- Read the rectangle's width from the horizontal axis. - Read its height from the vertical axis. - Multiply the rate by the elapsed time and check how the units combine.

1. Read the rectangle's width from the time axis: \(4\,\text{s}\). 2. Read its height from the rate axis: \(3\,\frac{\text{L}}{\text{s}}\). 3. The accumulated amount is \(4\,\text{s}\cdot 3\,\frac{\text{L}}{\text{s}}=12\,\text{L}\).

\(12\,\text{L}\)

- Read where the graph changes height and split the time interval there. - Treat each constant-rate piece as a rectangle. - Add the areas of the rectangles and keep the resulting quantity in gallons.

1. From \(t=0\) to \(t=2\), the rectangular area is \(2\,\text{min}\cdot 2\,\frac{\text{gal}}{\text{min}}=4\,\text{gal}\). 2. From \(t=2\) to \(t=3\), the rectangular area is \(1\,\text{min}\cdot 5\,\frac{\text{gal}}{\text{min}}=5\,\text{gal}\). 3. Add the accumulated amounts: \(4\,\text{gal}+5\,\text{gal}=9\,\text{gal}\).

\(9\,\text{gal}\)

- Read the height of the rate line and the length of the time interval. - How should a rate below zero affect the sign of the total change?

1. The horizontal rate is \(-3\), so the rectangle has width \(2\) and height \(3\) below the axis. 2. Area below the axis is counted as negative: \((-3)\cdot2=-6\).

\(-6\)

- How much does the freezer's temperature change during one hour? - What would three equal one-hour changes add up to?

1. Multiply the signed rate by elapsed time: \((-2)\cdot3=-6\). 2. The temperature has fallen by \(6\,\text{°F}\), so the signed change is \(-6\,\text{°F}\).

\(-6\,\text{°F}\)

- Match each time interval with the speed at its left endpoint. - Every rectangle has the same width here. - Multiply each speed by its interval width, then add the resulting distances.

1. The three intervals are \([0,1]\), \([1,2]\), and \([2,3]\), each with width \(1\,\text{s}\). 2. The left-endpoint speeds are \(2\), \(3\), and \(5\,\frac{\text{m}}{\text{s}}\). 3. The estimate is \(1\cdot 2+1\cdot 3+1\cdot 5=10\,\text{m}\).

\(10\,\text{m}\)

- Read the height of each shaded rectangle from the graph. - Every rectangle has width \(1\) minute. - Think about whether a decreasing graph lies above or below a rectangle whose height is chosen at the right edge.

a) The right-endpoint rectangle heights are \(5\), \(4\), \(3\), and \(2\,\frac{\text{L}}{\text{min}}\). With width \(1\,\text{min}\), the estimate is \(5+4+3+2=14\,\text{L}\). b) Because the rate decreases throughout the interval, each right-endpoint height is below the graph over most of its rectangle, while each left-endpoint height is above the graph over most of its rectangle. Therefore the right-endpoint estimate is smaller.

a) \(14\,\text{L}\) b) For a decreasing rate, right-endpoint rectangles lie below the graph while left-endpoint rectangles lie above it, so the right-endpoint estimate is smaller.

- How long does each rate last? - Multiply each rate by its interval length. - Add the three amounts after finding them separately.

1. In the first minute, the pump adds \(1\cdot4=4\,\text{L}\). 2. In the next \(1.5\) minutes, it adds \(1.5\cdot2=3\,\text{L}\). 3. In the final \(1.5\) minutes, it adds \(1.5\cdot5=7.5\,\text{L}\). 4. The total is \(4+3+7.5=14.5\,\text{L}\).

\(14.5\,\text{L}\)

- Split the region under the graph where its shape changes. - Identify the familiar geometric shape in each section. - Because velocity is nonnegative, each geometric area contributes positively to distance traveled.

1. From \(0\) to \(2\) seconds, the region is a triangle with area \(\frac{1}{2}\cdot 2\cdot 4=4\,\text{m}\). 2. From \(2\) to \(5\) seconds, the region is a rectangle with area \(3\cdot 4=12\,\text{m}\). 3. From \(5\) to \(6\) seconds, the region is a triangle with area \(\frac{1}{2}\cdot 1\cdot 4=2\,\text{m}\). 4. The total distance is \(4+12+2=18\,\text{m}\).

\(18\,\text{m}\)

- Read the two endpoint heights from the vertical axis. - Read the horizontal width of the interval from the time axis. - Use the trapezoid area formula and check the resulting units.

1. Read the endpoint rates from the graph: \(2\,\frac{\text{L}}{\text{min}}\) and \(6\,\frac{\text{L}}{\text{min}}\). 2. The time interval has width \(4\,\text{min}\), so the trapezoid area is \(\frac{2+6}{2}\cdot 4=16\). 3. Multiplying \(\frac{\text{L}}{\text{min}}\) by minutes gives liters, so the volume added is \(16\,\text{L}\).

