) c) An increase of \(2\,\text{L}\) d) \(50\,\text{L}\)

- Use signed area under the net-rate graph. - A negative net arrival rate means the number of visitors inside is decreasing. - Compare the positive and negative geometric areas. - Distinguish net change from final attendance.

1. From \(t=0\) to \(t=6\), the net increase in the number of visitors inside is the area above the axis: \(\frac12(2)(4)+(2)(4)+\frac12(2)(4)=16\) visitors. Thus, the number inside increases by \(16\), so statement a) is false. 2. On \(6<t<10\), \(v(t)<0\), so more visitors leave than enter and the number inside decreases. Statement b) is true. 3. The area below the axis from \(t=6\) to \(t=10\) has magnitude \(\frac12(4)(2)=4\) visitors. The net change is \(16-4=12\) visitors, which is positive. Statement c) is true. 4. The integral gives the change in the number of visitors, not the final number. The initial attendance would have to be added. Statement d) is false.

a) False b) True c) True d) False

- Inventory increases where the net rate is positive and decreases where it is negative. - Use signed geometric areas to find the accumulated inventory change. - After the rate becomes constant and negative, use linear accumulation to locate when the inventory returns to zero.

1. The inventory increases while \(r(t)>0\) and decreases while \(r(t)<0\). The rate changes from positive to negative at \(t=6\), so the greatest inventory occurs at \(t=6\,\text{h}\). 2. The signed geometric areas from \(0\) to \(10\) are \(16\), \(4\), \(-4\), and \(-8\) pallets, so \(I(10)=8\) pallets. 3. For \(t\ge8\), the net rate is \(-4\) pallets per hour. Starting from \(I(10)=8\), solve \(8-4(t-10)=0\), giving \(t=12\,\text{h}\).

a) \(t=6\,\text{h}\) b) An increase of \(8\) pallets c) \(t=12\,\text{h}\)

- The stored amount is constant when the net rate is zero. - Use the signs of the rate to determine when stored data increases or decreases. - Compute the signed areas of the triangles and rectangles.

1. The amount of stored data remains constant when \(q(t)=0\), which occurs from \(t=4\) to \(t=6\). 2. The rate is positive before \(t=4\), zero from \(t=4\) to \(t=6\), and negative after \(t=6\). Therefore, the stored amount is greatest throughout \(4\le t\le6\). 3. The positive signed area from \(t=0\) to \(t=4\) is \(2\cdot2+\frac12\cdot2\cdot2=6\,\text{MB}\). The negative signed area from \(t=6\) to \(t=12\) is \(-\left(\frac12\cdot2\cdot2+2\cdot2+\frac12\cdot2\cdot2\right)=-8\,\text{MB}\). 4. The total change is \(6\,\text{MB}-8\,\text{MB}=-2\,\text{MB}\). The system stores \(2\,\text{MB}\) less data than it did initially.

a) The stored amount is constant and greatest from \(t=4\) to \(t=6\) minutes. b) \(-2\,\text{MB}\); the system stores \(2\,\text{MB}\) less than at the beginning.

- Multiply the units represented by a grid square’s width and height. - Treat area below the time axis as negative. - The stored amount is greatest when the rate changes from positive to negative.

1. One grid square is \(1\,\text{h}\) wide and \(1\,\text{kg/h}\) high, so it represents \(1\,\text{h}\cdot1\,\frac{\text{kg}}{\text{h}}=1\,\text{kg}\). 2. The positive area between the zeros \(t=0.5\) and \(t=5.5\) is a trapezoid: \(\frac{1}{2}\cdot(5+2)\cdot3=10.5\,\text{kg}\). 3. The two negative triangles have combined area \(2\cdot\left(\frac{1}{2}\cdot0.5\cdot1\right)=0.5\,\text{kg}\). Thus, the net change is \(10.5\,\text{kg}-0.5\,\text{kg}=10\,\text{kg}\). 4. The stored amount increases while \(f(t)>0\), from \(t=0.5\) to \(t=5.5\), and decreases afterward. Therefore, the amount of resin in storage is greatest at \(t=5.5\) hours.

a) \(1\,\text{kg}\) b) \(10\,\text{kg}\) c) \(t=5.5\) hours

- Both quantities are accumulated inflow, so integrate the same rate over different upper limits. - The integrand is a product of \(t\) and an exponential function. - Compare the accumulated volumes only after evaluating both definite integrals.

