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Questions: Algebra BusinessCalculus
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Use Simpson's rule to find the arclength of the curve \(\displaystyle f(x)=x^{2}\) on \(\displaystyle (3,8)\) with \(\displaystyle n=36\).
\(\displaystyle \Delta x = \frac{ 8 - 3 }{ 36 }\). \(\displaystyle x_i = a +i\Delta x = 3 + i \frac{5}{36}\)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. \(\displaystyle x_k = 3 + (2k-1)\cdot \frac{5}{36}\) for \(\displaystyle k=1\) to \(\displaystyle k =18\) and \(\displaystyle x_j = 3 + (2j)\cdot \frac{5}{36}\) for \(\displaystyle j=1\) to \(\displaystyle j =17\). \(\displaystyle f(3) +f(8)+4\sum_{k=1}^{18}f\left(\frac{5 k}{18} + \frac{103}{36}\right) + 2\sum_{j=1}^{17}f\left(\frac{5 j}{18} + 3\right)\). The value is \(\displaystyle 55.244\)
\begin{question}Use Simpson's rule to find the arclength of the curve $f(x)=x^{2}$ on $(3,8)$ with $n=36$.
\soln{9cm}{$\Delta x = \frac{ 8 - 3 }{ 36 }$. $x_i = a +i\Delta x = 3 + i \frac{5}{36}$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. $x_k = 3 + (2k-1)\cdot \frac{5}{36}$ for $k=1$ to $k =18$ and $x_j = 3 + (2j)\cdot \frac{5}{36}$ for $j=1$ to $j =17$. $f(3) +f(8)+4\sum_{k=1}^{18}f\left(\frac{5 k}{18} + \frac{103}{36}\right) + 2\sum_{j=1}^{17}f\left(\frac{5 j}{18} + 3\right)$. The value is $55.244$}
\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)=x^{2} " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%3Dx%5E%7B2%7D%20" alt="LaTeX: \displaystyle f(x)=x^{2} " data-equation-content=" \displaystyle f(x)=x^{2} " /> on <img class="equation_image" title=" \displaystyle (3,8) " src="/equation_images/%20%5Cdisplaystyle%20%283%2C8%29%20" alt="LaTeX: \displaystyle (3,8) " data-equation-content=" \displaystyle (3,8) " /> with <img class="equation_image" title=" \displaystyle n=36 " src="/equation_images/%20%5Cdisplaystyle%20n%3D36%20" alt="LaTeX: \displaystyle n=36 " data-equation-content=" \displaystyle n=36 " /> . </p> </p><p> <p> <img class="equation_image" title=" \displaystyle \Delta x = \frac{ 8 - 3 }{ 36 } " src="/equation_images/%20%5Cdisplaystyle%20%5CDelta%20x%20%3D%20%5Cfrac%7B%208%20-%203%20%7D%7B%2036%20%7D%20" alt="LaTeX: \displaystyle \Delta x = \frac{ 8 - 3 }{ 36 } " data-equation-content=" \displaystyle \Delta x = \frac{ 8 - 3 }{ 36 } " /> . <img class="equation_image" title=" \displaystyle x_i = a +i\Delta x = 3 + i \frac{5}{36} " src="/equation_images/%20%5Cdisplaystyle%20x_i%20%3D%20a%20%2Bi%5CDelta%20x%20%3D%203%20%2B%20i%20%5Cfrac%7B5%7D%7B36%7D%20" alt="LaTeX: \displaystyle x_i = a +i\Delta x = 3 + i \frac{5}{36} " data-equation-content=" \displaystyle x_i = a +i\Delta x = 3 + i \frac{5}{36} " /> 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 = 3 + (2k-1)\cdot \frac{5}{36} " src="/equation_images/%20%5Cdisplaystyle%20x_k%20%3D%203%20%2B%20%282k-1%29%5Ccdot%20%5Cfrac%7B5%7D%7B36%7D%20" alt="LaTeX: \displaystyle x_k = 3 + (2k-1)\cdot \frac{5}{36} " data-equation-content=" \displaystyle x_k = 3 + (2k-1)\cdot \frac{5}{36} " /> 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 =18 " src="/equation_images/%20%5Cdisplaystyle%20k%20%3D18%20" alt="LaTeX: \displaystyle k =18 " data-equation-content=" \displaystyle k =18 " /> and <img class="equation_image" title=" \displaystyle x_j = 3 + (2j)\cdot \frac{5}{36} " src="/equation_images/%20%5Cdisplaystyle%20x_j%20%3D%203%20%2B%20%282j%29%5Ccdot%20%5Cfrac%7B5%7D%7B36%7D%20" alt="LaTeX: \displaystyle x_j = 3 + (2j)\cdot \frac{5}{36} " data-equation-content=" \displaystyle x_j = 3 + (2j)\cdot \frac{5}{36} " /> 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 =17 " src="/equation_images/%20%5Cdisplaystyle%20j%20%3D17%20" alt="LaTeX: \displaystyle j =17 " data-equation-content=" \displaystyle j =17 " /> . <img class="equation_image" title=" \displaystyle f(3) +f(8)+4\sum_{k=1}^{18}f\left(\frac{5 k}{18} + \frac{103}{36}\right) + 2\sum_{j=1}^{17}f\left(\frac{5 j}{18} + 3\right) " src="/equation_images/%20%5Cdisplaystyle%20f%283%29%20%2Bf%288%29%2B4%5Csum_%7Bk%3D1%7D%5E%7B18%7Df%5Cleft%28%5Cfrac%7B5%20k%7D%7B18%7D%20%2B%20%5Cfrac%7B103%7D%7B36%7D%5Cright%29%20%2B%202%5Csum_%7Bj%3D1%7D%5E%7B17%7Df%5Cleft%28%5Cfrac%7B5%20j%7D%7B18%7D%20%2B%203%5Cright%29%20" alt="LaTeX: \displaystyle f(3) +f(8)+4\sum_{k=1}^{18}f\left(\frac{5 k}{18} + \frac{103}{36}\right) + 2\sum_{j=1}^{17}f\left(\frac{5 j}{18} + 3\right) " data-equation-content=" \displaystyle f(3) +f(8)+4\sum_{k=1}^{18}f\left(\frac{5 k}{18} + \frac{103}{36}\right) + 2\sum_{j=1}^{17}f\left(\frac{5 j}{18} + 3\right) " /> . The value is <img class="equation_image" title=" \displaystyle 55.244 " src="/equation_images/%20%5Cdisplaystyle%2055.244%20" alt="LaTeX: \displaystyle 55.244 " data-equation-content=" \displaystyle 55.244 " /> </p> </p>