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AP Calculus AB : Integrals

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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 #301 : Integrals

Use the trapezoidal rule to approximate the following integral:

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

Possible Answers:

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

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

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

${9+9e^3}$

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 #302 : Integrals

Solve using the trapezoidal approximation:

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

Possible Answers:

16+e^2

32+e^2

33+e^2

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

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 #303 : Integrals

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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Example Question #304 : Integrals

Approximate the integral using the trapezoidal rule:

$\int_{0}^{6}x^2\sin(\frac{\pi x}{12})dx$

Possible Answers:

216

108

Correct answer:

108

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

$\int_{0}^{6}x^2\sin(\frac{\pi x}{12})dx \approx \frac{(6)(36+0)}{2}=108$

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Example Question #41 : Trapezoidal Sums

Using the Trapezoidal Rule with four subintervals, approximate to four decimal places:

$\int_{2}^{3} \log_{3}( x^{2} ) \mathrm{d} x$

Possible Answers:

2.0620

1.5620

1.7465

1.6558

1.6542

Correct answer:

1.6542

Explanation:

The Trapezoidal Rule states that a definite integral

$\int_{a}^{b} f(x) \textup{d}x$

can be approximated by computing the expression

$T_{N}=\frac{b-a}{2N} \left [f (x_{0} ) +2 f (x_{1}) +2 f ( x_{2} ) + ... 2f ( x_{N-1} )+ f (x_{N} ) \right ]$

where $x_{k} = a + \frac{b-a}{N} \cdot k$ for k = 0, 1, 2, 3, ...,N

Since we are dividing $\left [ 2,3 \right ]$ into four subintervals, set

a = 2, b =3, N = 4 ; the expression becomes

$T_{4}=\frac{1}{8} \left [ f(2 ) +2 f \left (2.25 \right ) +2 f \left (2.5 \right ) + 2f \left ( 2.75 \right ) + f (3 ) \right ]$

where

$f(x)= \log_{3}( x^{2} )$

$f(2)= \log_{3}( 2^{2} ) = \log_{3}4= \frac{\ln 4}{\ln 3} \approx 1.2619$

$f(2.25)= \log_{3}( 2.25^{2} ) = \log_{3}5.0625= \frac{\ln 5.0625}{\ln 3} \approx 1.4763$

$f(2.5)= \log_{3}( 2.5^{2} ) = \log_{3}6.25= \frac{\ln 6.25}{\ln 3} \approx 1.6681$

$f(2.75)= \log_{3}( 2.75^{2} ) = \log_{3}7.5625= \frac{\ln 7.5625}{\ln 3} \approx 1.8416$

$f(3)= \log_{3}( 3^{2} ) = \log_{3}9= 2$

$T_{4}=\frac{1}{8} \left [ 1.2619 +2 (1.4763 )+2 ( 1.6681)+ 2 ( 1.8416) +2 \right]$

$\approx 1.6542$

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Example Question #1111 : Ap Calculus Ab

Solve in the integral by method of trapezoidal sums

$\int_{3}^{5}\cos(e^{3x})dx$

Possible Answers:

$\cos(e^{15})+\cos(e^{9})$

$\cos(e^{5})-\cos(e^{9})$

$\cos(e^{15})-\cos(e^{9})$

$\cos(e^{15})-\sin(e^{9})$

Correct answer:

$\cos(e^{15})-\cos(e^{9})$

Explanation:

To use the trapezoidal rule, we apply the following formula:

$\int_{a}^{b}f(x)dx\approx(b-a)\left [ \frac{f(a)-f(b)}{2}\right ]$

Using the integral from our problem statement, we get

$(5-3)\frac{(\cos(e^{15})-\cos(e^{9}))}{2}=\cos(e^{15})-\cos(e^{9})$

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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 #31 : Trapezoidal Sums

Example Question #34 : Trapezoidal Sums

Example Question #35 : Trapezoidal Sums

Example Question #36 : Trapezoidal Sums

Example Question #301 : Integrals

Example Question #302 : Integrals

Example Question #303 : Integrals

Example Question #304 : Integrals

Example Question #41 : Trapezoidal Sums

Example Question #1111 : Ap Calculus Ab

All AP Calculus AB Resources

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