1. Integration by parts gives an antiderivative \(F(t)=-20(t+10)e^{-0.1t}\). 2. The first-10-hour volume is \(\int_0^{10}f(t)\,\text{d}t=200-400e^{-1}\approx52.85\) thousand cubic meters. 3. The first-20-hour volume is \(\int_0^{20}f(t)\,\text{d}t=200-600e^{-2}\approx118.80\) thousand cubic meters. 4. The 20-hour accumulated volume is more than twice the 10-hour accumulated volume.

a) \((200-400e^{-1})\times10^3\,\text{m}^3\approx52.85\times10^3\,\text{m}^3\) b) \((200-600e^{-2})\times10^3\,\text{m}^3\approx118.80\times10^3\,\text{m}^3\), more than twice the 10-hour amount

- Build the basin-volume function from the initial amount plus accumulated net flow. - A maximum occurs where the net rate changes from positive to negative. - Returning to the initial volume means the accumulated net change is zero. - Keep the final value exact until the requested rounding step.

1. Since \(V'(t)=f(t)\) and \(V(0)=80\), \(V(t)=-\frac{1}{3}t^3+3t^2+80\). 2. \(V(3)=98\,\text{m}^3\). 3. Since \(f(t)=t(6-t)\), the net rate changes from positive to negative at \(t=6\). Thus the maximum occurs there, with \(V(6)=116\,\text{m}^3\). 4. Solving \(V(t)=80\) gives \(t^2\left(3-\frac{t}{3}\right)=0\), so the first positive return time is \(t=9\,\text{h}\). 5. \(V(10)=\frac{140}{3}\,\text{m}^3\approx46.67\,\text{m}^3\).

a) \(98\,\text{m}^3\) b) At \(t=6\,\text{h}\); \(116\,\text{m}^3\) c) \(t=9\,\text{h}\) d) \(\frac{140}{3}\,\text{m}^3\approx46.67\,\text{m}^3\)

- Net battery power is solar input minus household use. - Integrate net power and add the initial stored energy. - The battery is charging exactly when net power is positive.

1. The battery energy is \(E(t)=5+\int_0^t(p(\tau)-c(\tau))\,\text{d}\tau\). 2. The accumulated net energy change by \(t=6\) is \(7\,\text{kWh}\), so \(E(6)=12\,\text{kWh}\). 3. The accumulated net change by \(t=12\) is \(14\,\text{kWh}\), so \(E(12)=19\,\text{kWh}\). 4. The battery charges when \(p(t)>c(t)\). Solving \(-\frac19t^2+\frac43t=1.5\) gives \(t=6\pm\frac{3\sqrt{10}}{2}\), so charging occurs between those times.

a) \(E(6)=12\,\text{kWh}\); \(E(12)=19\,\text{kWh}\) b) \(6-\frac{3\sqrt{10}}{2}<t<6+\frac{3\sqrt{10}}{2}\), approximately \(1.26<t<10.74\) hours after 6:00 a.m.

- Use the sign of the rate to determine when the stored volume rises or falls. - A minimum can occur when the rate changes from negative to positive. - Estimate the signed area between the graph and the time axis. - Add the estimated net change to the initial volume.

1. The rate is negative from \(t=0\) to \(t=2\), positive from \(t=2\) to \(t=8\), and negative again from \(t=8\) to \(t=10\). Thus the volume first decreases, then increases, and then decreases. The positive accumulation from \(t=2\) to \(t=8\) is greater than the later loss from \(t=8\) to \(t=10\), so the global minimum occurs at \(t=2\). 2. Estimating the signed area under the graph from \(t=0\) to \(t=8\) gives a net change of about \(5\,\text{L}\). Therefore, the estimated amount after \(8\) hours is \(100\,\text{L}+5\,\text{L}\approx105\,\text{L}\).

a) \(t=2\) hours b) Approximately \(105\,\text{L}\)

- A definite integral of the net flow gives signed volume change. - Add the accumulated change to the initial volume for a final amount. - Taking the absolute value of the rate prevents positive and negative changes from canceling.

1. An antiderivative is \(F(t)=0.005t^4-0.2t^3+2.25t^2-5t\). Thus \(\int_2^8f(t)\,\text{d}t=24.6\,\text{m}^3\), a net increase. 2. Since \(\int_0^{10}f(t)\,\text{d}t=25\,\text{m}^3\), the basin volume after \(10\) hours is \(125\,\text{m}^3\). 3. The integral of \(|f|\) gives the total variation in basin volume caused by the net flow over the \(15\) hours: increases and decreases are both counted by magnitude rather than canceling.

a) A net increase of \(24.6\,\text{m}^3\) b) \(125\,\text{m}^3\) c) The total amount of volume change caused by the net flow, with increases and decreases both counted positively.

