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AP Calculus AB : Numerical approximations to definite integrals

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Example Questions

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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Example Question #8 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

$\begin{align*}&\text{Approximate }\int_{-6}^{39}f(x)\\&\text{Using a right Riemann sum, if }f(x)\text{ has the values:}\\&f(-6)=-4\\&f(3)=-1\\&f(12)=11\\&f(21)=13\\&f(30)=-16\\&f(39)=-13\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(-6)=-4\\&f(3)=-1\\&f(12)=11\\&f(21)=13\\&f(30)=-16\\&f(39)=-13\\&\text{Although the total interval has a length of }45\\&\text{, notice the points are equidistant, with }5\text{ intervals between.}\\&\text{Each equidistant interval has length: }9\\&\text{Since we are using the right point of each interval, we disregard the first function value:}\\&\int_{-6}^{39}f(x)\approx9[(-1)+(11)+(13)+(-16)+(-13)]\\&\int_{-6}^{39}f(x)\approx-54\end{align*}$

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Example Question #191 : Integrals

$\begin{align*}&\text{Approximate }\int_{-5}^{13}f(x)\\&\text{Using a right Riemann sum, if }f(x)\text{ has the values:}\\&f(-5)=-13\\&f(1)=-12\\&f(7)=17\\&f(13)=7\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(-5)=-13\\&f(1)=-12\\&f(7)=17\\&f(13)=7\\&\text{Although the total interval has a length of }18\\&\text{, notice the points are equidistant, with }3\text{ intervals between.}\\&\text{Each equidistant interval has length: }6\\&\text{Since we are using the right point of each interval, we disregard the first function value:}\\&\int_{-5}^{13}f(x)\approx6[(-12)+(17)+(7)]\\&\int_{-5}^{13}f(x)\approx72\end{align*}$

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

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

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 #13 : 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 #14 : 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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AP Calculus AB : Numerical approximations to definite integrals

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

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

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

Example Question #8 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

Example Question #191 : Integrals

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

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

Example Question #13 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

Example Question #14 : 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)

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