\(16\,\text{L}\)

- Split the graph at the times where the velocity is \(0\). - Areas above and below the time axis have different signs when finding displacement. - For total distance, use the magnitude of each geometric area.

1. From \(t=0\) to \(t=4\), the graph forms a triangle above the time axis with area \(\frac{1}{2}\cdot4\cdot4=8\,\text{m}\). 2. From \(t=4\) to \(t=6\), the graph forms a triangle below the time axis with area magnitude \(\frac{1}{2}\cdot2\cdot2=2\,\text{m}\). 3. Displacement uses signed area, so the displacement is \(8-2=6\,\text{m}\). 4. Distance uses the magnitudes of both areas, so the total distance is \(8+2=10\,\text{m}\). 5. The negative velocity indicates motion in the opposite direction, which subtracts from displacement but still adds to distance traveled.

a) \(6\,\text{m}\) b) \(10\,\text{m}\) c) The below-axis area represents motion in the opposite direction, so it is negative for displacement but positive when counting total distance.

- Multiply each signed rate by how long it lasts. - Keep a negative rate negative when adding the changes. - Use the starting amount only after finding the total change.

1. The first interval adds \(2\cdot3=6\,\text{L}\). 2. The second interval changes the amount by \(3\cdot(-2)=-6\,\text{L}\). 3. The last interval adds \(1\cdot1=1\,\text{L}\). 4. The net change is \(6-6+1=1\,\text{L}\). Adding this to the starting amount gives \(20+1=21\,\text{L}\).

a) \(1\,\text{L}\) b) \(21\,\text{L}\)

- Check which portion of the velocity graph lies below the time axis. - In calculating the total distance, should backward travel cancel forward travel?

1. In the first two seconds, signed displacement is \(3\cdot2=6\) feet. 2. In the next two seconds, displacement is \((-2)\cdot2=-4\) feet. 3. Net position change is \(6-4=2\) feet; total distance is \(6+4=10\) feet.

Net change in position: \(2\,\text{ft}\). Total distance: \(10\,\text{ft}\).

- Where does the rate cross zero? What geometric regions lie on each side of that input? - Compare what backward motion contributes to displacement and to distance.

1. The triangular area above the axis has base \(2\) and height \(2\), so it contributes \(\frac{1}{2}\cdot2\cdot2=2\). 2. The equal triangle below the axis contributes \(-2\). 3. Net change in position is \(2-2=0\,\text{m}\), but total distance traveled is \(2+2=4\,\text{m}\).

Net change in position: \(0\,\text{m}\). Total distance traveled: \(4\,\text{m}\). Forward and backward changes cancel for net position, but both count toward distance.

- What ribbon feed rate can you read from the vertical axis? - How does the width of the rectangle change when the ending time is \(t\)?

1. Read a constant rate of \(4\,\text{in./s}\) from the graph. 2. The area under the rate graph from \(0\) to \(t\) is a rectangle of height \(4\) and width \(t\), so \(L(t)=4t\) inches. 3. At \(t=3\), the total ribbon length is \(L(3)=12\) inches.

\(L(t)=4t\) inches for \(0\le t\le3\); \(L(3)=12\) inches.

- Where does the graph lie above or below the time axis? - Work out the signed temperature change in each interval before combining it with the initial temperature.

1. From \(t=0\) to \(t=2\), the rate is \(2\,{}^\circ\text{C/h}\), so the temperature rises by \(4\,{}^\circ\text{C}\) to \(14\,{}^\circ\text{C}\). 2. From \(t=2\) to \(t=4\), the rate is \(-1\,{}^\circ\text{C/h}\), so the temperature falls by \(2\,{}^\circ\text{C}\). 3. The ending temperature is \(12\,{}^\circ\text{C}\).

Rising for \(0<t<2\); falling for \(2<t<4\). At \(t=2\): \(14\,{}^\circ\text{C}\). At \(t=4\): \(12\,{}^\circ\text{C}\).

- Convert each interval to its duration. - Express the number printed in the unknown-rate interval using \(r\). - The two interval amounts must add to the given total.

1. In the first two hours, it prints \(2\cdot3=6\) posters. 2. The last three hours produce \(3r\) posters. 3. Since the total is \(18\), solve \(6+3r=18\). Therefore, \(r=4\,\frac{\text{posters}}{\text{h}}\).

\(4\,\frac{\text{posters}}{\text{h}}\)

- Read each rectangle's width and height from the axes and grid. - Add rectangle areas, not just rectangle heights. - Compare the top of each left-endpoint rectangle with the increasing curve over the same subinterval.

1. In panel a), the two rectangles each have width \(1\) and heights \(1\) and \(2\). The estimate is \(1(1+2)=3\). 2. In panel b), the four rectangles each have width \(0.5\) and heights \(1\), \(1.25\), \(2\), and \(3.25\). The estimate is \(0.5(1+1.25+2+3.25)=3.75\). 3. The graph is increasing, so each left-endpoint rectangle lies below the curve over its interval. The refined partition uses later, larger left-endpoint heights on some subintervals, so its lower estimate is larger.

a) \(3\) b) \(3.75\) c) Because the rate is increasing, both are lower estimates, and the finer left-endpoint partition gives the larger one.

