Trapezoidal sums - AP Calculus AB

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Question

Solve using the trapezoidal approximation:

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

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Answer

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

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Question

Solve the integral using the trapezoidal approximation:

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

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Answer

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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Question

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

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Answer

$\begin{align}&$\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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Question

$\begin{align}&$\text{Using the method of trapezoidal sums approximate the integral:}$\&\int_${-4}^{1.4}$(11sin(17cos(x)))dx\&$\text{Using }$2$\text{ intervals.}$\end{align}$

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Answer

$\begin{align}&$\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}$$\text{ and variable paired heights :}$f(x_$i),f(x_{i+1}$)$\text{, which depend on the function values at }$x_$i,x_{i+1}$\&$\text{We're asked to approximate}$\int_${-4}^{1.4}$(11sin(17cos(x)))dx\&$\text{So the interval is }$[-4,1.4]$\text{ the subintervals have length }$$\frac{1.4-(-4)}{2}$=\frac{27}{10}$\&$\text{and the x-values are:}$\&[-4,-$\frac{13}{10}$,$\frac{7}{5}$]\&\int_${-4}^{1.4}$(11sin(17cos(x)))dx\approx$\frac{27}{20}$[(11sin(17cos((-4))))+2(11sin(17cos((-$\frac{13}{10}$))))+(11sin(17cos(($\frac{7}{5}$))))]\&\int_${-4}^{1.4}$(11sin(17cos(x)))dx\approx-10.84\end{align}$

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Question

$\begin{align}&$\text{Calculate the trapezoidal sums integral approximation of :}$\&\int_${0}^{2}$$(6cos(20e^{(x)}$))dx\&$\text{Using points over }$2$\text{ intervals.}$\end{align}$

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Answer

$\begin{align}&$\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}$$\text{ and variable paired heights :}$f(x_$i),f(x_{i+1}$)$\text{, which depend on the function values at }$x_$i,x_{i+1}$\&$\text{We're asked to approximate}$\int_${0}^{2}$$(6cos(20e^{(x)}$))dx\&$\text{So the interval is }$[0,2]$\text{ the subintervals have length }$$\frac{2-(0)}{2}$=1\&$\text{and the x-values are:}$\&[0,1,2]\&\int_${0}^{2}$$(6cos(20e^{(x)}$$))dx\approx$\frac{1}{2}$[(6cos(20e^{((0))}$$))+2(6cos(20e^{((1))}$$))+(6cos(20e^{((2))}$))]\&\int_${0}^{2}$$(6cos(20e^{(x)}$))dx\approx-5.20\end{align}$

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Question

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

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Answer

$\begin{align}&$\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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Question

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

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Answer

$\begin{align}&$\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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Question

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

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Answer

$\begin{align}&$\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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Question

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

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Answer

$\begin{align}&$\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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Question

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

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Answer

$\begin{align}&$\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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Question

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

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Answer

$\begin{align}&$\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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Question

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

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Answer

$\begin{align}&$\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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Question

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

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Answer

$\begin{align}&$\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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Question

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

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Answer

$\begin{align}&$\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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Question

$\begin{align}&$\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.}$\end{align}$

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Answer

$\begin{align}&$\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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Question

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

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Answer

$\begin{align}&$\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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Question

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

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Answer

$\begin{align}&$\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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Question

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

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$\begin{align}&$\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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$\begin{align}&$\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.}$\end{align}$

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

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$\begin{align}&$\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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