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AP Calculus AB : Integrals

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Example Questions

Example Question #31 : Applications Of Antidifferentiation

A particle at the origin has an initial velocity of . If its acceleration is given by a = -24t^2 + 6t - 8 , find the position of the particle after 1 second.

Possible Answers:

24m

18m

32m

29m

27m

Correct answer:

27m

Explanation:

In this problem, letting s(t) denote the position of the particle and v(t) denote the velocity, we know that s''(t) = v'(t) = a(t) . Integrating and working backwards we have,

$v(t) = \int a(t)dt = \int (-24t^2 + 6t - 8)dt = -8t^3 + 3t^2 -8t + C$

Plugging in our initial condition, v(0) = 32 , we see immediately that C = 32 .

Repeating the process again for s(t) , we find that

$s(t) = \int v(t)dt = \int (-8t^3 + 3t^2 -8t +32)dt = -2t^4 + t^3 - 4t^2 + 32t + C$

Plugging in our initial condition, s(0) = 0 (we started at the origin) we see that C = 0 . This gives us a final equation

s(t) = -2t^4 + t^3 - 4t^2 + 32t . The problem asks for s(1) which is simply s(t) = -2(1)^4 + (1)^3 - 4(1)^2 + 32(1) = -4 + 1 - 4 + 32 = 27

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Example Question #2 : Finding Specific Antiderivatives Using Initial Conditions, Including Applications To Motion Along A Line

Find the integral which satisfies the specific conditions of $(\frac{\pi}{2}, 4)$

$\int4sin(t)-12cos(t)+3t^2dt$

Possible Answers:

y=-4tcos(t)-12tsin(t)+t^3

y=4cos(t)+12sin(t)+t^3

y=-4cos(t)-12sin(t)+t^3+12.12

y=-16tan(t)+t^3+12.12

Correct answer:

y=-4cos(t)-12sin(t)+t^3+12.12

Explanation:

Find the integral which satisfies the specific conditions of $(\frac{\pi}{2}, 4)$

$\int4sin(t)-12cos(t)+3t^2dt$

To do this problem, we need to recall that integrals are also called anti-derivatives. This means that we can calculate integrals by reversing our integration rules.

Furthermore, to find the specific answer using initial conditions, we need to find our "c" at the end.

Thus, we can have the following rules.

$\int sin(t)dt = -cos(t)+c$

$\int cos(t)dt = sin(t)+c$

$\int t^ndt=\frac{t^{n+1}}{n+1}+c$

Using these rules, we can find our answer:

$\int4sin(t)-12cos(t)+3t^2dt$

Will become:

$-4cos(t)-12sin(t)+\frac{3t^3}{3}+c=-4cos(t)-12sin(t)+t^3+c$

And so our anti-derivative is:

-4cos(t)-12sin(t)+t^3+c

Now, let's find c. First set our above expression equal to y

y= -4cos(t)-12sin(t)+t^3+c

Next, plug in for y and t. Then solve for c

$4= -4cos(\frac{\pi}{2})-12sin(\frac{\pi}{2})+(\frac{\pi}{2})^3+c$

Looks a bit messy, but we can clean it up to get:

$4= 0-12+3.88+c\rightarrow 4=-8.12+c\rightarrow c=12.12$

Now, to solve, simply replace c with 12.12

y=-4cos(t)-12sin(t)+t^3+12.12

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Example Question #881 : Ap Calculus Ab

$\begin{align*}&\text{The function }f(x)\text{ has the following values on the interval }x=[-2,13]:\\&f(-2)=5\\&f(1)=-19\\&f(4)=5\\&f(7)=-6\\&f(10)=-18\\&f(13)=0\\&\\&\text{Approximate the integral of }f(x)\text{ on this interval using right Riemann sums.}\end{align*}$

Possible Answers:

495

Correct answer:

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(-2)=5\\&f(1)=-19\\&f(4)=5\\&f(7)=-6\\&f(10)=-18\\&f(13)=0\\&\\&\text{Although the total interval has a length of }15\\&\text{Notice the points are equidistant. This distance is our subinterval }3\\&\text{and since we are using the right point of each interval, we disregard the first function value:}\\&\int_{-2}^{13}f(x)\approx3[(-19)+(5)+(-6)+(-18)+(0)]\\&\int_{-2}^{13}f(x)\approx-114\end{align*}$

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Example Question #891 : Ap Calculus Ab

$\begin{align*}&\text{Approximate }\int_{-8}^{1}f(x)\\&\text{Using a right Riemann sum, if }f(x)\text{ has the values:}\\&f(-8)=-13\\&f(-5)=-19\\&f(-2)=6\\&f(1)=-9\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(-8)=-13\\&f(-5)=-19\\&f(-2)=6\\&f(1)=-9\\&\text{Although the total interval has a length of }9\\&\text{Notice the points are equidistant. This distance is our subinterval: }3\\&\text{Since we are using the right point of each interval, we disregard the first function value:}\\&\int_{-8}^{1}f(x)\approx3[(-19)+(6)+(-9)]\\&\int_{-8}^{1}f(x)\approx-66\end{align*}$

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Example Question #3 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