- Find each line equation from two points on the graph. - Accumulated volume is geometric area under the rate graph. - Split part c at \(t=2\), where the line formula changes. - Compare the area accumulated by \(t=2\) with half of the full triangular area.

1. a) On \([0,2)\), the line through \((0,0)\) and \((2,8)\) has slope \(4\), so \(r(t)=4t\). On \([2,5]\), the line through \((2,8)\) and \((5,0)\) has slope \(-\frac{8}{3}\), so \(r(t)=-\frac{8}{3}t+\frac{40}{3}\). Outside \([0,5]\), \(r(t)=0\). 2. b) The area from \(0\) to \(1\) is a triangle: \(\frac12(1)(4)=2\,\text{L}\). 3. c) The trapezoid area from \(1.5\) to \(2\) is \(\frac{6+8}{2}(0.5)=3.5\,\text{L}\). From \(2\) to \(3\), the trapezoid area is \(\frac{8+\frac{16}{3}}{2}(1)=\frac{20}{3}\,\text{L}\). Therefore, the total is \(\frac{7}{2}+\frac{20}{3}=\frac{61}{6}\,\text{L}\approx10.17\,\text{L}\). 4. d) The total area is a triangle with base \(5\) and height \(8\), so the total volume is \(20\,\text{L}\). By \(t=2\), the accumulated area is only \(\frac12(2)(8)=8\,\text{L}\), which is less than half of \(20\,\text{L}\). Therefore, the halfway time is greater than \(2\) hours.

a) \(r(t)=\begin{cases}4t & 0\le t<2\\-\frac{8}{3}t+\frac{40}{3} & 2\le t\le5\\0 & \text{otherwise}\end{cases}\) b) \(2\,\text{L}\) c) \(\frac{61}{6}\,\text{L}\approx10.17\,\text{L}\) d) Only \(8\,\text{L}\) of the total \(20\,\text{L}\) has accumulated by \(t=2\), so the halfway time must be greater than \(2\) hours.

- What value does the density have when the wait time is zero? How does it change as time increases? - Which part of the area under the curve represents waits no longer than five minutes? - How should the cumulative probability behave for negative wait times? - For the mean, first evaluate the integral with a finite upper bound. What happens as that bound grows?

1. For \(x\ge0\), the graph meets the y-axis at \((0,0.2)\), decreases, and approaches \(0\) as \(x\to\infty\). 2. \(P(X\le5)=\int_0^5 0.2e^{-0.2x}\,\text{d}x=1-e^{-1}\). 3. The CDF is \(F(x)=0\) for \(x<0\), and \(F(x)=1-e^{-0.2x}\) for \(x\ge0\). 4. An antiderivative of \(0.2xe^{-0.2x}\) is \(-e^{-0.2x}(x+5)\). Therefore, \(\mu=[-e^{-0.2x}(x+5)]_0^\infty=5\,\text{minutes}\).

a) For \(x\ge0\), the graph meets the y-axis at \((0,0.2)\), decreases, and approaches the x-axis. b) \(P(X\le5)=1-e^{-1}\) c) \(F(x)=0\) for \(x<0\) and \(F(x)=1-e^{-0.2x}\) for \(x\ge0\) d) \(\mu=5\,\text{minutes}\)

- Check nonnegativity and total area. - Use the stated mean formula rather than averaging the two endpoint values. - Consider which interval corresponds to the event in part c). - For the CDF, let the upper integration limit vary.

1. The function is nonnegative on \([0,1]\), and \(\int_0^1(6x-6x^2)\,\text{d}x=1\). 2. The mean is \(E(X)=\int_0^1x(6x-6x^2)\,\text{d}x=0.5\), so the average obscured fraction is \(50\%\). 3. The probability is \(P(0.25\le X\le0.75)=\int_{0.25}^{0.75}6x(1-x)\,\text{d}x=\frac{11}{16}=0.6875\). 4. For \(0\le x\le1\), \(F(x)=\int_0^x(6t-6t^2)\,\text{d}t=3x^2-2x^3\).

a) \(f(x)\ge0\) and \(\int_0^1f(x)\,\text{d}x=1\) b) \(\mu=0.5\), meaning an average obscured fraction of \(50\%\) c) \(P(0.25\le X\le0.75)=\frac{11}{16}=0.6875\) d) \(F(x)=3x^2-2x^3\) for \(0\le x\le1\)

- Verify both defining conditions of a probability density. - Find each tail probability with an integral over the relevant interval. - Use the supplied mean formula separately for each company. - Use the graph to anticipate which model favors shorter waits.

