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AP Calculus AB : Numerical approximations to definite integrals

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Example Question #30 : Trapezoidal Sums

$\begin{align*}&f(-9)=9\\&f(1)=6\\&f(9)=14\\&f(13)=-4\\&f(23)=10\\&\text{Approximate, using a trapezoidal sum, }\int_{-9}^{23}f(x)\end{align*}$

Possible Answers:

204

205

25.5

816

Correct answer:

205

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're given a set of values:}\\&f(-9)=9\\&f(1)=6\\&f(9)=14\\&f(13)=-4\\&f(23)=10\\&\text{Notice that the points are not equidistant, with varying interval lengths between}\\&\text{These lengths in sequence are: }10,8,4,10\\&\text{This means we should calculate each trapezoid separately: }\sum_{i=1}^{n}(x_{i+1}-x_i)\frac{f(x_{i+1})+f(x_i)}{2}\\&\int_{-9}^{23}f(x)\approx(10)\frac{9+(6)}{2}+(8)\frac{6+(14)}{2}+(4)\frac{14+(-4)}{2}+(10)\frac{-4+(10)}{2}\\&\int_{-9}^{23}f(x)\approx205\end{align*}$

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Example Question #31 : Trapezoidal Sums

$\begin{align*}&\text{Approximate }\int_{-5}^{17}f(x)\\&\text{Using the trapezoidal rule, if }f(x)\text{ has the values:}\\&f(-5)=5\\&f(1)=-6\\&f(11)=11\\&f(17)=-4\end{align*}$

Possible Answers:

121

5.5

40.3333

Correct answer:

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're given a set of values:}\\&f(-5)=5\\&f(1)=-6\\&f(11)=11\\&f(17)=-4\\&\text{Notice that the points are not equidistant, with varying interval lengths between}\\&\text{These lengths in sequence are: }6,10,6\\&\text{This means we should calculate each trapezoid separately: }\sum_{i=1}^{n}(x_{i+1}-x_i)\frac{f(x_{i+1})+f(x_i)}{2}\\&\int_{-5}^{17}f(x)\approx(6)\frac{5+(-6)}{2}+(10)\frac{-6+(11)}{2}+(6)\frac{11+(-4)}{2}\\&\int_{-5}^{17}f(x)\approx43\end{align*}$

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Example Question #32 : Trapezoidal Sums

$\begin{align*}&\text{The function }f(x)\text{ has the following values on the interval }x=[1,32]:\\&f(1)=15\\&f(3)=-9\\&f(7)=9\\&f(13)=-15\\&f(17)=14\\&\text{Approximate the integral of }f(x)\text{ on this interval using the trapezoidal rule.}\end{align*}$

Possible Answers:

-3.875

-23.101

-0.74519

Correct answer:

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're given a set of values:}\\&f(1)=15\\&f(3)=-9\\&f(7)=9\\&f(13)=-15\\&f(17)=14\\&\text{Notice that the points are not equidistant, with varying interval lengths between}\\&\text{These lengths in sequence are: }2,4,6,4\\&\text{This means we should calculate each trapezoid separately: }\sum_{i=1}^{n}(x_{i+1}-x_i)\frac{f(x_{i+1})+f(x_i)}{2}\\&\int_{1}^{32}f(x)\approx(2)\frac{15+(-9)}{2}+(4)\frac{-9+(9)}{2}+(6)\frac{9+(-15)}{2}+(4)\frac{-15+(14)}{2}\\&\int_{1}^{32}f(x)\approx-14\end{align*}$

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Example Question #31 : Trapezoidal Sums

$\begin{align*}&f(-5)=-2\\&f(-1)=19\\&f(9)=-19\\&f(15)=19\\&f(25)=-13\\&\text{Approximate, using a trapezoidal sum, }\int_{-5}^{25}f(x)\end{align*}$

Possible Answers:

345

86.25

11.5

Correct answer:

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're given a set of values:}\\&f(-5)=-2\\&f(-1)=19\\&f(9)=-19\\&f(15)=19\\&f(25)=-13\\&\text{Notice that the points are not equidistant, with varying interval lengths between}\\&\text{These lengths in sequence are: }4,10,6,10\\&\text{This means we should calculate each trapezoid separately: }\sum_{i=1}^{n}(x_{i+1}-x_i)\frac{f(x_{i+1})+f(x_i)}{2}\\&\int_{-5}^{25}f(x)\approx(4)\frac{-2+(19)}{2}+(10)\frac{19+(-19)}{2}+(6)\frac{-19+(19)}{2}+(10)\frac{19+(-13)}{2}\\&\int_{-5}^{25}f(x)\approx64\end{align*}$

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Example Question #34 : Trapezoidal Sums

$\begin{align*}&\text{The function }f(x)\text{ has the following values on the interval }x=[-10,20]:\\&f(-10)=-17\\&f(-2)=19\\&f(2)=6\\&f(10)=-11\\&f(20)=-4\\&\text{Approximate the integral of }f(x)\text{ on this interval using the trapezoidal rule.}\end{align*}$

Possible Answers:

105

3.5

26.25

Correct answer:

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're given a set of values:}\\&f(-10)=-17\\&f(-2)=19\\&f(2)=6\\&f(10)=-11\\&f(20)=-4\\&\text{Notice that the points are not equidistant, with varying interval lengths between}\\&\text{These lengths in sequence are: }8,4,8,10\\&\text{This means we should calculate each trapezoid separately: }\sum_{i=1}^{n}(x_{i+1}-x_i)\frac{f(x_{i+1})+f(x_i)}{2}\\&\int_{-10}^{20}f(x)\approx(8)\frac{-17+(19)}{2}+(4)\frac{19+(6)}{2}+(8)\frac{6+(-11)}{2}+(10)\frac{-11+(-4)}{2}\\&\int_{-10}^{20}f(x)\approx-37\end{align*}$

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Example Question #35 : Trapezoidal Sums

$\begin{align*}&\text{Approximate }\int_{4}^{11}f(x)\\&\text{Using the trapezoidal rule, if }f(x)\text{ has the values:}\\&f(4)=-17\\&f(6)=18\\&f(10)=-4\\&f(11)=-19\end{align*}$

Possible Answers:

17.5

-9.3333

Correct answer:

17.5

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're given a set of values:}\\&f(4)=-17\\&f(6)=18\\&f(10)=-4\\&f(11)=-19\\&\text{Notice that the points are not equidistant, with varying interval lengths between}\\&\text{These lengths in sequence are: }2,4,1\\&\text{This means we should calculate each trapezoid separately: }\sum_{i=1}^{n}(x_{i+1}-x_i)\frac{f(x_{i+1})+f(x_i)}{2}\\&\int_{4}^{11}f(x)\approx(2)\frac{-17+(18)}{2}+(4)\frac{18+(-4)}{2}+(1)\frac{-4+(-19)}{2}\\&\int_{4}^{11}f(x)\approx17.5\end{align*}$

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Example Question #36 : Trapezoidal Sums

$\begin{align*}&\text{The function }f(x)\text{ has the following values on the interval }x=[2,22]:\\&f(2)=7\\&f(5)=-15\\&f(7)=-20\\&f(11)=2\\&\text{Approximate the integral of }f(x)\text{ on this interval using the trapezoidal rule.}\end{align*}$

Possible Answers:

-203.3333

-1129.6296

-56.4815

Correct answer:

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're given a set of values:}\\&f(2)=7\\&f(5)=-15\\&f(7)=-20\\&f(11)=2\\&\text{Notice that the points are not equidistant, with varying interval lengths between}\\&\text{These lengths in sequence are: }3,2,4\\&\text{This means we should calculate each trapezoid separately: }\sum_{i=1}^{n}(x_{i+1}-x_i)\frac{f(x_{i+1})+f(x_i)}{2}\\&\int_{2}^{22}f(x)\approx(3)\frac{7+(-15)}{2}+(2)\frac{-15+(-20)}{2}+(4)\frac{-20+(2)}{2}\\&\int_{2}^{22}f(x)\approx-83\end{align*}$

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Example Question #71 : Numerical Approximations To Definite Integrals

Use the trapezoidal rule to approximate the following integral:

$\int_{0}^{3}(\frac{x^2}{3}+xe^x)dx$

Possible Answers:

${9+9e^3}$

$\frac{3+3e^3}{2}$

$\frac{9+9e^3}{2}$

$\frac{9-9e^3}{2}$

Correct answer:

$\frac{9+9e^3}{2}$

Explanation:

The trapezoidal rule for approximating a definite integral is given by

$\int_{a}^{b}f(x)dx\approx (b-a)(\frac{f(a)+f(b)}{2})$

Using the above formula to approximate our integral, we get

$\int_{0}^{3}(\frac{x^2}{3}+xe^x)dx \approx (3-0)(\frac{3+3e^3}{2})=\frac{9+9e^3}{2}$

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Example Question #31 : Trapezoidal Sums

Solve using the trapezoidal approximation:

$\int_{0}^{2}(x^5+e^x)dx$

Possible Answers:

32+e^2

$\frac{33+e^2}{2}$

16+e^2

33+e^2

Correct answer:

33+e^2

Explanation:

The trapezoidal approximation of a definite integral is given by the following:

$\int_{a}^{b}f(x)dx \approx \frac{f(b)+f(a)}{2}(b-a)$

Using this approximation for our integral, we get

$(2-0)(\frac{32+e^2+1}{2})=33+e^2$

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Example Question #31 : Trapezoidal Sums

Solve the integral using the trapezoidal approximation:

$\int_{0}^{4}xe^{2x}dx$

Possible Answers:

8e^8

Correct answer:

8e^8

Explanation:

To approximate the definite integral using the trapezoidal rule, we use the following approximation:

$\int_{a}^{b}f(x)dx\approx (b-a)(\frac{f(b)+f(a)}{2})$

For our integral, we get

$\int_{0}^{4}xe^{2x}dx \approx (4-0)(\frac{4e^8}{2}+0)=8e^8$

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AP Calculus AB : Numerical approximations to definite integrals

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #30 : Trapezoidal Sums

Example Question #31 : Trapezoidal Sums

Example Question #32 : Trapezoidal Sums

Example Question #31 : Trapezoidal Sums

Example Question #34 : Trapezoidal Sums

Example Question #35 : Trapezoidal Sums

Example Question #36 : Trapezoidal Sums

Example Question #71 : Numerical Approximations To Definite Integrals

Example Question #31 : Trapezoidal Sums

Example Question #31 : Trapezoidal Sums

All AP Calculus AB Resources

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