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

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

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

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

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

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 #6 : 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 #7 : 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:

180

155

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 #1 : 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 #9 : 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 #10 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

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

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

Study concepts, example questions & explanations for AP Calculus AB

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

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

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 #6 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

Example Question #7 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

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

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