1. Both functions are nonnegative on \([0,2]\), and each integrates to \(1\) over that interval. 2. For SprintBus, \(P(X>1)=1-\int_0^1\frac38x^2\,\text{d}x=\frac78\). For SteadyLine, \(P(X>1)=\int_1^2\frac38(x-2)^2\,\text{d}x=\frac18\). 3. SprintBus has mean \(E(X)=\int_0^2x\frac38x^2\,\text{d}x=1.5\,\text{minutes}\), while SteadyLine has mean \(E(X)=\int_0^2x\frac38(x-2)^2\,\text{d}x=0.5\,\text{minute}\). SteadyLine has the smaller mean.

a) Both are nonnegative and have integral \(1\) on \([0,2]\). b) SprintBus: \(\frac78\); SteadyLine: \(\frac18\) c) SprintBus: \(1.5\,\text{minutes}\); SteadyLine: \(0.5\,\text{minute}\). Choose SteadyLine for the smaller expected wait.

- Accumulate the failure rate from the start of the test to time \(t\). - Use differences of cumulative values for a subinterval count. - Determine half of the total expected failures before solving the threshold equation.

1. For \(0\le t\le5\), \(H(t)=\int_0^t3(5-s)^2\,\text{d}s=125-(5-t)^3\). Thus \(H(t)=0\) for \(t<0\) and \(H(t)=125\) for \(t>5\). 2. The expected number from hour \(1\) to hour \(2\) is \(H(2)-H(1)=37\) sensors. 3. Half of the total \(125\) expected failures is \(62.5\). Solving \(125-(5-t)^3=62.5\) gives \(t=5-\sqrt[3]{62.5}\approx1.031\) hours.

a) \(H(t)=\begin{cases}0 & t<0\\125-(5-t)^3 & 0\le t\le5\\125 & t>5\end{cases}\) b) \(37\) sensors c) \(t=5-\sqrt[3]{62.5}\approx1.031\) hours, so half of the expected failures occur a little over one hour into the test.

- Use the entire triangular region to determine how much water is transferred over the full interval. - Decide whether the target amount is reached before or after the peak rate. - On the decreasing side, it may be easier to work with the small amount that remains to be transferred.

1. The total transferred volume is the area of the triangle under the rate graph: \(\frac{1}{2}(4)(5)=10\,\text{L}\). Because \(8.75\,\text{L}\) is more than half of the total, \(k\) lies on the decreasing side of the graph. 2. After time \(k\), \(10-8.75=1.25\,\text{L}\) remains to be transferred. On the decreasing segment, the rate is \(q(t)=\frac52(4-t)\), so the region from \(k\) to \(4\) is a triangle with base \(4-k\) and height \(\frac52(4-k)\). 3. Set its area equal to the remaining volume: \(\frac{1}{2}(4-k)\left[\frac{5}{2}(4-k)\right]=1.25\). Thus \((4-k)^2=1\). Since \(k<4\), \(4-k=1\), so \(k=3\,\text{min}\).

\(k=3\,\text{min}\)

- Check whether the density function is even or odd before integrating both halves separately. - For the numerical estimate, evaluate the density at equally spaced points from \(0\) to \(1.5\). - Use concavity to decide whether trapezoids lie above or below the density curve. - Use symmetry again to extend the estimate from the right half to the whole central section.

1. The density is even. The total mass is \(\int_{-2}^{2}d(x)\,\text{d}x=1\,\text{kg}\), so symmetry gives \(0.5\,\text{kg}\) for the part with \(x>0\). 2. On \([0,1.5]\), the density values at spacing \(0.5\) are \(d(0)=0.375\), \(d(0.5)=0.3515625\), \(d(1)=0.28125\), and \(d(1.5)=0.1640625\). Since \(d''(x)=-\frac{3}{16}<0\), the trapezoidal estimate is a lower estimate. 3. By symmetry, the trapezoidal estimate on \([-1.5,1.5]\) is \(2\cdot0.5\left[\frac{1}{2}(0.375)+0.3515625+0.28125+\frac{1}{2}(0.1640625)\right]=0.90234375\,\text{kg}\). Therefore the actual mass is greater than \(0.90\,\text{kg}\).

a) \(0.5\,\text{kg}\) b) The composite trapezoidal lower estimate is \(0.90234375\,\text{kg}\), so the central section has mass greater than \(0.90\,\text{kg}\).

