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AP Calculus AB : Integrals

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3 Diagnostic Tests 164 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #3 : Trapezoidal Sums

$\begin{align*}&\text{Using }3\text{ intervals, and the method of trapezoidal sums approximate the integral:}\\&\int_{-1}^{8.9}(5sin(9e^{(x)}))dx\end{align*}$

Possible Answers:

35.68

6.37

4.20

114.16

Correct answer:

35.68

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{-1}^{8.9}(5sin(9e^{(x)}))dx\\&\text{So the interval is }[-1,8.9]\text{ the subintervals have length }\frac{8.9-(-1)}{3}=\frac{33}{10}\\&\text{and the x-values are:}\\&[-1,\frac{23}{10},\frac{28}{5},\frac{89}{10}]\\&\int_{-1}^{8.9}(5sin(9e^{(x)}))dx\approx\frac{33}{20}[(5sin(9e^{((-1))}))+2(5sin(9e^{((\frac{23}{10}))}))+2(5sin(9e^{((\frac{28}{5}))}))+(5sin(9e^{((\frac{89}{10}))}))]\\&\int_{-1}^{8.9}(5sin(9e^{(x)}))dx\approx35.68\end{align*}$

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Example Question #4 : Trapezoidal Sums

$\begin{align*}&\text{Using the method of trapezoidal sums approximate the integral:}\\&\int_{1}^{6.4}(-6cos(19x^{2}))dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

-16.73

-33.45

-93.67

-138.84

Correct answer:

-16.73

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{1}^{6.4}(-6cos(19x^{2}))dx\\&\text{So the interval is }[1,6.4]\text{ the subintervals have length }\frac{6.4-(1)}{3}=\frac{9}{5}\\&\text{and the x-values are:}\\&[1,\frac{14}{5},\frac{23}{5},\frac{32}{5}]\\&\int_{1}^{6.4}(-6cos(19x^{2}))dx\approx\frac{9}{10}[(-6cos(19(1)^{2}))+2(-6cos(19(\frac{14}{5})^{2}))+2(-6cos(19(\frac{23}{5})^{2}))+(-6cos(19(\frac{32}{5})^{2}))]\\&\int_{1}^{6.4}(-6cos(19x^{2}))dx\approx-16.73\end{align*}$

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Example Question #5 : Trapezoidal Sums

$\begin{align*}&\text{Using the method of trapezoidal sums approximate the integral:}\\&\int_{2}^{6.5}(17cos(2sin(2x)))dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

5.51

3.13

7.65

27.55

Correct answer:

27.55

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{2}^{6.5}(17cos(2sin(2x)))dx\\&\text{So the interval is }[2,6.5]\text{ the subintervals have length }\frac{6.5-(2)}{3}=\frac{3}{2}\\&\text{and the x-values are:}\\&[2,\frac{7}{2},5,\frac{13}{2}]\\&\int_{2}^{6.5}(17cos(2sin(2x)))dx\approx\frac{3}{4}[(17cos(2sin(2(2))))+2(17cos(2sin(2(\frac{7}{2}))))+2(17cos(2sin(2(5))))+(17cos(2sin(2(\frac{13}{2}))))]\\&\int_{2}^{6.5}(17cos(2sin(2x)))dx\approx27.55\end{align*}$

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Example Question #6 : Trapezoidal Sums

$\begin{align*}&\text{Using the method of trapezoidal sums approximate the integral:}\\&\int_{0}^{4.4}(-196cos(3x)^{2})dx\\&\text{Using }2\text{ intervals.}\end{align*}$

Possible Answers:

-2160.41

-88.69

-130.70

-744.97

Correct answer:

-744.97

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{0}^{4.4}(-196cos(3x)^{2})dx\\&\text{So the interval is }[0,4.4]\text{ the subintervals have length }\frac{4.4-(0)}{2}=\frac{11}{5}\\&\text{and the x-values are:}\\&[0,\frac{11}{5},\frac{22}{5}]\\&\int_{0}^{4.4}(-196cos(3x)^{2})dx\approx\frac{11}{10}[(-196cos(3(0))^{2})+2(-196cos(3(\frac{11}{5}))^{2})+(-196cos(3(\frac{22}{5}))^{2})]\\&\int_{0}^{4.4}(-196cos(3x)^{2})dx\approx-744.97\end{align*}$

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Example Question #7 : Trapezoidal Sums

$\begin{align*}&\text{Using }3\text{ intervals, and the method of trapezoidal sums approximate the integral:}\\&\int_{3}^{7.2}(11cos(6e^{(4x)}))dx\end{align*}$

Possible Answers:

-0.56

-18.44

-1.86

-0.29

Correct answer:

-1.86

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{3}^{7.2}(11cos(6e^{(4x)}))dx\\&\text{So the interval is }[3,7.2]\text{ the subintervals have length }\frac{7.2-(3)}{3}=\frac{7}{5}\\&\text{and the x-values are:}\\&[3,\frac{22}{5},\frac{29}{5},\frac{36}{5}]\\&\int_{3}^{7.2}(11cos(6e^{(4x)}))dx\approx\frac{7}{10}[(11cos(6e^{(4(3))}))+2(11cos(6e^{(4(\frac{22}{5}))}))+2(11cos(6e^{(4(\frac{29}{5}))}))+(11cos(6e^{(4(\frac{36}{5}))}))]\\&\int_{3}^{7.2}(11cos(6e^{(4x)}))dx\approx-1.86\end{align*}$

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Example Question #8 : Trapezoidal Sums

$\begin{align*}&\text{Calculate the trapezoidal sums integral approximation of :}\\&\int_{-5}^{-3.8}(-16sin(9e^{(5x)}))dx\\&\text{Using points over }2\text{ intervals.}\end{align*}$

