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AP Calculus AB : Riemann sums (left, right, and midpoint evaluation points)

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Example Questions

Example Question #891 : Ap Calculus Ab

$\begin{align*}&\text{The function }f(x)\text{ has the following values on the interval }x=[1,11]:\\&f(1)=10\\&f(3)=3\\&f(5)=-13\\&f(7)=4\\&f(9)=-8\\&f(11)=-15\\&\text{Approximate the integral of }f(x)\text{ on this interval using right Riemann sums.}\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(1)=10\\&f(3)=3\\&f(5)=-13\\&f(7)=4\\&f(9)=-8\\&f(11)=-15\\&\text{Although the total interval has a length of }10\\&\text{, notice the points are equidistant, with }5\text{ intervals between.}\\&\text{Each equidistant interval has length: }2\\&\text{Since we are using the right point of each interval, we disregard the first function value:}\\&\int_{1}^{11}f(x)\approx2[(3)+(-13)+(4)+(-8)+(-15)]\\&\int_{1}^{11}f(x)\approx-58\end{align*}$

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Example Question #12 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

$\begin{align*}&\text{The function }f(x)\text{ has the following values on the interval }x=[-7,3]:\\&f(-7)=-18\\&f(-5)=-2\\&f(-3)=-20\\&f(-1)=16\\&f(1)=-12\\&f(3)=-17\\&\text{Approximate the integral of }f(x)\text{ on this interval using right Riemann sums.}\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(-7)=-18\\&f(-5)=-2\\&f(-3)=-20\\&f(-1)=16\\&f(1)=-12\\&f(3)=-17\\&\text{Although the total interval has a length of }10\\&\text{, notice the points are equidistant, with }5\text{ intervals between.}\\&\text{Each equidistant interval has length: }2\\&\text{Since we are using the right point of each interval, we disregard the first function value:}\\&\int_{-7}^{3}f(x)\approx2[(-2)+(-20)+(16)+(-12)+(-17)]\\&\int_{-7}^{3}f(x)\approx-70\end{align*}$

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Example Question #11 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

$\begin{align*}&\text{Approximate }\int_{5}^{13}f(x)\\&\text{Using a left Riemann sum, if }f(x)\text{ has the values:}\\&f(5)=-7\\&f(7)=-8\\&f(9)=-18\\&f(11)=-8\\&f(13)=-19\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(5)=-7\\&f(7)=-8\\&f(9)=-18\\&f(11)=-8\\&f(13)=-19\\&\text{Although the total interval has a length of }8\\&\text{, notice the points are equidistant, with }4\text{ intervals between.}\\&\text{Each equidistant interval has length: }2\\&\text{Since we are using the left point of each interval, we disregard the larst function value:}\\&\int_{5}^{13}f(x)\approx2[(-7)+(-8)+(-18)+(-8)]\\&\int_{5}^{13}f(x)\approx-82\end{align*}$

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Example Question #12 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

$\begin{align*}&\text{Approximate }\int_{-9}^{26}f(x)\\&\text{Using a left Riemann sum, if }f(x)\text{ has the values:}\\&f(-9)=-17\\&f(-2)=11\\&f(5)=17\\&f(12)=1\\&f(19)=-16\\&f(26)=13\end{align*}$

Possible Answers:

315

Correct answer:

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(-9)=-17\\&f(-2)=11\\&f(5)=17\\&f(12)=1\\&f(19)=-16\\&f(26)=13\\&\text{Although the total interval has a length of }35\\&\text{, notice the points are equidistant, with }5\text{ intervals between.}\\&\text{Each equidistant interval has length: }7\\&\text{Since we are using the left point of each interval, we disregard the larst function value:}\\&\int_{-9}^{26}f(x)\approx7[(-17)+(11)+(17)+(1)+(-16)]\\&\int_{-9}^{26}f(x)\approx-28\end{align*}$

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Example Question #15 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

$\begin{align*}&f(-10)=-19\\&f(-3)=7\\&f(4)=4\\&f(11)=1\\&\text{Approximate, using a left Riemann sum, }\int_{-10}^{11}f(x)\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(-10)=-19\\&f(-3)=7\\&f(4)=4\\&f(11)=1\\&\text{Although the total interval has a length of }21\\&\text{Notice the points are equidistant, with }3\text{ intervals between.}\\&\text{Each equidistant interval has length: }7\\&\text{Since we are using the left point of each interval, we disregard the larst function value:}\\&\int_{-10}^{11}f(x)\approx7[(-19)+(7)+(4)]\\&\int_{-10}^{11}f(x)\approx-56\end{align*}$

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Example Question #16 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