- Accumulated volume is area under the feed-rate graph. - Use a rectangle in part a and a trapezoid in part b. - First locate the graph section containing \(k\). - Write an accumulation equation using the linear rate on \([4,6]\).

1. a) The rate is constant at \(1\,\text{L/min}\) from \(t=1\) to \(t=3\), so the volume is \((3-1)(1)=2\,\text{L}\). 2. b) From \(5\) to \(6\), the area is a trapezoid with heights \(r(5)=3\) and \(r(6)=5\): \(\frac{3+5}{2}(1)=4\,\text{L}\). 3. c) The accumulated volume through \(t=4\) is \(4\,\text{L}\), so \(k\in[4,6]\). On this interval, \(r(t)=2t-7\). Solve \(4+\int_4^k(2t-7)\,\text{d}t=7\). This simplifies to \(k^2-7k+9=0\). The solution in \([4,6]\) is \(k=\frac{7+\sqrt{13}}{2}\approx5.30\,\text{min}\).

a) \(2\,\text{L}\) b) \(4\,\text{L}\) c) \(k=\frac{7+\sqrt{13}}{2}\approx5.30\,\text{min}\)

- Energy generated is area under a power-versus-time graph. - Read or calculate the graph’s heights at the requested times. - First find half of the total triangular area. - Determine which side of the peak contains the halfway time.

1. The area from \(0\) to \(2\) is a triangle with height \(r(2)=5\): \(\frac12(2)(5)=5\,\text{kWh}\). 2. On \([4,10]\), the line has equation \(r(t)=\frac{5}{3}(10-t)\). Thus, \(r(6)=\frac{20}{3}\) and \(r(8)=\frac{10}{3}\). The trapezoid area is \(\frac12\left(\frac{20}{3}+\frac{10}{3}\right)(2)=10\,\text{kWh}\). 3. The total triangular area is \(\frac12(10)(10)=50\,\text{kWh}\), so half is \(25\,\text{kWh}\). The energy generated through \(t=4\) is \(\frac12(4)(10)=20\,\text{kWh}\), so \(t_m\in[4,10]\). 4. Solve \(\int_4^{t_m}\frac{5}{3}(10-t)\,\mathrm{d}t=5\). This gives \(t_m^2-20t_m+70=0\), so the solution in the interval is \(t_m=10-\sqrt{30}\approx4.52\,\text{h}\).

a) \(5\,\text{kWh}\) b) \(10\,\text{kWh}\) c) \(t_m=10-\sqrt{30}\approx4.52\,\text{h}\)

- Interpret the sign of the rate before describing what it means for the water level. - A maximum of the accumulated water level occurs when its rate changes from positive to negative. - A definite integral of the rate gives net change in height. - Add accumulated change to the initial water level when an actual level is requested.

1. \(v(2)=0.8\cos\left(\frac{\pi}{3}\right)=0.4\,\text{m/h}\). At 10:00 a.m., the water level is rising at \(0.4\,\text{m/h}\). 2. The rate is positive for \(0<t<3\) and negative for \(3<t<9\). Thus the water level changes from increasing to decreasing at \(t=3\), so it is highest then on the interval. 3. \(\int_0^3v(t)\,\text{d}t=\left[\frac{4.8}{\pi}\sin\left(\frac{\pi}{6}t\right)\right]_0^3=\frac{4.8}{\pi}\,\text{m}\approx1.53\,\text{m}\). This is the rise in water level from 8:00 a.m. to 11:00 a.m. 4. At 2:00 p.m., \(t=6\). Since \(\int_0^6v(t)\,\text{d}t=0\), the water level has no net change from its initial value and is \(3.50\,\text{m}\).

a) \(0.4\,\text{m/h}\); the water level is rising at that rate at 10:00 a.m. b) \(t=3\), or 11:00 a.m. c) \(\frac{4.8}{\pi}\,\text{m}\approx1.53\,\text{m}\), the increase in water level from 8:00 a.m. to 11:00 a.m. d) \(3.50\,\text{m}\)

- Relate the slope of the accumulation function to the value of the integrand. - Check what happens to the slope where the integrand reaches zero. - Use how the integrand itself increases or decreases to determine the accumulation graph's concavity. - Do not rely only on the fact that the accumulation function passes through its lower-limit point; all candidates share that feature.

