Difference between revisions of "009B Sample Midterm 3, Problem 1"
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| − | |'''1.''' The height of each rectangle in the right-hand Riemann sum is given by choosing the right endpoint of the interval. | + | | |
| + | ::'''1.''' The height of each rectangle in the right-hand Riemann sum is given by choosing the right endpoint of the interval. | ||
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| − | |'''2.''' See the Riemann sums (insert link) for more information. | + | | |
| + | ::'''2.''' See the Riemann sums (insert link) for more information. | ||
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!Step 1: | !Step 1: | ||
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| − | |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 | + | |Let <math style="vertical-align: -5px">f(x)=\sin(x).</math> Each interval has length <math>\frac{\pi}{4}.</math> |
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| + | |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(\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>\begin{array}{rcl} |
| + | \displaystyle{\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)} & = & \displaystyle{\frac{\pi}{4}\bigg(\frac{\sqrt{2}}{2}+1+\frac{\sqrt{2}}{2}+0\bigg)}\\ | ||
| + | &&\\ | ||
| + | & = & \displaystyle{\frac{\pi}{4}(\sqrt{2}+1).}\\ | ||
| + | \end{array}</math> | ||
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!Final Answer: | !Final Answer: | ||
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| − | |<math>\frac{\pi}{4}(\sqrt{2}+1)</math> | + | | <math>\frac{\pi}{4}(\sqrt{2}+1)</math> |
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[[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']] | [[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']] | ||
Revision as of 17:54, 18 April 2016
Divide the interval into four subintervals of equal length and compute the right-endpoint Riemann sum of
| Foundations: |
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| Recall: |
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Solution:
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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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