Trapezoidal sums - AP Calculus AB

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Question

Solve using the trapezoidal approximation:

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

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

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

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Question

Solve the integral using the trapezoidal approximation:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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