1. By the Fundamental Theorem of Calculus, \(F'(x)=f(x)=4-\frac14x^2\). Thus \(F'(x)>0\) for \(-4<x<4\), while \(F'(-4)=F'(4)=0\). The correct graph must increase throughout the interval and flatten at both endpoints. 2. Also, \(F''(x)=f'(x)=-\frac{1}{2}x\). Therefore \(F\) is concave up for \(x<0\) and concave down for \(x>0\). 3. All three candidates pass through the origin and share the same displayed endpoint values, so those features alone do not distinguish them. Candidate 1 has the required increasing behavior, horizontal endpoint tangents, and concavity change at \(x=0\). Candidate 2 has constant slope, and Candidate 3 has the opposite concavity pattern.

Candidate 1

- Signed area under a power graph gives the change in stored energy. - Do not use the endpoint values alone; all three candidates share them. - Use the sign of \(g\) to decide where \(G\) must increase or decrease. - A sign change of \(g\) from negative to positive gives a local minimum of \(G\); the opposite sign change gives a local maximum.

1. The signed-area balance consists of four triangles with areas \(-2\), \(2\), \(2\), and \(-2\) kilowatt-hours. Their sum is \(0\), so the total stored-energy change is \(0\,\text{kWh}\). 2. The accumulation function satisfies \(G(-2)=0\) and \(G'(x)=g(x)\). All three candidate graphs also end at \(G(6)=0\), so the endpoint values alone do not determine the answer. 3. The rate \(g\) is negative on \((-2,0)\), positive on \((0,4)\), and negative on \((4,6)\). Therefore, \(G\) must decrease, then increase, then decrease. 4. At \(x=0\), \(g\) changes from negative to positive, so \(G\) has a local minimum. At \(x=4\), \(g\) changes from positive to negative, so \(G\) has a local maximum. 5. Only Graph 1 has the required monotonicity and extrema while sharing the same endpoint values as the other candidates.

a) \(0\,\text{kWh}\) b) Graph 1; all three candidates satisfy \(G(-2)=G(6)=0\), but only Graph 1 decreases on \((-2,0)\), increases on \((0,4)\), decreases on \((4,6)\), has a local minimum at \(x=0\), and has a local maximum at \(x=4\).

- Total production is represented by area under a production-rate graph. - For Leon's statement, compare the full \(10\)-hour areas. - For Sarah's statement, use only the interval from \(0\) to \(4\). - A higher graph indicates a higher instantaneous production rate.

1. Leon: Total production is the area under each rate graph. Machine 1 produces \(\frac{60+10}{2}(10)=350\) parts. Machine 2 produces \(\frac{1}{2}(4)(40)+(6)(40)=80+240=320\) parts. Leon is incorrect. 2. Sarah: During the first \(4\) hours, Machine 1 produces \(\frac{60+40}{2}(4)=200\) parts, while Machine 2 produces \(\frac{1}{2}(4)(40)=80\) parts. Sarah is correct. 3. Jan: For \(t>4\), \(r_2(t)=40\), while \(r_1(t)<40\). Therefore, Machine 2 has the higher production rate after hour \(4\), so Jan is correct.

Leon: Incorrect; Machine 1 produces \(350\) parts, and Machine 2 produces \(320\) parts. Sarah: Correct; Machine 1 produces \(200\) parts, and Machine 2 produces \(80\) parts during the first \(4\) hours. Jan: Correct; for \(t>4\), Machine 2 has the higher hourly production rate.

- Where is the net rate \(q(t)\) positive? - What does a definite integral of a rate represent in context? - How does accumulated net change combine with the initial volume? - What sign change in \(q(t)\) produces a local maximum of volume?

1. Factor \(q(t)=-0.5(t-2)(t-6)\). Since the parabola opens downward, \(q(t)>0\) on \((2,6)\), so the volume increases during that interval. 2. The integral \(\int_2^6q(t)\,\text{d}t\) is the net change in water volume from \(t=2\) to \(t=6\). Its value is \(\frac{16}{3}\,\text{m}^3\). 3. Use \(V(6)=150+\int_0^6q(t)\,\text{d}t\). Since \(\int_0^6q(t)\,\text{d}t=\left[-\frac{1}{6}t^3+2t^2-6t\right]_0^6=0\), the volume is \(150\,\text{m}^3\). 4. The rate changes from positive to negative at \(t=6\), so the volume changes from increasing to decreasing there.

a) \(2<t<6\) b) The net change in water volume from hour \(2\) to hour \(6\), equal to \(\frac{16}{3}\,\text{m}^3\) c) \(150\,\text{m}^3\) d) The rate changes from positive to negative at \(t=6\).

