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Questions: Algebra BusinessCalculus
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Use Simpson's rule to find the arclength of the curve \(\displaystyle f(x)=\sin{\left(x \right)}\) on \(\displaystyle (4,9)\) with \(\displaystyle n=40\).
\(\displaystyle \Delta x = \frac{ 9 - 4 }{ 40 }\). \(\displaystyle x_i = a +i\Delta x = 4 + i \frac{1}{8}\)Using the 1,4,2,...,2,4,1 pattern the sum can be written as \(\displaystyle x_i\) can be written split into the even and odd terms. \(\displaystyle x_k = 4 + (2k-1)\cdot \frac{1}{8}\) for \(\displaystyle k=1\) to \(\displaystyle k =20\) and \(\displaystyle x_j = 4 + (2j)\cdot \frac{1}{8}\) for \(\displaystyle j=1\) to \(\displaystyle j =19\). \(\displaystyle f(4) +f(9)+4\sum_{k=1}^{20}f\left(\frac{k}{4} + \frac{31}{8}\right) + 2\sum_{j=1}^{19}f\left(\frac{j}{4} + 4\right)\). The value is \(\displaystyle 5.9021\)
\begin{question}Use Simpson's rule to find the arclength of the curve $f(x)=\sin{\left(x \right)}$ on $(4,9)$ with $n=40$.
\soln{9cm}{$\Delta x = \frac{ 9 - 4 }{ 40 }$. $x_i = a +i\Delta x = 4 + i \frac{1}{8}$Using the 1,4,2,...,2,4,1 pattern the sum can be written as $x_i$ can be written split into the even and odd terms. $x_k = 4 + (2k-1)\cdot \frac{1}{8}$ for $k=1$ to $k =20$ and $x_j = 4 + (2j)\cdot \frac{1}{8}$ for $j=1$ to $j =19$. $f(4) +f(9)+4\sum_{k=1}^{20}f\left(\frac{k}{4} + \frac{31}{8}\right) + 2\sum_{j=1}^{19}f\left(\frac{j}{4} + 4\right)$. The value is $5.9021$}
\end{question}
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\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>Use Simpson's rule to find the arclength of the curve <img class="equation_image" title=" \displaystyle f(x)=\sin{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%3D%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle f(x)=\sin{\left(x \right)} " data-equation-content=" \displaystyle f(x)=\sin{\left(x \right)} " /> on <img class="equation_image" title=" \displaystyle (4,9) " src="/equation_images/%20%5Cdisplaystyle%20%284%2C9%29%20" alt="LaTeX: \displaystyle (4,9) " data-equation-content=" \displaystyle (4,9) " /> with <img class="equation_image" title=" \displaystyle n=40 " src="/equation_images/%20%5Cdisplaystyle%20n%3D40%20" alt="LaTeX: \displaystyle n=40 " data-equation-content=" \displaystyle n=40 " /> . </p> </p><p> <p> <img class="equation_image" title=" \displaystyle \Delta x = \frac{ 9 - 4 }{ 40 } " src="/equation_images/%20%5Cdisplaystyle%20%5CDelta%20x%20%3D%20%5Cfrac%7B%209%20-%204%20%7D%7B%2040%20%7D%20" alt="LaTeX: \displaystyle \Delta x = \frac{ 9 - 4 }{ 40 } " data-equation-content=" \displaystyle \Delta x = \frac{ 9 - 4 }{ 40 } " /> . <img class="equation_image" title=" \displaystyle x_i = a +i\Delta x = 4 + i \frac{1}{8} " src="/equation_images/%20%5Cdisplaystyle%20x_i%20%3D%20a%20%2Bi%5CDelta%20x%20%3D%204%20%2B%20i%20%5Cfrac%7B1%7D%7B8%7D%20" alt="LaTeX: \displaystyle x_i = a +i\Delta x = 4 + i \frac{1}{8} " data-equation-content=" \displaystyle x_i = a +i\Delta x = 4 + i \frac{1}{8} " /> Using the 1,4,2,...,2,4,1 pattern the sum can be written as <img class="equation_image" title=" \displaystyle x_i " src="/equation_images/%20%5Cdisplaystyle%20x_i%20" alt="LaTeX: \displaystyle x_i " data-equation-content=" \displaystyle x_i " /> can be written split into the even and odd terms. <img class="equation_image" title=" \displaystyle x_k = 4 + (2k-1)\cdot \frac{1}{8} " src="/equation_images/%20%5Cdisplaystyle%20x_k%20%3D%204%20%2B%20%282k-1%29%5Ccdot%20%5Cfrac%7B1%7D%7B8%7D%20" alt="LaTeX: \displaystyle x_k = 4 + (2k-1)\cdot \frac{1}{8} " data-equation-content=" \displaystyle x_k = 4 + (2k-1)\cdot \frac{1}{8} " /> for <img class="equation_image" title=" \displaystyle k=1 " src="/equation_images/%20%5Cdisplaystyle%20k%3D1%20" alt="LaTeX: \displaystyle k=1 " data-equation-content=" \displaystyle k=1 " /> to <img class="equation_image" title=" \displaystyle k =20 " src="/equation_images/%20%5Cdisplaystyle%20k%20%3D20%20" alt="LaTeX: \displaystyle k =20 " data-equation-content=" \displaystyle k =20 " /> and <img class="equation_image" title=" \displaystyle x_j = 4 + (2j)\cdot \frac{1}{8} " src="/equation_images/%20%5Cdisplaystyle%20x_j%20%3D%204%20%2B%20%282j%29%5Ccdot%20%5Cfrac%7B1%7D%7B8%7D%20" alt="LaTeX: \displaystyle x_j = 4 + (2j)\cdot \frac{1}{8} " data-equation-content=" \displaystyle x_j = 4 + (2j)\cdot \frac{1}{8} " /> for <img class="equation_image" title=" \displaystyle j=1 " src="/equation_images/%20%5Cdisplaystyle%20j%3D1%20" alt="LaTeX: \displaystyle j=1 " data-equation-content=" \displaystyle j=1 " /> to <img class="equation_image" title=" \displaystyle j =19 " src="/equation_images/%20%5Cdisplaystyle%20j%20%3D19%20" alt="LaTeX: \displaystyle j =19 " data-equation-content=" \displaystyle j =19 " /> . <img class="equation_image" title=" \displaystyle f(4) +f(9)+4\sum_{k=1}^{20}f\left(\frac{k}{4} + \frac{31}{8}\right) + 2\sum_{j=1}^{19}f\left(\frac{j}{4} + 4\right) " src="/equation_images/%20%5Cdisplaystyle%20f%284%29%20%2Bf%289%29%2B4%5Csum_%7Bk%3D1%7D%5E%7B20%7Df%5Cleft%28%5Cfrac%7Bk%7D%7B4%7D%20%2B%20%5Cfrac%7B31%7D%7B8%7D%5Cright%29%20%2B%202%5Csum_%7Bj%3D1%7D%5E%7B19%7Df%5Cleft%28%5Cfrac%7Bj%7D%7B4%7D%20%2B%204%5Cright%29%20" alt="LaTeX: \displaystyle f(4) +f(9)+4\sum_{k=1}^{20}f\left(\frac{k}{4} + \frac{31}{8}\right) + 2\sum_{j=1}^{19}f\left(\frac{j}{4} + 4\right) " data-equation-content=" \displaystyle f(4) +f(9)+4\sum_{k=1}^{20}f\left(\frac{k}{4} + \frac{31}{8}\right) + 2\sum_{j=1}^{19}f\left(\frac{j}{4} + 4\right) " /> . The value is <img class="equation_image" title=" \displaystyle 5.9021 " src="/equation_images/%20%5Cdisplaystyle%205.9021%20" alt="LaTeX: \displaystyle 5.9021 " data-equation-content=" \displaystyle 5.9021 " /> </p> </p>