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AP Calculus AB : AP Calculus AB

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Example Questions

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 #7 : 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:

9698.21

53340.17

4849.11

91163.20

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 #8 : 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:

84.43

5349.44

810.52

3079.98

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 #51 : Numerical Approximations To Definite Integrals

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

Possible Answers:

-2.41

-21.69

-67.24

-143.16

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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Example Question #281 : Integrals

$\begin{align*}&\text{The function }f(x)\text{ has the following values on the interval }x=[-7,9]:\\&f(-7)=-17\\&f(-3)=17\\&f(1)=8\\&f(5)=2\\&f(9)=-8\\&\text{Approximate the integral of }f(x)\text{ on this interval using the trapezoidal rule.}\end{align*}$

Possible Answers:

464

116

Correct answer:

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're given a set of values:}\\&f(-7)=-17\\&f(-3)=17\\&f(1)=8\\&f(5)=2\\&f(9)=-8\\&\text{The total interval has a length of }16\\&\text{Notice the points are equidistant, with }4\text{ intervals between.}\\&\text{Applying the rules above, we can write:}\\&\int_{-7}^{9}f(x)\approx\frac{16}{2\cdot4}[(-17)+2(17)+2(8)+2(2)+(-8)]\\&\int_{-7}^{9}f(x)\approx58\end{align*}$

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Example Question #282 : Integrals

$\begin{align*}&f(-8)=-10\\&f(1)=-4\\&f(10)=-17\\&f(19)=8\\&f(28)=-4\\&\text{Approximate, using a trapezoidal sum, }\int_{-8}^{28}f(x)\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're given a set of values:}\\&f(-8)=-10\\&f(1)=-4\\&f(10)=-17\\&f(19)=8\\&f(28)=-4\\&\text{The total interval has a length of }36\\&\text{Notice the points are equidistant, with }4\text{ intervals between.}\\&\text{Applying the rules above, we can write:}\\&\int_{-8}^{28}f(x)\approx\frac{36}{2\cdot4}[(-10)+2(-4)+2(-17)+2(8)+(-4)]\\&\int_{-8}^{28}f(x)\approx-180\end{align*}$

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Example Question #283 : Integrals

$\begin{align*}&f(-8)=-5\\&f(-1)=-14\\&f(6)=11\\&f(13)=15\\&f(20)=-6\\&\text{Approximate, using a trapezoidal sum, }\int_{-8}^{20}f(x)\end{align*}$

Possible Answers:

45.5

364

Correct answer:

45.5

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're given a set of values:}\\&f(-8)=-5\\&f(-1)=-14\\&f(6)=11\\&f(13)=15\\&f(20)=-6\\&\text{The total interval has a length of }28\\&\text{Notice the points are equidistant, with }4\text{ intervals between.}\\&\text{Applying the rules above, we can write:}\\&\int_{-8}^{20}f(x)\approx\frac{28}{2\cdot4}[(-5)+2(-14)+2(11)+2(15)+(-6)]\\&\int_{-8}^{20}f(x)\approx45.5\end{align*}$

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Example Question #284 : Integrals

$\begin{align*}&\text{Approximate }\int_{3}^{21}f(x)\\&\text{Using the trapezoidal rule, if }f(x)\text{ has the values:}\\&f(3)=4\\&f(9)=-7\\&f(15)=-8\\&f(21)=-2\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're given a set of values:}\\&f(3)=4\\&f(9)=-7\\&f(15)=-8\\&f(21)=-2\\&\text{The total interval has a length of }18\\&\text{Notice the points are equidistant, with }3\text{ intervals between.}\\&\text{Applying the rules above, we can write:}\\&\int_{3}^{21}f(x)\approx\frac{18}{2\cdot3}[(4)+2(-7)+2(-8)+(-2)]\\&\int_{3}^{21}f(x)\approx-84\end{align*}$

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Example Question #285 : Integrals

$\begin{align*}&\text{The function }f(x)\text{ has the following values on the interval }x=[1,25]:\\&f(1)=-3\\&f(7)=-3\\&f(13)=-15\\&f(19)=-19\\&f(25)=-9\\&\text{Approximate the integral of }f(x)\text{ on this interval using the trapezoidal rule.}\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're given a set of values:}\\&f(1)=-3\\&f(7)=-3\\&f(13)=-15\\&f(19)=-19\\&f(25)=-9\\&\text{The total interval has a length of }24\\&\text{Notice the points are equidistant, with }4\text{ intervals between.}\\&\text{Applying the rules above, we can write:}\\&\int_{1}^{25}f(x)\approx\frac{24}{2\cdot4}[(-3)+2(-3)+2(-15)+2(-19)+(-9)]\\&\int_{1}^{25}f(x)\approx-258\end{align*}$

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AP Calculus AB : AP Calculus AB

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #8 : Trapezoidal Sums

Example Question #7 : Trapezoidal Sums

Example Question #8 : Trapezoidal Sums

Example Question #1081 : Ap Calculus Ab

Example Question #51 : Numerical Approximations To Definite Integrals

Example Question #281 : Integrals

Example Question #282 : Integrals

Example Question #283 : Integrals

Example Question #284 : Integrals

Example Question #285 : Integrals

All AP Calculus AB Resources

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