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AP Calculus BC : Integrals

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Example Questions

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Example Question #11 : Numerical Approximations To Definite Integrals

$\begin{align*}&\text{Using the method of left point Riemann sums approximate the integral:}\\&\int_{2}^{19.2}(-3cos(6ln(3x)))dx\\&\text{Using }4\text{ intervals.}\end{align*}$

Possible Answers:

1.46

11.80

1.81

3.19

Correct answer:

11.80

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{2}^{19.2}(-3cos(6ln(3x)))dx\\&\text{So the interval is }[2,19.2]\text{ the subintervals have length }\frac{19.2-(2)}{4}=\frac{43}{10}\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[2,\frac{63}{10},\frac{53}{5},\frac{149}{10}]\\&\int_{2}^{19.2}(-3cos(6ln(3x)))dx=\frac{43}{10}[(-3cos(6ln(6)))+(-3cos(6ln(\frac{189}{10})))+(-3cos(6ln(\frac{159}{5})))+(-3cos(6ln(\frac{447}{10})))]\\&\int_{2}^{19.2}(-3cos(6ln(3x)))dx=11.80\end{align*}$

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Example Question #11 : Integrals

$\begin{align*}&\text{Using }3\text{ intervals, and the method of left point Riemann sums approximate the integral:}\\&\int_{-1}^{10.7}(-275tan(4x)^{2})dx\end{align*}$

Possible Answers:

-7340.85

-22756.63

-1048.69

-60194.95

Correct answer:

-7340.85

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-1}^{10.7}(-275tan(4x)^{2})dx\\&\text{So the interval is }[-1,10.7]\text{ the subintervals have length }\frac{10.7-(-1)}{3}=\frac{39}{10}\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[-1,\frac{29}{10},\frac{34}{5}]\\&\int_{-1}^{10.7}(-275tan(4x)^{2})dx=\frac{39}{10}[(-275tan(4)^{2})+(-275tan(\frac{58}{5})^{2})+(-275tan(\frac{136}{5})^{2})]\\&\int_{-1}^{10.7}(-275tan(4x)^{2})dx=-7340.85\end{align*}$

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Example Question #11 : Riemann Sum: Left Evaluation

$\begin{align*}&\text{Using the method of left point Riemann sums approximate the integral:}\\&\int_{1}^{6.4}(14tan(14cos(6x)))dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

-394.53

-41.53

-146.12

-2012.10

Correct answer:

-394.53

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{1}^{6.4}(14tan(14cos(6x)))dx\\&\text{So the interval is }[1,6.4]\text{ the subintervals have length }\frac{6.4-(1)}{3}=\frac{9}{5}\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[1,\frac{14}{5},\frac{23}{5}]\\&\int_{1}^{6.4}(14tan(14cos(6x)))dx=\frac{9}{5}[(14tan(14cos(6)))+(14tan(14cos(\frac{84}{5})))+(14tan(14cos(\frac{138}{5})))]\\&\int_{1}^{6.4}(14tan(14cos(6x)))dx=-394.53\end{align*}$

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Example Question #11 : Integrals

$\begin{align*}&\text{Using }4\text{ intervals, and the method of left point Riemann sums approximate the integral:}\\&\int_{3}^{22.6}(10tan(8e^{(4x)}))dx\end{align*}$

Possible Answers:

-19.64

-168.87

-506.60

-25.59

Correct answer:

-168.87

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{3}^{22.6}(10tan(8e^{(4x)}))dx\\&\text{So the interval is }[3,22.6]\text{ the subintervals have length }\frac{22.6-(3)}{4}=\frac{49}{10}\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[3,\frac{79}{10},\frac{64}{5},\frac{177}{10}]\\&\int_{3}^{22.6}(10tan(8e^{(4x)}))dx=\frac{49}{10}[(10tan(8e^{(12)}))+(10tan(8e^{(\frac{158}{5})}))+(10tan(8e^{(\frac{256}{5})}))+(10tan(8e^{(\frac{354}{5})}))]\\&\int_{3}^{22.6}(10tan(8e^{(4x)}))dx=-168.87\end{align*}$

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Example Question #11 : Integrals

$\begin{align*}&\text{Using }3\text{ intervals, and the method of left point Riemann sums approximate the integral:}\\&\int_{5}^{6.5}(6tan(19cos(5x)))dx\end{align*}$

Possible Answers:

-5.54

-1.38

-0.63

-0.91

Correct answer:

-5.54

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{5}^{6.5}(6tan(19cos(5x)))dx\\&\text{So the interval is }[5,6.5]\text{ the subintervals have length }\frac{6.5-(5)}{3}=\frac{1}{2}\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[5,\frac{11}{2},6]\\&\int_{5}^{6.5}(6tan(19cos(5x)))dx=\frac{1}{2}[(6tan(19cos(25)))+(6tan(19cos(\frac{55}{2})))+(6tan(19cos(30)))]\\&\int_{5}^{6.5}(6tan(19cos(5x)))dx=-5.54\end{align*}$

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Example Question #1 : Riemann Sum: Right Evaluation

Given a function $y=\frac{5}{x}$ , find the Right Riemann Sum of the function on the interval [0,4] divided into four sub-intervals.

Possible Answers:

$\frac{122}{12}$

$\frac{124}{12}$

$\frac{123}{12}$

$\frac{125}{12}$

$\frac{126}{12}$

Correct answer:

$\frac{125}{12}$

Explanation:

In order to find the Riemann Sum of a given function, we need to approximate the area under the line or curve resulting from the function using rectangles spaced along equal sub-intervals of a given interval. Since we have an interval [0,4] divided into sub-intervals, we'll be using rectangles with vertices at x=0,1,2,3,4 .

