Difference between revisions of "009B Sample Midterm 3, Problem 1"
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
!Foundations: | !Foundations: | ||
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'''1.''' The height of each rectangle in the right-hand Riemann sum | '''1.''' The height of each rectangle in the right-hand Riemann sum | ||
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| − | | is given by choosing the right endpoint of the interval. | + | | is given by choosing the right endpoint of the interval. |
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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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| − | <math>\begin{array}{rcl} | + | <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}\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)}\\ | ||
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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 18:29, 7 February 2017
Divide the interval into four subintervals of equal length and compute the right-endpoint Riemann sum of
| Foundations: |
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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. |
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2. See the Riemann sums (insert link) for more information. |
Solution:
| Step 1: |
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| Let |
| Each interval has length |
| Therefore, 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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