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Accumulation of change

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52489512
Suppose \(\int_0^6f(x)\,\text{d}x=24\). Interpret the value of the integral in each context. a) The function \(f\) models the instantaneous rate of change of the number of bacteria in a petri dish, where \(x\) is measured in hours and \(f(x)\) is measured in millions of bacteria per hour. b) The function \(f\) models the electrical power used by a factory, where \(x\) is measured in hours and \(f(x)\) is measured in megawatts. c) A flat component is bounded by the x-axis, the lines \(x=0\) and \(x=6\), and the graph of \(f\). The graph lies above the x-axis on \([0,6]\), and both coordinates are measured in centimeters.

Hints

- Multiply the units of the function values by the units of the independent variable. - Integrating a rate over time gives an accumulated change. - Power integrated over time gives energy. - A definite integral can represent geometric area when the graph lies above the axis.

Solution

1. a) The integral gives the net change in the bacteria population during the first \(6\) hours. The population increases by \(24\) million bacteria. 2. b) Integrating power over time gives energy. The factory uses \(24\,\text{MWh}\) during the \(6\)-hour period. 3. c) Because the graph is above the x-axis, the integral gives the geometric area of the component, which is \(24\,\text{cm}^2\).

Answer

a) The bacteria population increases by \(24\) million during the first \(6\) hours. b) The factory uses \(24\,\text{MWh}\) of electrical energy. c) The component has area \(24\,\text{cm}^2\).
52489612
Suppose \(\int_2^5g(x)\,\text{d}x=-1.5\). Explain the meaning of the equation in each context. a) The function \(g\) is the rate of change of the water level in a reservoir, where \(x\) is measured in hours and \(g(x)\) is measured in meters per hour. b) The function \(g\) is a diver's vertical velocity, where \(x\) is measured in seconds and \(g(x)\) is measured in meters per second. Positive values indicate upward motion. c) The function \(g\) is the rate of change of air pressure during a weather shift, where \(x\) is measured in hours and \(g(x)\) is measured in hectopascals per hour.

Hints

- Use the sign of the integral to determine the direction of the net change. - Velocity integrated over time gives displacement. - Multiply the rate units by the units of the independent variable. - The integral describes net change, not necessarily total variation.

Solution

1. a) The integral gives the net change in water level from hour \(2\) to hour \(5\). The negative value means the level decreases by \(1.5\,\text{m}\). 2. b) The integral gives the diver's vertical displacement from \(2\) to \(5\) seconds. The value \(-1.5\) means the diver ends \(1.5\,\text{m}\) lower. 3. c) The integral gives the net change in air pressure from hour \(2\) to hour \(5\). The pressure decreases by \(1.5\,\text{hPa}\).

Answer

a) The water level decreases by \(1.5\,\text{m}\). b) The diver's vertical position decreases by \(1.5\,\text{m}\). c) The air pressure decreases by \(1.5\,\text{hPa}\).
52660912
The function \(f\) is the instantaneous rate of change of the mass of a pollutant in a lake. Time \(t\) is measured in weeks, and \(f(t)\) is measured in kilograms per week. At \(t=0\), the lake contains \(500\,\text{kg}\) of pollutant. Interpret each statement in context. a) \(f(t)<0\) for \(t\in[5,8]\) b) \(\int_0^4f(t)\,\text{d}t=-100\) c) \(500+\int_0^Tf(t)\,\text{d}t=250\)

Hints

- Interpret the sign of the rate as increase or decrease. - The integral of a rate gives net change. - Identify the initial value in part c). - Use initial value plus accumulated change to interpret the final value.

Solution

1. a) A negative rate means the pollutant mass is decreasing, so it decreases throughout weeks \(5\) through \(8\). 2. b) The integral is the net change during the first \(4\) weeks. The pollutant mass decreases by \(100\,\text{kg}\). 3. c) The expression on the left is the pollutant mass at time \(T\). The equation says that the mass is \(250\,\text{kg}\), half the initial amount, at time \(T\).

