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AP Calculus AB : Integrals

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Example Questions

Example Question #12 : Interpretations And Properties Of Definite Integrals

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

Possible Answers:

5.76 $\displaystyle 5.76$

31.12 $\displaystyle 31.12$

0.66 $\displaystyle 0.66$

12.68 $\displaystyle 12.68$

Correct answer:

5.76 $\displaystyle 5.76$

Explanation:

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

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Example Question #13 : Interpretations And Properties Of Definite Integrals

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

Possible Answers:

-4.87 $\displaystyle -4.87$

-2.12 $\displaystyle -2.12$

-19.92 $\displaystyle -19.92$

-0.35 $\displaystyle -0.35$

Correct answer:

-2.12 $\displaystyle -2.12$

Explanation:

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

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Example Question #14 : Interpretations And Properties Of Definite Integrals

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

Possible Answers:

17.94 $\displaystyle 17.94$

37.07 $\displaystyle 37.07$

0.68 $\displaystyle 0.68$

5.98 $\displaystyle 5.98$

Correct answer:

5.98 $\displaystyle 5.98$

Explanation:

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

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Example Question #15 : Interpretations And Properties Of Definite Integrals

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

Possible Answers:

-98.53 $\displaystyle -98.53$

-24.63 $\displaystyle -24.63$

-62.12 $\displaystyle -62.12$

-10.71 $\displaystyle -10.71$

Correct answer:

-10.71 $\displaystyle -10.71$

Explanation:

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

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Example Question #16 : Interpretations And Properties Of Definite Integrals

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

Possible Answers:

0.62 $\displaystyle 0.62$

0.18 $\displaystyle 0.18$

0.07 $\displaystyle 0.07$

3.16 $\displaystyle 3.16$

Correct answer:

0.62 $\displaystyle 0.62$

Explanation:

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

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Example Question #17 : Interpretations And Properties Of Definite Integrals

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

Possible Answers:

39.36 $\displaystyle 39.36$

1147.00 $\displaystyle 1147.00$

370.00 $\displaystyle 370.00$

58.73 $\displaystyle 58.73$

Correct answer:

370.00 $\displaystyle 370.00$

Explanation:

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

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Example Question #21 : Interpretations And Properties Of Definite Integrals

$\begin{align*}&\text{The velocity of a particle is given as:}\\&v(t)=\frac{9}{17}\\&\text{Find its change in position over the time interval }t=[3,6.5]\end{align*}$ $\displaystyle \begin{align*}&\text{The velocity of a particle is given as:}\\&v(t)=\frac{9}{17}\\&\text{Find its change in position over the time interval }t=[3,6.5]\end{align*}$

Possible Answers:

1.85 $\displaystyle 1.85$

11.67 $\displaystyle 11.67$

0.19 $\displaystyle 0.19$

6.67 $\displaystyle 6.67$

Correct answer:

1.85 $\displaystyle 1.85$

Explanation:

$\begin{align*}&\text{If the function defining a rate of change is known, the total}\\&\text{change over a given interval of time is only a matter of integrating}\\&\text{this rate function over this same time interval. Considering}\\&\text{the function given:}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int_{3}^{6.5}(\frac{9}{17})dt=\frac{(9t)}{17}|_{3}^{6.5}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{3}^{6.5}(\frac{9}{17})dt=\frac{(9(6.5))}{17}-(\frac{(9(3))}{17})\\&\int_{3}^{6.5}(\frac{9}{17})dt=1.85\end{align*}$ $\displaystyle \begin{align*}&\text{If the function defining a rate of change is known, the total}\\&\text{change over a given interval of time is only a matter of integrating}\\&\text{this rate function over this same time interval. Considering}\\&\text{the function given:}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int_{3}^{6.5}(\frac{9}{17})dt=\frac{(9t)}{17}|_{3}^{6.5}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{3}^{6.5}(\frac{9}{17})dt=\frac{(9(6.5))}{17}-(\frac{(9(3))}{17})\\&\int_{3}^{6.5}(\frac{9}{17})dt=1.85\end{align*}$

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Example Question #21 : Interpretations And Properties Of Definite Integrals

$\begin{align*}&\text{The rate of change of the length of an elastic is:}\\&\frac{dl(t)}{dt}=\frac{(50e^{(t)})}{7}\\&\text{Determine the change in length over the time interval }t=[0.5,5.5]\end{align*}$ $\displaystyle \begin{align*}&\text{The rate of change of the length of an elastic is:}\\&\frac{dl(t)}{dt}=\frac{(50e^{(t)})}{7}\\&\text{Determine the change in length over the time interval }t=[0.5,5.5]\end{align*}$

Possible Answers:

5728.88 $\displaystyle 5728.88$

186.67 $\displaystyle 186.67$

1736.02 $\displaystyle 1736.02$

275.56 $\displaystyle 275.56$

Correct answer:

1736.02 $\displaystyle 1736.02$

Explanation:

