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AP Calculus AB : Interpretations and properties of definite integrals

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Example Questions

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Example Question #9 : Definite Integral Of The Rate Of Change Of A Quantity Over An Interval Interpreted As The Change Of The Quantity Over The Interval

$\begin{align*}&\text{The velocity of a particle is given as:}\\&v(t)=\frac{(26sin(t + 2))}{3}\\&\text{Find its change in position over the time interval }t=4,5.5\end{align*}$

Possible Answers:

5.32

35.09

1.97

0.54

Correct answer:

5.32

Explanation:

$\begin{align*}&\text{If the velocity function of a particle is known, finding the}\\&\text{distance it travels over a given interval of time is only a}\\&\text{matter of integrating the velocity function over this same interval.}\\&\text{Considering the function given:}\\&\text{Utilizing integral rules:}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\int_{4}^{5.5}(\frac{(26sin(t + 2))}{3})dt=-\frac{(26cos(t + 2))}{3}|_{4}^{5.5}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{4}^{5.5}(\frac{(26sin(t + 2))}{3})dt=-\frac{(26cos((5.5) + 2))}{3}-(-\frac{(26cos((4) + 2))}{3})\\&\int_{4}^{5.5}(\frac{(26sin(t + 2))}{3})dt=5.32\end{align*}$

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Example Question #51 : Integrals

$\begin{align*}&\text{The velocity of a particle is given as:}\\&v(t)=\frac{21}{5}\\&\text{Find its change in position over the time interval }t=4,7\end{align*}$

Possible Answers:

3.50

1.85

1.45

12.60

Correct answer:

12.60

Explanation:

$\begin{align*}&\text{If the function defining a rate of change is known, the total}\\&\text{change over a given interval of time is only a matter of integrating}\\&\text{this rate function over this same time interval. Considering}\\&\text{the function given:}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int_{4}^{7}(\frac{21}{5})dt=\frac{(21t)}{5}|_{4}^{7}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{4}^{7}(\frac{21}{5})dt=\frac{(21(7))}{5}-(\frac{(21(4))}{5})\\&\int_{4}^{7}(\frac{21}{5})dt=12.60\end{align*}$

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Example Question #11 : Definite Integral Of The Rate Of Change Of A Quantity Over An Interval Interpreted As The Change Of The Quantity Over The Interval

$\begin{align*}&\text{The flow of water into a tank is given as:}\\&\frac{dV(t)}{dt}=\frac{(79e^{(-t)})}{9}\\&\text{Find the change in volume over the time interval }t=[0.5,1.5]\end{align*}$

Possible Answers:

3.37

10.10

31.97

0.50

Correct answer:

3.37

Explanation:

$\begin{align*}&\text{If the function defining a rate of change is known, the total}\\&\text{change over a given interval of time is only a matter of integrating}\\&\text{this rate function over this same time interval. Considering}\\&\text{the function given:}\\&\text{Utilizing integral rules:}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\\&\int_{0.5}^{1.5}(\frac{(79e^{(-t)})}{9})dt=-\frac{(79e^{(-t)})}{9}|_{0.5}^{1.5}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{0.5}^{1.5}(\frac{(79e^{(-t)})}{9})dt=-\frac{(79e^{(-(1.5))})}{9}-(-\frac{(79e^{(-(0.5))})}{9})\\&\int_{0.5}^{1.5}(\frac{(79e^{(-t)})}{9})dt=3.37\end{align*}$

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Example Question #11 : Definite Integral Of The Rate Of Change Of A Quantity Over An Interval Interpreted As The Change Of The Quantity Over The Interval

$\begin{align*}&\text{The flow of water into a tank is given as:}\\&\frac{dV(t)}{dt}=\frac{(7\cdot 3^{(\frac{t}{3})})}{2}\\&\text{Find the change in volume over the time interval }t=[0.5,2]\end{align*}$

Possible Answers:

8.40

68.06

1.25

19.33

Correct answer:

8.40

Explanation:

$\begin{align*}&\text{If the function defining a rate of change is known, the total}\\&\text{change over a given interval of time is only a matter of integrating}\\&\text{this rate function over this same time interval. Considering}\\&\text{the function given:}\\&\text{Utilizing integral rules:}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\int_{0.5}^{2}(\frac{(7\cdot 3^{(\frac{t}{3})})}{2})dt=\frac{(21\cdot 3^{(\frac{t}{3})})}{(2ln(3))}|_{0.5}^{2}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{0.5}^{2}(\frac{(7\cdot 3^{(\frac{t}{3})})}{2})dt=\frac{(21\cdot 3^{(\frac{(2)}{3})})}{(2ln(3))}-(\frac{(21\cdot 3^{(\frac{(0.5)}{3})})}{(2ln(3))})\\&\int_{0.5}^{2}(\frac{(7\cdot 3^{(\frac{t}{3})})}{2})dt=8.40\end{align*}$

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Example Question #11 : Interpretations And Properties Of Definite Integrals

$\begin{align*}&\text{The rate of change of the length of an elastic is:}\\&\frac{dl(t)}{dt}=\frac{(17sin(t + 2))}{4}\\&\text{Determine the change in length over the time interval }t=[5,6.5]\end{align*}$

Possible Answers:

12.68

0.66

5.76

31.12

Correct answer:

5.76

Explanation:

$\begin{align*}&\text{If the function defining a rate of change is known, the total}\\&\text{change over a given interval of time is only a matter of integrating}\\&\text{this rate function over this same time interval. Considering}\\&\text{the function given:}\\&\text{Utilizing integral rules:}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\int_{5}^{6.5}(\frac{(17sin(t + 2))}{4})dt=-\frac{(17cos(t + 2))}{4}|_{5}^{6.5}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{5}^{6.5}(\frac{(17sin(t + 2))}{4})dt=-\frac{(17cos((6.5) + 2))}{4}-(-\frac{(17cos((5) + 2))}{4})\\&\int_{5}^{6.5}(\frac{(17sin(t + 2))}{4})dt=5.76\end{align*}$

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Example Question #11 : Definite Integral Of The Rate Of Change Of A Quantity Over An Interval Interpreted As The Change Of The Quantity Over The Interval

$\begin{align*}&\text{The flow of water into a tank is given as:}\\&\frac{dV(t)}{dt}=\frac{(57cos(3t))}{8}\\&\text{Find the change in volume over the time interval }t=[1,6]\end{align*}$

Possible Answers:

-19.92

-0.35

-4.87

-2.12

Correct answer:

-2.12

Explanation:

$\begin{align*}&\text{If the function defining a rate of change is known, the total}\\&\text{change over a given interval of time is only a matter of integrating}\\&\text{this rate function over this same time interval. Considering}\\&\text{the function given:}\\&\text{Utilizing integral rules:}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\int_{1}^{6}(\frac{(57cos(3t))}{8})dt=\frac{(19sin(3t))}{8}|_{1}^{6}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{1}^{6}(\frac{(57cos(3t))}{8})dt=\frac{(19sin(3(6)))}{8}-(\frac{(19sin(3(1)))}{8})\\&\int_{1}^{6}(\frac{(57cos(3t))}{8})dt=-2.12\end{align*}$

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Example Question #63 : Integrals

$\begin{align*}&\text{The velocity of a particle is given as:}\\&v(t)=\frac{(12e^{(t)})}{37}\\&\text{Find its change in position over the time interval }t=[0.5,3]\end{align*}$

Possible Answers:

5.98

0.68

17.94

37.07

Correct answer:

5.98

Explanation:

$\begin{align*}&\text{If the function defining a rate of change is known, the total}\\&\text{change over a given interval of time is only a matter of integrating}\\&\text{this rate function over this same time interval. Considering}\\&\text{the function given:}\\&\text{Utilizing integral rules:}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\\&\int_{0.5}^{3}(\frac{(12e^{(t)})}{37})dt=\frac{(12e^{(t)})}{37}|_{0.5}^{3}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{0.5}^{3}(\frac{(12e^{(t)})}{37})dt=\frac{(12e^{((3))})}{37}-(\frac{(12e^{((0.5))})}{37})\\&\int_{0.5}^{3}(\frac{(12e^{(t)})}{37})dt=5.98\end{align*}$

