Difference between revisions of "009B Sample Midterm 3, Problem 1"
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| − | <span class="exam">Divide the interval <math>[0,\pi]</math> into four subintervals of equal length <math>\frac{\pi}{4}</math> and compute the right-endpoint Riemann sum of <math>y=\sin (x)</math> | + | <span class="exam">Divide the interval <math style="vertical-align: -5px">[0,\pi]</math> into four subintervals of equal length <math>\frac{\pi}{4}</math> and compute the right-endpoint Riemann sum of <math style="vertical-align: -5px">y=\sin (x).</math> |
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| − | |Let <math>f(x)=\sin(x)</math> | + | |Let <math style="vertical-align: -5px">f(x)=\sin(x).</math> Each interval has length <math>\frac{\pi}{4}.</math> So, the right-endpoint Riemann sum of <math style="vertical-align: -5px">f(x)</math> on the interval <math style="vertical-align: -5px">[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> | + | | |
| + | ::<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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|Thus, the right-endpoint Riemann sum is | |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> | + | | |
| + | ::<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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Revision as of 16:31, 29 March 2016
Divide the interval into four subintervals of equal length and compute the right-endpoint Riemann sum of
![{\displaystyle [0,\pi ]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3e2a912eda6ef1afe46a81b518fe9da64a832751)
| Foundations: |
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| Recall: |
| 1. The height of each rectangle in the right-hand Riemann sum is given by choosing the right endpoint of the interval. |
| 2. See the Riemann sums (insert link) for more information. |
Solution:
| Step 1: |
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| Let Each interval has length So, the right-endpoint Riemann sum of on the interval is |
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| Step 2: |
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| Thus, the right-endpoint Riemann sum is |
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| Final Answer: |
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