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AP Calculus AB : Trapezoidal sums

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Example Questions

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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Example Question #1083 : Ap Calculus Ab

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

Possible Answers:

116

464

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 #1084 : Ap Calculus Ab

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

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 #1085 : Ap Calculus Ab

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

Possible Answers:

364

45.5

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 #1086 : Ap Calculus Ab

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

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 #1087 : Ap Calculus Ab

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

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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Example Question #1091 : Ap Calculus Ab

$\begin{align*}&f(4)=-2\\&f(13)=-11\\&f(22)=11\\&f(31)=11\\&f(40)=10\\&\text{Approximate, using a trapezoidal sum, }\int_{4}^{40}f(x)\end{align*}$

Possible Answers:

135

1080

270

Correct answer:

135

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're given a set of values:}\\&f(4)=-2\\&f(13)=-11\\&f(22)=11\\&f(31)=11\\&f(40)=10\\&\text{The total interval has a length of }36\\&\text{Notice the points are equidistant, with }4\text{ intervals between.}\\&\text{Applying the rules above, we can write:}\\&\int_{4}^{40}f(x)\approx\frac{36}{2\cdot4}[(-2)+2(-11)+2(11)+2(11)+(10)]\\&\int_{4}^{40}f(x)\approx135\end{align*}$

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Example Question #1092 : Ap Calculus Ab

$\begin{align*}&\text{The function }f(x)\text{ has the following values on the interval }x=[-3,18]:\\&f(-3)=-12\\&f(4)=-16\\&f(11)=13\\&f(18)=-13\\&\text{Approximate the integral of }f(x)\text{ on this interval using the trapezoidal rule.}\end{align*}$

Possible Answers:

-108.5

Correct answer:

-108.5

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're given a set of values:}\\&f(-3)=-12\\&f(4)=-16\\&f(11)=13\\&f(18)=-13\\&\text{The total interval has a length of }21\\&\text{Notice the points are equidistant, with }3\text{ intervals between.}\\&\text{Applying the rules above, we can write:}\\&\int_{-3}^{18}f(x)\approx\frac{21}{2\cdot3}[(-12)+2(-16)+2(13)+(-13)]\\&\int_{-3}^{18}f(x)\approx-108.5\end{align*}$

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Example Question #1093 : Ap Calculus Ab

$\begin{align*}&\text{The function }f(x)\text{ has the following values on the interval }x=[-2,34]:\\&f(-2)=8\\&f(7)=-14\\&f(16)=19\\&f(25)=2\\&f(34)=7\\&\text{Approximate the integral of }f(x)\text{ on this interval using the trapezoidal rule.}\end{align*}$

Possible Answers:

1044

130.5

261

Correct answer:

130.5

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're given a set of values:}\\&f(-2)=8\\&f(7)=-14\\&f(16)=19\\&f(25)=2\\&f(34)=7\\&\text{The total interval has a length of }36\\&\text{Notice the points are equidistant, with }4\text{ intervals between.}\\&\text{Applying the rules above, we can write:}\\&\int_{-2}^{34}f(x)\approx\frac{36}{2\cdot4}[(8)+2(-14)+2(19)+2(2)+(7)]\\&\int_{-2}^{34}f(x)\approx130.5\end{align*}$

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AP Calculus AB : Trapezoidal sums

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #1081 : Ap Calculus Ab

Example Question #1082 : Ap Calculus Ab

Example Question #1083 : Ap Calculus Ab

Example Question #1084 : Ap Calculus Ab

Example Question #1085 : Ap Calculus Ab

Example Question #1086 : Ap Calculus Ab

Example Question #1087 : Ap Calculus Ab

Example Question #1091 : Ap Calculus Ab

Example Question #1092 : Ap Calculus Ab

Example Question #1093 : Ap Calculus Ab

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