Answer

a) The pollutant mass decreases from week \(5\) through week \(8\). b) The pollutant mass decreases by \(100\,\text{kg}\) during the first \(4\) weeks. c) At time \(T\), the lake contains \(250\,\text{kg}\) of pollutant.
53392912
The graph of the derivative \(f^{\prime}\) is shown. The graph of \(f\) passes through \(P(0,1)\). Use geometry and the graph to find \(f(2)\).
Figure for problem 533929

Hints

- The signed area under a derivative gives the change in the original function. - Identify the geometric shape under the graph from \(x=0\) to \(x=2\). - Add the change to the given initial value.

Solution

1. The change in \(f\) from \(x=0\) to \(x=2\) is \(f(2)-f(0)=\int_0^2 f^{\prime}(x)\,dx\). 2. The region under \(f^{\prime}\) is a trapezoid with parallel sides of lengths \(1\) and \(2\) and width \(2\). Its area is \(\frac{1+2}{2}\cdot2=3\). 3. Therefore, \(f(2)=f(0)+3=1+3=4\).

Answer

\(f(2)=4\)
52658712
A technical process is being monitored. The function \(f\) gives the temperature of a component, in degrees Celsius, as a function of time \(t\), in minutes. Match each description with all applicable expressions from the list. a) After \(20\) minutes, the temperature is exactly \(65\,^\circ\text{C}\). b) At \(t=20\), the component is heating at its greatest local rate. c) From \(t=20\) to \(t=50\), the temperature increases by \(15\,^\circ\text{C}\). d) The average temperature during the first \(50\) minutes is \(72\,^\circ\text{C}\). Expressions: A. \(f(20)=65\) B. \(f'(20)>0\) C. \(f''(20)=0\) D. \(f'''(20)<0\) E. \(f(50)-f(20)=15\) F. \(\int_{20}^{50}f'(t)\,\text{d}t=15\) G. \(\frac{1}{50}\int_0^{50}f(t)\,\text{d}t=72\) H. \(\int_0^{50}f(t)\,\text{d}t=3600\)

Hints

- A locally greatest heating rate is a local maximum of the derivative. - The fundamental theorem connects a change in function values with the integral of the derivative. - The average value of a function is its integral divided by the interval length. - Equivalent equations may describe the same statement.

Solution

1. a) Statement a) directly gives the function value at \(t=20\), so A applies. 2. b) The heating rate is \(f'\). A local maximum of \(f'\) at \(t=20\) requires \(f''(20)=0\) and \(f'''(20)<0\). Because the component is heating, \(f'(20)>0\). Thus, B, C, and D apply. 3. c) The temperature change can be written as a difference of function values or as the integral of the derivative, so E and F apply. 4. d) The average-value equation is \(\frac{1}{50}\int_0^{50}f(t)\,\text{d}t=72\), so G applies. Multiplying by \(50\) gives \(\int_0^{50}f(t)\,\text{d}t=3600\), so H also applies.

Answer

a) A b) B, C, D c) E, F d) G, H
52658812
The net flow rate of water in a reservoir is modeled for \(0\le t\le8\) by \(k(t)=100(t^2-10t+21)\), where \(t\) is measured in hours and \(k(t)\) is measured in cubic meters per hour. A negative value means water is leaving the reservoir. At \(t=0\), the reservoir contains \(5000\,\text{m}^3\) of water. a) Explain the meaning of \(\int_3^7k(t)\,\text{d}t\) in context. b) Evaluate \(\int_0^3k(t)\,\text{d}t\) and interpret the result. c) Write an expression for the water volume \(V(8)\) after \(8\) hours.

Hints

- A negative net flow rate decreases the stored volume. - Integrating a flow rate over time gives a volume change. - Add accumulated change to the initial volume. - Check the units obtained from \(\text{m}^3/\text{h}\) multiplied by hours.