To approximate the area under the curve, we need to find the areas of each rectangle in the sub-intervals. We already know the width or base of each rectangle is b=1 because the rectangles are spaced unit apart. Since we're looking for the Right Riemann Sum of $y=\frac{5}{x}$ , we want to find the heights of each rectangle by taking the values of each rightmost function value on each sub-interval, as follows:

$f(4)=\frac{5}{4}$

$f(3)=\frac{5}{3}$

$f(2)=\frac{5}{2}$

$f(1)=\frac{5}{1}=5$

Putting it all together, the Right Riemann Sum is

$A=bh=1(\frac{5}{4}+\frac{5}{3}+\frac{5}{2}+5)=\frac{15}{12}+\frac{20}{12}+\frac{30}{12}+\frac{60}{12}=\frac{125}{12}$ .

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Example Question #1 : Riemann Sum: Right Evaluation

$\begin{align*}&\text{Using the method of right Riemann sums approximate the integral:}\\&\int_{1}^{15.8}(-15sin(14e^{(4x)}))dx\\&\text{Using }4\text{ intervals.}\end{align*}$

Possible Answers:

191.30

20.79

5.20

3.35

Correct answer:

20.79

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{1}^{15.8}(-15sin(14e^{(4x)}))dx\\&\text{So the interval is }[1,15.8]\text{ the subintervals have length }\frac{15.8-(1)}{4}=\frac{37}{10}\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[\frac{47}{10},\frac{42}{5},\frac{121}{10},\frac{79}{5}]\\&\int_{1}^{15.8}(-15sin(14e^{(4x)}))dx=\frac{37}{10}[(-15sin(14e^{(4(\frac{47}{10}))}))+(-15sin(14e^{(4(\frac{42}{5}))}))+(-15sin(14e^{(4(\frac{121}{10}))}))+(-15sin(14e^{(4(\frac{79}{5}))}))]\\&\int_{1}^{15.8}(-15sin(14e^{(4x)}))dx=20.79\end{align*}$

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Example Question #1 : Riemann Sum: Right Evaluation

$\begin{align*}&\text{Using the method of right Riemann sums approximate the integral:}\\&\int_{4}^{14}(-8sin(12e^{(x)}))dx\\&\text{Using }4\text{ intervals.}\end{align*}$

Possible Answers:

-143.01

-89.59

-51.69

-17.23

Correct answer:

-17.23

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{4}^{14}(-8sin(12e^{(x)}))dx\\&\text{So the interval is }[4,14]\text{ the subintervals have length }\frac{14-(4)}{4}=\frac{5}{2}\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[\frac{13}{2},9,\frac{23}{2},14]\\&\int_{4}^{14}(-8sin(12e^{(x)}))dx=\frac{5}{2}[(-8sin(12e^{((\frac{13}{2}))}))+(-8sin(12e^{((9))}))+(-8sin(12e^{((\frac{23}{2}))}))+(-8sin(12e^{((14))}))]\\&\int_{4}^{14}(-8sin(12e^{(x)}))dx=-17.23\end{align*}$

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Example Question #1 : Riemann Sum: Right Evaluation

$\begin{align*}&\text{Using the method of right Riemann sums approximate the integral:}\\&\int_{0}^{6}(5x - 1000sin(3x)^{2})dx\\&\text{Using }4\text{ intervals.}\end{align*}$

Possible Answers:

-403.65

-3390.69

-1412.79

-21361.33

Correct answer:

-3390.69

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{0}^{6}(5x - 1000sin(3x)^{2})dx\\&\text{So the interval is }[0,6]\text{ the subintervals have length }\frac{6-(0)}{4}=\frac{3}{2}\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[\frac{3}{2},3,\frac{9}{2},6]\\&\int_{0}^{6}(5x - 1000sin(3x)^{2})dx=\frac{3}{2}[(5(\frac{3}{2}) - 1000sin(3(\frac{3}{2}))^{2})+(5(3) - 1000sin(3(3))^{2})+(5(\frac{9}{2}) - 1000sin(3(\frac{9}{2}))^{2})+(5(6) - 1000sin(3(6))^{2})]\\&\int_{0}^{6}(5x - 1000sin(3x)^{2})dx=-3390.69\end{align*}$

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Example Question #1 : Riemann Sum: Right Evaluation

$\begin{align*}&\text{Using the method of right Riemann sums approximate the integral:}\\&\int_{-4}^{-1.6}(15cos(12sin(6x)) + 5x^{3})dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

-1732.45

-208.73

-52.18

-31.63

Correct answer:

-208.73

Explanation:

$\begin{align*}\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[-\frac{16}{5},-\frac{12}{5},-\frac{8}{5}]\\&\int_{-4}^{-1.6}(15cos(12sin(6x)) + 5x^{3})dx\approx\frac{4}{5}[(15cos(12sin(6(-\frac{16}{5}))) + 5(-\frac{16}{5})^{3})+(15cos(12sin(6(-\frac{12}{5}))) + 5(-\frac{12}{5})^{3})+(15cos(12sin(6(-\frac{8}{5}))) + 5(-\frac{8}{5})^{3})]\\&\int_{-4}^{-1.6}(15cos(12sin(6x)) + 5x^{3})dx\approx-208.73\end{align*}$

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AP Calculus BC : Integrals

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #11 : Numerical Approximations To Definite Integrals

Example Question #11 : Integrals

Example Question #11 : Riemann Sum: Left Evaluation

Example Question #11 : Integrals

Example Question #11 : Integrals

Example Question #1 : Riemann Sum: Right Evaluation

Example Question #1 : Riemann Sum: Right Evaluation

Example Question #1 : Riemann Sum: Right Evaluation

Example Question #1 : Riemann Sum: Right Evaluation

Example Question #1 : Riemann Sum: Right Evaluation

All AP Calculus BC Resources

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