$\begin{align*}&\text{Approximate }\int_{-6}^{34}f(x)\\&\text{Using a left Riemann sum, if }f(x)\text{ has the values:}\\&f(-6)=8\\&f(2)=2\\&f(10)=-4\\&f(18)=-18\\&f(26)=11\\&f(34)=-7\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(-6)=8\\&f(2)=2\\&f(10)=-4\\&f(18)=-18\\&f(26)=11\\&f(34)=-7\\&\text{Although the total interval has a length of }40\\&\text{Notice the points are equidistant, with }5\text{ intervals between.}\\&\text{Each equidistant interval has length: }8\\&\text{Since we are using the left point of each interval, we disregard the larst function value:}\\&\int_{-6}^{34}f(x)\approx8[(8)+(2)+(-4)+(-18)+(11)]\\&\int_{-6}^{34}f(x)\approx-8\end{align*}$

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Example Question #191 : Integrals

$\begin{align*}&\text{The function }f(x)\text{ has the following values on the interval }x=[-8,2]:\\&f(-8)=2\\&f(-6)=-1\\&f(-4)=16\\&f(-2)=12\\&f(0)=10\\&f(2)=-18\\&\text{Approximate the integral of }f(x)\text{ on this interval using left Riemann sums.}\end{align*}$

Possible Answers:

210

Correct answer:

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(-8)=2\\&f(-6)=-1\\&f(-4)=16\\&f(-2)=12\\&f(0)=10\\&f(2)=-18\\&\text{Although the total interval has a length of }10\\&\text{Notice the points are equidistant, with }5\text{ intervals between.}\\&\text{Each equidistant interval has length: }2\\&\text{Since we are using the left point of each interval, we disregard the larst function value:}\\&\int_{-8}^{2}f(x)\approx2[(2)+(-1)+(16)+(12)+(10)]\\&\int_{-8}^{2}f(x)\approx78\end{align*}$

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Example Question #192 : Integrals

$\begin{align*}&\text{The function }f(x)\text{ has the following values on the interval }x=[5,29]:\\&f(5)=8\\&f(13)=-15\\&f(21)=9\\&f(29)=-16\\&\text{Approximate the integral of }f(x)\text{ on this interval using left Riemann sums.}\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(5)=8\\&f(13)=-15\\&f(21)=9\\&f(29)=-16\\&\text{Although the total interval has a length of }24\\&\text{Notice the points are equidistant, with }3\text{ intervals between.}\\&\text{Each equidistant interval has length: }8\\&\text{Since we are using the left point of each interval, we disregard the larst function value:}\\&\int_{5}^{29}f(x)\approx8[(8)+(-15)+(9)]\\&\int_{5}^{29}f(x)\approx16\end{align*}$

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Example Question #201 : Integrals

$\begin{align*}&f(0)=10\\&f(4)=3\\&f(8)=10\\&f(12)=-11\\&f(16)=10\\&\text{Approximate, using a left Riemann sum, }\int_{0}^{16}f(x)\end{align*}$

Possible Answers:

352

Correct answer:

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(0)=10\\&f(4)=3\\&f(8)=10\\&f(12)=-11\\&f(16)=10\\&\text{Although the total interval has a length of }16\\&\text{Notice the points are equidistant, with }4\text{ intervals between.}\\&\text{Each equidistant interval has length: }4\\&\text{Since we are using the left point of each interval, we disregard the larst function value:}\\&\int_{0}^{16}f(x)\approx4[(10)+(3)+(10)+(-11)]\\&\int_{0}^{16}f(x)\approx48\end{align*}$

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Example Question #901 : Ap Calculus Ab

$\begin{align*}&\text{The function }f(x)\text{ has the following values on the interval }x=[-5,5]:\\&f(-5)=-5\\&f(-3)=8\\&f(-1)=4\\&f(1)=12\\&f(3)=-5\\&f(5)=-12\\&\text{Approximate the integral of }f(x)\text{ on this interval using left Riemann sums.}\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(-5)=-5\\&f(-3)=8\\&f(-1)=4\\&f(1)=12\\&f(3)=-5\\&f(5)=-12\\&\text{Although the total interval has a length of }10\\&\text{Notice the points are equidistant, with }5\text{ intervals between.}\\&\text{Each equidistant interval has length: }2\\&\text{Since we are using the left point of each interval, we disregard the larst function value:}\\&\int_{-5}^{5}f(x)\approx2[(-5)+(8)+(4)+(12)+(-5)]\\&\int_{-5}^{5}f(x)\approx28\end{align*}$

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AP Calculus AB : Riemann sums (left, right, and midpoint evaluation points)

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #891 : Ap Calculus Ab

Example Question #12 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

Example Question #11 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

Example Question #12 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

Example Question #15 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

Example Question #16 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

Example Question #191 : Integrals

Example Question #192 : Integrals

Example Question #201 : Integrals

Example Question #901 : Ap Calculus Ab

All AP Calculus AB Resources

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