- Build the stored-energy function from the initial energy and accumulated net power. - A minimum occurs where the net power changes from negative to positive. - Average net power equals the energy change divided by elapsed time. - Factor out the known solution \(t=0\) when solving for the return time.

1. The stored energy is \(E(t)=40+\int_0^tr(x)\,\mathrm{d}x=40-0.02\left(\frac{t^3}{3}-10t^2+64t\right)\). 2. \(E(4)=\frac{2824}{75}\approx37.65\,\text{kWh}\), and \(E(10)=\frac{608}{15}\approx40.53\,\text{kWh}\). 3. The net power is zero at \(t=4\) and \(t=16\). It changes from negative to positive at \(t=4\), so the minimum during the first \(10\) hours occurs at \(t=4\), with stored energy approximately \(37.65\,\text{kWh}\). 4. \(E(16)=\frac{3256}{75}\approx43.41\,\text{kWh}\). The average net power is \(\frac{E(16)-E(0)}{16}\approx0.213\,\text{kW}\). 5. Set \(E(t)=40\). After factoring out the known solution \(t=0\), solve \(t^2-30t+192=0\). The first positive solution is \(t=15-\sqrt{33}\approx9.26\,\text{h}\).

a) After \(4\) hours: \(\frac{2824}{75}\,\text{kWh}\approx37.65\,\text{kWh}\); after \(10\) hours: \(\frac{608}{15}\,\text{kWh}\approx40.53\,\text{kWh}\) b) At \(t=4\,\text{h}\); minimum stored energy \(\frac{2824}{75}\,\text{kWh}\approx37.65\,\text{kWh}\) c) Approximately \(0.213\,\text{kW}\) d) \(t=15-\sqrt{33}\approx9.26\,\text{h}\)

- Differentiate the power function only for the maximum-power comparison. - Stored energy equals initial energy plus accumulated power. - Keep the exponential expressions exact until the requested rounding step. - Average charging power equals energy gained divided by elapsed time.

1. \(P'(t)=(40-20t)e^{-0.5t}\), so the derivative changes from positive to negative at \(t=2\). The maximum power is \(P(2)=\frac{80}{e}\,\text{W}\). 2. Integration by parts gives \(\int40te^{-0.5t}\,\text{d}t=-80(t+2)e^{-0.5t}+C\). 3. Using \(E(0)=200\) gives \(E(t)=360-80(t+2)e^{-0.5t}\). 4. Thus \(E(4)=360-480e^{-2}\,\text{Wh}\approx295.04\,\text{Wh}\). 5. The average charging power on \([0,4]\) is \(\frac{E(4)-E(0)}{4}=40-120e^{-2}\,\text{W}\approx23.76\,\text{W}\). 6. Since \((t+2)e^{-0.5t}\to0\), \(E(t)\to360\,\text{Wh}\).

a) At \(t=2\,\text{h}\); maximum power \(\frac{80}{e}\,\text{W}\approx29.43\,\text{W}\) b) \(E(t)=360-80(t+2)e^{-0.5t}\) c) \(360-480e^{-2}\,\text{Wh}\approx295.04\,\text{Wh}\) d) \(40-120e^{-2}\,\text{W}\approx23.76\,\text{W}\) e) \(360\,\text{Wh}\)

- The area under the entire density must equal one. - Think about which part of the density corresponds to the allowed error sizes. - Use each supplied formula with the normalized density, not the unscaled polynomial. - Check that the standard deviation has units of millimeters, not square millimeters.

1. The total-area condition gives \(c\int_0^2(4-x^2)\,\text{d}x=c[4x-\frac{x^3}{3}]_0^2=\frac{16c}{3}=1\). Hence \(c=\frac{3}{16}\). 2. The required probability is \(\int_0^{1.5}\frac{3}{16}(4-x^2)\,\text{d}x=\frac{117}{128}\approx0.9141\). 3. The mean is \(E(X)=\int_0^2xf(x)\,\text{d}x=\frac{3}{4}\,\text{mm}\). 4. Also, \(E(X^2)=\int_0^2x^2f(x)\,\text{d}x=\frac45\,\text{mm}^2\), so \(E(X^2)-[E(X)]^2=\frac{19}{80}\,\text{mm}^2\). 5. Therefore \(\sigma(X)=\sqrt{\frac{19}{80}}\,\text{mm}\approx0.4873\,\text{mm}\).

a) \(c=\frac{3}{16}\) b) \(P(X\le1.5)=\frac{117}{128}\approx0.9141\) c) \(E(X)=\frac34\,\text{mm}\); \(\sigma(X)=\sqrt{\frac{19}{80}}\,\text{mm}\approx0.4873\,\text{mm}\)

- Normalize the density by setting its total integral equal to one. - Use an integral over the relevant interval for the probability. - Use the graph to estimate where the density peaks, then differentiate to check. - Put the mean on a number line. How would you use the given spread to locate the two cutoffs?