- Read the four measured rate values from the marked points. - Decide which endpoint gives the smaller rectangle height on each interval when the rate is increasing. - Use the other endpoint heights for the upper bound.

1. Read the marked rates as \(1\), \(2\), \(4\), and \(7\) units per minute. 2. Because the rate is increasing, left-endpoint rectangles give a lower bound: \(1+2+4=7\) units. 3. Right-endpoint rectangles give an upper bound: \(2+4+7=13\) units. 4. Thus the accumulated amount lies between \(7\) and \(13\) units under the stated monotonicity assumption.

Lower bound: \(7\) units Upper bound: \(13\) units

- Equal starting and ending rates do not determine what happens between those times. - Break each region under the rate graph into familiar geometric shapes. - Compare the two total areas rather than only the endpoint heights.

1. For machine a), the rate is constantly \(2\) units per minute, so the rectangular area is \(4\cdot 2=8\) units. 2. For machine b), view the region as the same \(4\)-by-\(2\) rectangle plus a triangle above it with base \(4\) minutes and height \(2\) units per minute. 3. Machine b) accumulates \(8+\frac{1}{2}\cdot 4\cdot 2=12\) units. 4. Therefore, machine b) accumulates \(12-8=4\) more units, even though the two machines have the same initial and final rates.

Machine a) accumulates \(8\) units. Machine b) accumulates \(12\) units. Machine b) accumulates \(4\) more units.

- Relate the sign of the rate to whether the accumulated amount is rising or falling. - A zero rate adds no new amount over time. - To compare \(A(2)\) and \(A(5)\), account for the signed area added between those inputs.

1. From \(0\) to \(2\), the rate is positive, so \(A\) increases. 2. From \(2\) to \(3\), the rate is \(0\), so \(A\) is constant. 3. From \(3\) to \(5\), the rate is negative, so \(A\) decreases. 4. From \(5\) to \(6\), the rate is positive, so \(A\) increases again. 5. The accumulated amount at \(x=2\) is \(2\cdot2=4\) units. From \(2\) to \(5\), the zero-rate interval adds nothing and the negative rectangle contributes \(-1\cdot2=-2\) units, so \(A(5)=2\). Therefore, \(A(2)>A(5)\).

a) \(A\) increases on \((0,2)\) and \((5,6)\), is constant on \((2,3)\), and decreases on \((3,5)\). b) \(A(2)>A(5)\); specifically, \(A(2)=4\) and \(A(5)=2\).

- How do the regions above and below the time axis affect the height change? - You know the ending height. How can you work backward using the net change? - At what point does the lift stop moving up and start moving down?

1. From \(t=0\) to \(t=1\), the lift rises \(2\cdot1=2\,\text{ft}\). From \(t=1\) to \(t=3\), the triangular area adds \(\frac{1}{2}\cdot2\cdot2=2\,\text{ft}\). 2. From \(t=3\) to \(t=4\), the triangular area below the time axis gives a change of \(-\frac{1}{2}\cdot1\cdot1=-0.5\,\text{ft}\). The net change is \(2+2-0.5=3.5\,\text{ft}\). 3. Since the final height is \(11.5\,\text{ft}\), the starting height was \(11.5-3.5=8\,\text{ft}\). 4. The velocity changes from positive to negative at \(t=3\), so the lift is highest then. Its height is \(8+2+2=12\,\text{ft}\).

Starting height: \(8\,\text{ft}\). Highest at \(t=3\,\text{s}\), with height \(12\,\text{ft}\).

- Read the marked power values from the graph before calculating any areas. - For each trapezoid, average its two endpoint heights and multiply by its width. - Keep the four-trapezoid and two-trapezoid calculations separate. - Compare which visible peaks are represented by each partition.

1. Read the power values \(1\), \(3\), \(2\), \(5\), and \(3\,\text{kW}\) at \(t=0,1,2,3,4\). 2. For part a), the four trapezoid areas are \(\frac{1+3}{2}(1)=2\), \(\frac{3+2}{2}(1)=2.5\), \(\frac{2+5}{2}(1)=3.5\), and \(\frac{5+3}{2}(1)=4\). Their total is \(12\,\text{kWh}\). 3. For part b), the two trapezoids have width \(2\) hours: \(2\cdot\frac{1+2}{2}+2\cdot\frac{2+3}{2}=8\,\text{kWh}\). 4. The coarse estimate skips the higher intermediate power values at \(t=1\) and \(t=3\), so its straight segments cut below substantial parts of the displayed model. Power in kilowatts multiplied by time in hours gives kilowatt-hours.

a) \(12\,\text{kWh}\) b) \(8\,\text{kWh}\) c) The coarse estimate misses the intermediate peaks, so it is smaller. The accumulated energy unit is kilowatt-hours.