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Example Question #64 : Integrals

$\begin{align*}&\text{The rate of change of the length of an elastic is:}\\&\frac{dl(t)}{dt}=\frac{(73sin(t + 1))}{11}\\&\text{Determine the change in length over the time interval }t=[3,5]\end{align*}$

Possible Answers:

-10.71

-62.12

-24.63

-98.53

Correct answer:

-10.71

Explanation:

$\begin{align*}&\text{If the function defining a rate of change is known, the total}\\&\text{change over a given interval of time is only a matter of integrating}\\&\text{this rate function over this same time interval. Considering}\\&\text{the function given:}\\&\text{Utilizing integral rules:}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\int_{3}^{5}(\frac{(73sin(t + 1))}{11})dt=-\frac{(73cos(t + 1))}{11}|_{3}^{5}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{3}^{5}(\frac{(73sin(t + 1))}{11})dt=-\frac{(73cos((5) + 1))}{11}-(-\frac{(73cos((3) + 1))}{11})\\&\int_{3}^{5}(\frac{(73sin(t + 1))}{11})dt=-10.71\end{align*}$

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Example Question #65 : Integrals

$\begin{align*}&\text{The rate of change of the length of an elastic is:}\\&\frac{dl(t)}{dt}=\frac{(71sin(3t))}{7}\\&\text{Determine the change in length over the time interval }t=[4,6]\end{align*}$

Possible Answers:

0.62

3.16

0.18

0.07

Correct answer:

0.62

Explanation:

$\begin{align*}&\text{If the function defining a rate of change is known, the total}\\&\text{change over a given interval of time is only a matter of integrating}\\&\text{this rate function over this same time interval. Considering}\\&\text{the function given:}\\&\text{Utilizing integral rules:}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\int_{4}^{6}(\frac{(71sin(3t))}{7})dt=-\frac{(71cos(3t))}{21}|_{4}^{6}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{4}^{6}(\frac{(71sin(3t))}{7})dt=-\frac{(71cos(3(6)))}{21}-(-\frac{(71cos(3(4)))}{21})\\&\int_{4}^{6}(\frac{(71sin(3t))}{7})dt=0.62\end{align*}$

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Example Question #66 : Integrals

$\begin{align*}&\text{The rate of change of the length of an elastic is:}\\&\frac{dl(t)}{dt}=\frac{(37t)}{2}\\&\text{Determine the change in length over the time interval }t=[3,7]\end{align*}$

Possible Answers:

370.00

58.73

39.36

1147.00

Correct answer:

370.00

Explanation:

$\begin{align*}&\text{If the function defining a rate of change is known, the total}\\&\text{change over a given interval of time is only a matter of integrating}\\&\text{this rate function over this same time interval. Considering}\\&\text{the function given:}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int_{3}^{7}(\frac{(37t)}{2})dt=\frac{(37t^{2})}{4}|_{3}^{7}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{3}^{7}(\frac{(37t)}{2})dt=\frac{(37(7)^{2})}{4}-(\frac{(37(3)^{2})}{4})\\&\int_{3}^{7}(\frac{(37t)}{2})dt=370.00\end{align*}$

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AP Calculus AB : Interpretations and properties of definite integrals

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #9 : Definite Integral Of The Rate Of Change Of A Quantity Over An Interval Interpreted As The Change Of The Quantity Over The Interval

Example Question #51 : Integrals

Example Question #11 : Definite Integral Of The Rate Of Change Of A Quantity Over An Interval Interpreted As The Change Of The Quantity Over The Interval

Example Question #11 : Definite Integral Of The Rate Of Change Of A Quantity Over An Interval Interpreted As The Change Of The Quantity Over The Interval

Example Question #11 : Interpretations And Properties Of Definite Integrals

Example Question #11 : Definite Integral Of The Rate Of Change Of A Quantity Over An Interval Interpreted As The Change Of The Quantity Over The Interval

Example Question #63 : Integrals

Example Question #64 : Integrals

Example Question #65 : Integrals

Example Question #66 : Integrals

All AP Calculus AB Resources

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