Solution

1. a) Since \(k(t)=100(t-3)(t-7)\), the rate is negative on \((3,7)\). Therefore, \(\int_3^7k(t)\,\text{d}t\) is the negative net change in volume during that interval, and its absolute value is the volume that leaves the reservoir. 2. b) \(\int_0^3k(t)\,\text{d}t=100\left[\frac{1}{3}t^3-5t^2+21t\right]_0^3=2700\,\text{m}^3\). Thus, the reservoir gains \(2700\,\text{m}^3\) during the first \(3\) hours. 3. c) Add the accumulated net change to the initial volume: \(V(8)=5000+\int_0^8k(t)\,\text{d}t\).

Answer

a) The net volume change from hour \(3\) to hour \(7\); it is negative, and its absolute value is the outflow volume. b) \(2700\,\text{m}^3\), a net increase during the first \(3\) hours c) \(V(8)=5000+\int_0^8k(t)\,\text{d}t\)
52661012
A solar farm supplies electrical energy to the grid. The function \(P(t)\) gives its power output in kilowatts, where \(t\) is measured in hours after midnight. Write an appropriate mathematical expression, equation, or inequality for each question. a) How much energy is supplied between 8:00 a.m. and noon? b) At what time \(t\) does the instantaneous power equal the average power from 6:00 a.m. to 6:00 p.m.? c) Which \(2\)-hour intervals contain more than \(400\,\text{kWh}\) of supplied energy?

Hints

- Integrating power over time gives energy. - Use the average-value formula on \([6,18]\). - Represent a general \(2\)-hour interval as \([t,t+2]\). - Distinguish an instantaneous quantity from an accumulated quantity.

Solution

1. a) Integrating power over time gives energy, so the expression is \(\int_8^{12}P(t)\,\text{d}t\). 2. b) The average power on \([6,18]\) is \(\frac{1}{12}\int_6^{18}P(x)\,\text{d}x\). The required equation is \(P(t)=\frac{1}{12}\int_6^{18}P(x)\,\text{d}x\). 3. c) A \(2\)-hour interval beginning at \(t\) is \([t,t+2]\). The condition is \(\int_t^{t+2}P(x)\,\text{d}x>400\), with \(0\le t\le22\).

Answer

a) \(\int_8^{12}P(t)\,\text{d}t\) b) \(P(t)=\frac{1}{12}\int_6^{18}P(x)\,\text{d}x\) c) \(\int_t^{t+2}P(x)\,\text{d}x>400\), with \(0\le t\le22\)
53391612
The graph of the derivative \(f^{\prime}\) of a continuous function \(f\) is shown, and \(f(0)=2\). 1. At what x-value on \([0,6]\) does \(f\) attain its absolute maximum? 2. At what x-value is the slope of \(f\) greatest? What is that greatest slope? 3. Use areas in the graph of \(f^{\prime}\) to find \(f(4)\).
Figure for problem 533916

Hints

- The sign of \(f^{\prime}\) tells where \(f\) increases or decreases. - The greatest value of \(f^{\prime}\) is the greatest slope of \(f\). - The signed area under \(f^{\prime}\) gives the change in \(f\).

Solution

1. The derivative is positive on \((0,4)\) and negative on \((4,6)\). Thus, \(f\) increases through \(x=4\) and decreases afterward, so its absolute maximum on \([0,6]\) occurs at \(x=4\). 2. The slope of \(f\) is \(f^{\prime}(x)\). The graph of \(f^{\prime}\) reaches its greatest value, \(2\), at \(x=2\). 3. By accumulation of change, \(f(4)-f(0)=\int_0^4 f^{\prime}(x)\,dx\). The region consists of two triangles, each with area \(\frac12\cdot2\cdot2=2\), so the total change is \(4\). Therefore, \(f(4)=2+4=6\).

Answer

1. \(x=4\) 2. At \(x=2\); the greatest slope is \(2\). 3. \(f(4)=6\)

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