$\begin{align*}&\text{Approximate }\int_{-2}^{23}f(x)\\&\text{Using a right Riemann sum, if }f(x)\text{ has the values:}\\&f(-2)=-2\\&f(3)=-15\\&f(8)=14\\&f(13)=15\\&f(18)=-9\\&f(23)=-12\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(-2)=-2\\&f(3)=-15\\&f(8)=14\\&f(13)=15\\&f(18)=-9\\&f(23)=-12\\&\text{Although the total interval has a length of }25\\&\text{Notice the points are equidistant, with }5\text{ intervals between. Each equidistant interval has length: }5\\&\text{Since we are using the right point of each interval, we disregard the first function value:}\\&\int_{-2}^{23}f(x)\approx5[(-15)+(14)+(15)+(-9)+(-12)]\\&\int_{-2}^{23}f(x)\approx-35\end{align*}$

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Example Question #1 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

$\begin{align*}&\text{Approximate }\int_{-7}^{13}f(x)\\&\text{Using a right Riemann sum, if }f(x)\text{ has the values:}\\&f(-7)=18\\&f(-2)=-17\\&f(3)=-16\\&f(8)=-15\\&f(13)=-14\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(-7)=18\\&f(-2)=-17\\&f(3)=-16\\&f(8)=-15\\&f(13)=-14\\&\text{Although the total interval has a length of }20\\&\text{, notice the points are equidistant, with }4\text{ intervals between.}\\&\text{Each equidistant interval has length: }5\\&\text{Since we are using the right point of each interval, we disregard the first function value:}\\&\int_{-7}^{13}f(x)\approx5[(-17)+(-16)+(-15)+(-14)]\\&\int_{-7}^{13}f(x)\approx-310\end{align*}$

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Example Question #5 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

$\begin{align*}&f(4)=9\\&f(6)=10\\&f(8)=-18\\&f(10)=15\\&f(12)=18\\&\text{Approximate, using a right Riemann sum, }\int_{4}^{12}f(x)\end{align*}$

Possible Answers:

272

Correct answer:

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(4)=9\\&f(6)=10\\&f(8)=-18\\&f(10)=15\\&f(12)=18\\&\text{Although the total interval has a length of }8\\&\text{, notice the points are equidistant, with }4\text{ intervals between.}\\&\text{Each equidistant interval has length: }2\\&\text{Since we are using the right point of each interval, we disregard the first function value:}\\&\int_{4}^{12}f(x)\approx2[(10)+(-18)+(15)+(18)]\\&\int_{4}^{12}f(x)\approx50\end{align*}$

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Example Question #1 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

$\begin{align*}&\text{The function }f(x)\text{ has the following values on the interval }x=[-2,38]:\\&f(-2)=-13\\&f(6)=-4\\&f(14)=-15\\&f(22)=-19\\&f(30)=18\\&f(38)=-8\\&\text{Approximate the integral of }f(x)\text{ on this interval using right Riemann sums.}\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(-2)=-13\\&f(6)=-4\\&f(14)=-15\\&f(22)=-19\\&f(30)=18\\&f(38)=-8\\&\text{Although the total interval has a length of }40\\&\text{, notice the points are equidistant, with }5\text{ intervals between.}\\&\text{Each equidistant interval has length: }8\\&\text{Since we are using the right point of each interval, we disregard the first function value:}\\&\int_{-2}^{38}f(x)\approx8[(-4)+(-15)+(-19)+(18)+(-8)]\\&\int_{-2}^{38}f(x)\approx-224\end{align*}$

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Example Question #1 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

$\begin{align*}&\text{Approximate }\int_{0}^{15}f(x)\\&\text{Using a right Riemann sum, if }f(x)\text{ has the values:}\\&f(0)=-19\\&f(5)=14\\&f(10)=2\\&f(15)=15\end{align*}$

Possible Answers:

155

180

Correct answer:

155

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(0)=-19\\&f(5)=14\\&f(10)=2\\&f(15)=15\\&\text{Although the total interval has a length of }15\\&\text{, notice the points are equidistant, with }3\text{ intervals between.}\\&\text{Each equidistant interval has length: }5\\&\text{Since we are using the right point of each interval, we disregard the first function value:}\\&\int_{0}^{15}f(x)\approx5[(14)+(2)+(15)]\\&\int_{0}^{15}f(x)\approx155\end{align*}$

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Example Question #2 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

$\begin{align*}&\text{Approximate }\int_{-8}^{0}f(x)\\&\text{Using a right Riemann sum, if }f(x)\text{ has the values:}\\&f(-8)=7\\&f(-6)=-7\\&f(-4)=16\\&f(-2)=-16\\&f(0)=20\end{align*}$

Possible Answers:

160

Correct answer:

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(-8)=7\\&f(-6)=-7\\&f(-4)=16\\&f(-2)=-16\\&f(0)=20\\&\text{Although the total interval has a length of }8\\&\text{, notice the points are equidistant, with }4\text{ intervals between.}\\&\text{Each equidistant interval has length: }2\\&\text{Since we are using the right point of each interval, we disregard the first function value:}\\&\int_{-8}^{0}f(x)\approx2[(-7)+(16)+(-16)+(20)]\\&\int_{-8}^{0}f(x)\approx26\end{align*}$

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AP Calculus AB : Integrals

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #31 : Applications Of Antidifferentiation

Example Question #2 : Finding Specific Antiderivatives Using Initial Conditions, Including Applications To Motion Along A Line

Example Question #881 : Ap Calculus Ab

Example Question #891 : Ap Calculus Ab

Example Question #3 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

Example Question #1 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

Example Question #5 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

Example Question #1 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

Example Question #1 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

Example Question #2 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

All AP Calculus AB Resources

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