Difference between revisions of "009B Sample Final 3, Problem 1"
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!Step 1: | !Step 1: | ||
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− | | | + | |Since our interval is <math style="vertical-align: -5px">[-1,1]</math> and we are using <math style="vertical-align: -1px">4</math> rectangles, each rectangle has width <math style="vertical-align: -13px">\frac{1}{2}.</math> |
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− | | | + | |Let <math style="vertical-align: -6px">f(x)=1-x^2.</math> |
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− | | | + | |So, the left-endpoint Riemann sum is |
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− | | | + | | <math style="vertical-align: 0px">S=\frac{1}{2}\bigg(f(-1)+f\bigg(-\frac{1}{2}\bigg)+f(0)+f\bigg(\frac{1}{2}\bigg)\bigg).</math> |
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!Step 2: | !Step 2: | ||
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− | | | + | |Thus, the left-endpoint Riemann sum is |
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− | + | <math>\begin{array}{rcl} | |
− | + | \displaystyle{S} & = & \displaystyle{\frac{1}{2}\bigg(0+\frac{3}{4}+1+\frac{3}{4}\bigg)}\\ | |
− | + | &&\\ | |
− | + | & = & \displaystyle{\frac{5}{4}.} | |
+ | \end{array}</math> | ||
|} | |} | ||
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!Final Answer: | !Final Answer: | ||
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− | | | + | | <math>\frac{5}{4}</math> |
|} | |} | ||
[[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] | [[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] |
Revision as of 17:29, 28 February 2017
Divide the interval into four subintervals of equal length and compute the left-endpoint Riemann sum of
Foundations: |
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The height of each rectangle in the left-endpoint Riemann sum is given by choosing the left endpoint of the interval. |
Solution:
Step 1: |
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Since our interval is and we are using rectangles, each rectangle has width |
Let |
So, the left-endpoint Riemann sum is |
Step 2: |
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Thus, the left-endpoint Riemann sum is |
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Final Answer: |
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