009B Sample Midterm 2, Problem 1
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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.
| Foundations: |
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| Link to Riemann sums page |
Solution:
(a)
| Step 1: |
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| Let |
| Since our interval is and we are using 4 rectangles, each rectangle has width 1. So, the left-endpoint Riemann sum is |
| Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 1(f(1)+f(2)+f(3)+f(4))} . |
| Step 2: |
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| Thus, the left-endpoint Riemann sum is |
| Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 1(f(1)+f(2)+f(3)+f(4))={\bigg (}1+{\frac {1}{4}}+{\frac {1}{9}}+{1}{16}{\bigg )}={\frac {205}{144}}} . |
| The left-endpoint Riemann sum overestimates the area of . |
(b)
| Step 1: |
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| Let be the number of rectangles used in the left-endpoint Riemann sum for . |
| The width of each rectangle is . |
| Step 2: |
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| So, the left-endpoint Riemann sum is |
| . |
| Now, we let go to infinity to get a limit. |
| So, the area of is equal to . |
| Final Answer: |
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| (a) Left-endpoint Riemann sum: , The left-endpoint Riemann sum overestimates the area of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S} . |
| (b) Using left-endpoint Riemann sums: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n\to\infty} \frac{4}{n}\sum_{i=0}^{n-1}f\bigg(1+i\frac{4}{n}\bigg)} |