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− | <span class="exam"> Consider the region <math style="vertical-align: 0px">S</math> bounded by <math style="vertical-align: -13px">x=1,x=5,y=\frac{1}{x^2}</math> and the <math>x</math>-axis. | + | <span class="exam"> Consider the region <math style="vertical-align: 0px">S</math> bounded by <math style="vertical-align: -13px">x=1,x=5,y=\frac{1}{x^2}</math>  and the <math>x</math>-axis. |
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| ::<span class="exam">a) Use four rectangles and a Riemann sum to approximate the area of the region <math style="vertical-align: 0px">S</math>. Sketch the region <math style="vertical-align: 0px">S</math> and the rectangles and indicate whether your rectangles overestimate or underestimate the area of <math style="vertical-align: 0px">S</math>. | | ::<span class="exam">a) Use four rectangles and a Riemann sum to approximate the area of the region <math style="vertical-align: 0px">S</math>. Sketch the region <math style="vertical-align: 0px">S</math> and the rectangles and indicate whether your rectangles overestimate or underestimate the area of <math style="vertical-align: 0px">S</math>. |
Revision as of 08:16, 3 February 2016
Consider the region bounded by and the -axis.
- a) Use four rectangles and a Riemann sum to approximate the area of the region . Sketch the region and the rectangles and indicate whether your rectangles overestimate or underestimate the area of .
- b) Find an expression for the area of the region as a limit. Do not evaluate the limit.
Approximation of integral with left endpoints is an overestimate.
Solution:
(a)
Step 1:
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Let . Since our interval is and we are using 4 rectangles, each rectangle has width 1. Since the problem doesn't specify, we can choose either right- or left-endpoints. Choosing left-endpoints, the Riemann sum is
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Step 2:
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Thus, the left-endpoint Riemann sum is
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The left-endpoint Riemann sum overestimates the area of .
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(b)
Step 1:
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Let be the number of rectangles used in the left-endpoint Riemann sum for .
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The width of each rectangle is .
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Step 2:
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So, the left-endpoint Riemann sum is
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Now, we let go to infinity to get a limit.
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So, the area of is equal to .
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Final Answer:
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(a) The left-endpoint Riemann sum is , which overestimates the area of .
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(b) Using left-endpoint Riemann sums:
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