1. Normalization requires \(\int_0^4kx(4-x)^2\,\text{d}x=1\). Since the integral without \(k\) is \(\frac{64}{3}\), \(k=\frac3{64}\). 2. An antiderivative is \(F(x)=\frac3{64}\left(8x^2-\frac83x^3+\frac14x^4\right)\). Hence \(P(1\le X\le3)=F(3)-F(1)=\frac{11}{16}\). 3. The graph suggests a mode near \(1.3\). Differentiating gives \(f'(x)=\frac3{64}(16-16x+3x^2)\). The critical points are \(x=\frac43\) and \(x=4\); comparison with endpoints gives the mode \(x=\frac43\). 4. One standard deviation from the mean gives \([0.8,2.4]\), and \(P(0.8\le X\le2.4)=\int_{0.8}^{2.4}f(x)\,\text{d}x=0.64\).

a) \(k=\frac3{64}\) b) \(P(1\le X\le3)=\frac{11}{16}\) c) Estimate: about \(1.3\,\text{g}\); exact mode: \(\frac43\,\text{g}\) d) \(P(0.8\le X\le2.4)=0.64\)

- The stored volume increases when inflow exceeds outflow. - Find the times when the two rate formulas are equal. - The maximum stored volume occurs when the net rate changes from positive to negative. - Integrate inflow minus outflow over the interval where that difference is positive.

1. The reservoir volume increases when \(i(t)>o(t)\). Solving \(-0.1t^2+2t+5=12\) gives \(t=10\pm\sqrt{30}\). 2. Because the inflow parabola exceeds the constant outflow between the two intersections, volume increases on \((10-\sqrt{30},10+\sqrt{30})\). 3. The net rate changes from positive to negative at \(t=10+\sqrt{30}\), so the maximum volume occurs then. 4. The net volume added while the volume is increasing is \(\int_{10-\sqrt{30}}^{10+\sqrt{30}}(i(t)-o(t))\,\text{d}t=4\sqrt{30}\) thousand cubic meters, approximately \(21{,}909\,\text{m}^3\).

a) \(10-\sqrt{30}<t<10+\sqrt{30}\), approximately \(4.52<t<15.48\) hours b) \(t=10+\sqrt{30}\,\text{h}\approx15.48\,\text{h}\) c) \(4\sqrt{30}\times10^3\,\text{m}^3\approx21{,}909\,\text{m}^3\)

- Interpret area under a rate graph as accumulated volume. - Use triangle or trapezoid areas where possible. - For part c, write an accumulation equation after \(t=8\). - The pumping process begins only after the inflow ends.

1. The integral is the total volume entering during the \(20\)-minute storm. The graph forms a triangle, so \(\int_0^{20}r(t)\,\mathrm{d}t =\frac{1}{2}\cdot20\cdot40 =400\,\text{m}^3\). 2. During the first \(8\) minutes, the area is \(\frac{1}{2}\cdot8\cdot40=160\,\text{m}^3\). 3. After \(t=8\), the rate is \(r(t)=40-\frac{10}{3}(t-8)\). Set the accumulated volume equal to \(340\): \(160+\int_8^t\left(40-\frac{10}{3}(s-8)\right)\,\mathrm{d}s=340\). Let \(u=t-8\). Then \(40u-\frac{5}{3}u^2=180\). The solution in \(0\le u\le12\) is \(u=6\), so \(t=14\) minutes. 4. At \(t=20\), the basin contains \(400\,\text{m}^3\). Pumping at \(20\,\text{m}^3/\text{min}\) takes \(\frac{400}{20}=20\) minutes. Therefore, the basin is empty at \(t=40\) minutes.

a) The total inflow during the storm; \(400\,\text{m}^3\) b) \(160\,\text{m}^3\) c) \(t=14\) minutes d) \(t=40\) minutes