$\begin{align*}&\text{If the function defining a rate of change is known, the total}\\&\text{change over a given interval of time is only a matter of integrating}\\&\text{this rate function over this same time interval. Considering}\\&\text{the function given:}\\&\text{Utilizing integral rules:}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\\&\int_{0.5}^{5.5}(\frac{(50e^{(t)})}{7})dt=\frac{(50e^{(t)})}{7}|_{0.5}^{5.5}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{0.5}^{5.5}(\frac{(50e^{(t)})}{7})dt=\frac{(50e^{((5.5))})}{7}-(\frac{(50e^{((0.5))})}{7})\\&\int_{0.5}^{5.5}(\frac{(50e^{(t)})}{7})dt=1736.02\end{align*}$ $\displaystyle \begin{align*}&\text{If the function defining a rate of change is known, the total}\\&\text{change over a given interval of time is only a matter of integrating}\\&\text{this rate function over this same time interval. Considering}\\&\text{the function given:}\\&\text{Utilizing integral rules:}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\\&\int_{0.5}^{5.5}(\frac{(50e^{(t)})}{7})dt=\frac{(50e^{(t)})}{7}|_{0.5}^{5.5}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{0.5}^{5.5}(\frac{(50e^{(t)})}{7})dt=\frac{(50e^{((5.5))})}{7}-(\frac{(50e^{((0.5))})}{7})\\&\int_{0.5}^{5.5}(\frac{(50e^{(t)})}{7})dt=1736.02\end{align*}$

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Example Question #23 : Interpretations And Properties Of Definite Integrals

A pot of water begins at a temperature of $20^\circ C$ $\displaystyle 20^\circ C$ and is heated at a rate of $\tfrac{10}{(t+1)^2}$ $\displaystyle \tfrac{10}{(t+1)^2}$ degrees Celsius per minute. What will the temperature of the water be after 4 minutes?

Possible Answers:

$28^\circ C$ $\displaystyle 28^\circ C$

$32^\circ C$ $\displaystyle 32^\circ C$

$30.5^\circ C$ $\displaystyle 30.5^\circ C$

$25^\circ C$ $\displaystyle 25^\circ C$

$24^\circ C$ $\displaystyle 24^\circ C$

Correct answer:

$28^\circ C$ $\displaystyle 28^\circ C$

Explanation:

Let $\displaystyle T(t)$ denote the temperature of the pot after $\displaystyle t$ minutes.

The first thing to realize is that the quantity $\frac{-10}{(t+1)^2}$ $\displaystyle \frac{-10}{(t+1)^2}$ is the derivative of T(t) $\displaystyle T(t)$ . Using the fundamental theorem of Calculus, we know that

$\int_{0}^{4}\frac{10}{(t+1)^2} = T(4) - T(0)$ $\displaystyle \int_{0}^{4}\frac{10}{(t+1)^2} = T(4) - T(0)$

From there, we just need to solve the integral. Letting u = t+1, du=dt, we have the following:

$\int_{0}^{4}\frac{10}{(t+1)^2}dx = 10\int_{0}^{4} \frac{1}{u^2}du = \frac{-10}{(t+1)}]_0^4 = -2 + 10 = 8$ $\displaystyle \int_{0}^{4}\frac{10}{(t+1)^2}dx = 10\int_{0}^{4} \frac{1}{u^2}du = \frac{-10}{(t+1)}]_0^4 = -2 + 10 = 8$

Where the second equality follows by the power rule, and re-substituting t+1 = u.

Thus, we now have the equation $\displaystyle 8 = T(4) - T(0)$ . Because we know that the water started at $20^\circ C$ $\displaystyle 20^\circ C$ , all we need to do is rearrange and substitute.

$\displaystyle 8 + T(0) = T(4)$

$\displaystyle 8 + 20 = T(4)$

$\displaystyle 28 = T(4)$

Yielding our final answer, $28^\circ C$ $\displaystyle 28^\circ C$

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Example Question #61 : Integrals

A ball is thrown into the air. It's height, after t seconds is modeled by the formula:

h(t)=-15t^2+30t feet.

At what time will the velocity equal zero?

Possible Answers:

1.5s

Correct answer:

Explanation:

In order to find where the velocity is equal to zero, take the derivative of the function and set it equal to zero.

h(t) = –15t²+ 30t

h'(t) = –30t + 30

0 = –30t + 30

Then solve for "t".

–30 = –30t

t = 1

The velocity will be 0 at 1 second.

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AP Calculus AB : Integrals

Study concepts, example questions & explanations for AP Calculus AB

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Example Questions

Example Question #12 : Interpretations And Properties Of Definite Integrals

Example Question #13 : Interpretations And Properties Of Definite Integrals

Example Question #14 : Interpretations And Properties Of Definite Integrals

Example Question #15 : Interpretations And Properties Of Definite Integrals

Example Question #16 : Interpretations And Properties Of Definite Integrals

Example Question #17 : Interpretations And Properties Of Definite Integrals

Example Question #21 : Interpretations And Properties Of Definite Integrals

Example Question #21 : Interpretations And Properties Of Definite Integrals

Example Question #23 : Interpretations And Properties Of Definite Integrals

Example Question #61 : Integrals

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