Difference between revisions of "009B Sample Midterm 3, Problem 1"

Revision as of 14:08, 31 January 2016

Divide the interval $[0,\pi ]$ into four subintervals of equal length ${\frac {\pi }{4}}$ and compute the right-endpoint Riemann sum of $y=\sin(x)$

Foundations:
Link to Riemann sums page

Solution:

Step 1:
Let $f(x)=\sin(x)$ . Each interval has length ${\frac {\pi }{4}}$ . So, the right-endpoint Riemann sum of $f(x)$ on the interval $[0,\pi ]$ is
${\frac {\pi }{4}}{\bigg (}f{\bigg (}{\frac {\pi }{4}}{\bigg )}+f{\bigg (}{\frac {\pi }{2}}{\bigg )}+f{\bigg (}{\frac {3\pi }{4}}{\bigg )}+f(\pi ){\bigg )}$ .

Step 2:
Thus, the right-endpoint Riemann sum is
${\frac {\pi }{4}}{\bigg (}\sin {\bigg (}{\frac {\pi }{4}}{\bigg )}+\sin {\bigg (}{\frac {\pi }{2}}{\bigg )}+\sin {\bigg (}{\frac {3\pi }{4}}{\bigg )}+\sin(\pi ){\bigg )}={\frac {\pi }{4}}{\bigg (}{\frac {\sqrt {2}}{2}}+1+{\frac {\sqrt {2}}{2}}+0{\bigg )}={\frac {\pi }{4}}({\sqrt {2}}+1)$

Final Answer:
${\frac {\pi }{4}}({\sqrt {2}}+1)$

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 !Step 1: &nbsp;
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+|Let <math>f(x)=\sin(x)</math>. Each interval has length <math>\frac{\pi}{4}</math>. So, the right-endpoint Riemann sum of <math>f(x)</math> on the interval <math>[0,\pi]</math> is
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+|<math>\frac{\pi}{4}\bigg(f\bigg(\frac{\pi}{4}\bigg)+f\bigg(\frac{\pi}{2}\bigg)+f\bigg(\frac{3\pi}{4}\bigg)+f(\pi)\bigg)</math>.
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 !Step 2: &nbsp;
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+|Thus, the right-endpoint Riemann sum is
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+|<math>\frac{\pi}{4}\bigg(\sin\bigg(\frac{\pi}{4}\bigg)+\sin\bigg(\frac{\pi}{2}\bigg)+\sin\bigg(\frac{3\pi}{4}\bigg)+\sin(\pi)\bigg)=\frac{\pi}{4}\bigg(\frac{\sqrt{2}}{2}+1+\frac{\sqrt{2}}{2}+0\bigg)=\frac{\pi}{4}(\sqrt{2}+1)</math>
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 !Final Answer: &nbsp;
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+|<math>\frac{\pi}{4}(\sqrt{2}+1)</math>
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 [[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]