Possible Answers:

-12.19

-3.05

-0.38

-0.59

Correct answer:

-3.05

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{-5}^{-3.8}(-16sin(9e^{(5x)}))dx\\&\text{So the interval is }[-5,-3.8]\text{ the subintervals have length }\frac{-3.8-(-5)}{2}=\frac{3}{5}\\&\text{and the x-values are:}\\&[-5,-\frac{22}{5},\frac{11}{5}]\\&\int_{-5}^{-3.8}(-16sin(9e^{(5x)}))dx\approx\frac{3}{10}[(-16sin(9e^{(5(-5))}))+2(-16sin(9e^{(5(-\frac{22}{5}))}))+(-16sin(9e^{(5(\frac{11}{5}))}))]\\&\int_{-5}^{-3.8}(-16sin(9e^{(5x)}))dx\approx-3.05\end{align*}$

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Example Question #9 : Trapezoidal Sums

$\begin{align*}&\text{Using }3\text{ intervals, and the method of trapezoidal sums approximate the integral:}\\&\int_{-1}^{6.5}(2700sin(3x)^{2})dx\end{align*}$

Possible Answers:

53340.17

91163.20

4849.11

9698.21

Correct answer:

9698.21

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{-1}^{6.5}(2700sin(3x)^{2})dx\\&\text{So the interval is }[-1,6.5]\text{ the subintervals have length }\frac{6.5-(-1)}{3}=\frac{5}{2}\\&\text{and the x-values are:}\\&[-1,\frac{3}{2},4,\frac{13}{2}]\\&\int_{-1}^{6.5}(2700sin(3x)^{2})dx\approx\frac{5}{4}[(2700sin(3(-1))^{2})+2(2700sin(3(\frac{3}{2}))^{2})+2(2700sin(3(4))^{2})+(2700sin(3(\frac{13}{2}))^{2})]\\&\int_{-1}^{6.5}(2700sin(3x)^{2})dx\approx9698.21\end{align*}$

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Example Question #10 : Trapezoidal Sums

$\begin{align*}&\text{Calculate the trapezoidal sums integral approximation of :}\\&\int_{2}^{9}(17\cdot 2^ln(3x))dx\\&\text{Using points over }2\text{ intervals.}\end{align*}$

Possible Answers:

3079.98

810.52

84.43

5349.44

Correct answer:

810.52

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{2}^{9}(17\cdot 2^ln(3x))dx\\&\text{So the interval is }[2,9]\text{ the subintervals have length }\frac{9-(2)}{2}=\frac{7}{2}\\&\text{and the x-values are:}\\&[2,\frac{11}{2},9]\\&\int_{2}^{9}(17\cdot 2^ln(3x))dx\approx\frac{7}{4}[(17\cdot 2^ln(3(2)))+2(17\cdot 2^ln(3(\frac{11}{2})))+(17\cdot 2^ln(3(9)))]\\&\int_{2}^{9}(17\cdot 2^ln(3x))dx\approx810.52\end{align*}$

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Example Question #1081 : Ap Calculus Ab

$\begin{align*}&\text{Calculate the trapezoidal sums integral approximation of :}\\&\int_{5}^{13}(-4cos(14e^{(6x)}))dx\\&\text{Using points over }2\text{ intervals.}\end{align*}$

Possible Answers:

79.78

9.61

2.40

1.41

Correct answer:

9.61

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{5}^{13}(-4cos(14e^{(6x)}))dx\\&\text{So the interval is }[5,13]\text{ the subintervals have length }\frac{13-(5)}{2}=4\\&\text{and the x-values are:}\\&[5,9,13]\\&\int_{5}^{13}(-4cos(14e^{(6x)}))dx\approx2[(-4cos(14e^{(6(5))}))+2(-4cos(14e^{(6(9))}))+(-4cos(14e^{(6(13))}))]\\&\int_{5}^{13}(-4cos(14e^{(6x)}))dx\approx9.61\end{align*}$

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Example Question #1082 : Ap Calculus Ab

$\begin{align*}&\text{Calculate the trapezoidal sums integral approximation of :}\\&\int_{1}^{11.5}(13tan(18cos(x)))dx\\&\text{Using points over }3\text{ intervals.}\end{align*}$

Possible Answers:

-21.69

-143.16

-67.24

-2.41

Correct answer:

-21.69

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{1}^{11.5}(13tan(18cos(x)))dx\\&\text{So the interval is }[1,11.5]\text{ the subintervals have length }\frac{11.5-(1)}{3}=\frac{7}{2}\\&\text{and the x-values are:}\\&[1,\frac{9}{2},8,\frac{23}{2}]\\&\int_{1}^{11.5}(13tan(18cos(x)))dx\approx\frac{7}{4}[(13tan(18cos((1))))+2(13tan(18cos((\frac{9}{2}))))+2(13tan(18cos((8))))+(13tan(18cos((\frac{23}{2}))))]\\&\int_{1}^{11.5}(13tan(18cos(x)))dx\approx-21.69\end{align*}$

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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 #3 : Trapezoidal Sums

Example Question #4 : Trapezoidal Sums

Example Question #5 : Trapezoidal Sums

Example Question #6 : Trapezoidal Sums

Example Question #7 : Trapezoidal Sums

Example Question #8 : Trapezoidal Sums

Example Question #9 : Trapezoidal Sums

Example Question #10 : Trapezoidal Sums

Example Question #1081 : Ap Calculus Ab

Example Question #1082 : Ap Calculus Ab

All AP Calculus AB Resources

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