Difference between revisions of "009B Sample Midterm 2, Problem 1"
Jump to navigation
Jump to search
Line 58: | Line 58: | ||
|Now, we let <math style="vertical-align: 0px">n</math> go to infinity to get a limit. | |Now, we let <math style="vertical-align: 0px">n</math> go to infinity to get a limit. | ||
|- | |- | ||
− | |So, the area of <math style="vertical-align: 0px">S</math> is equal to <math style="vertical-align: -20px">\lim_{n\to\infty} \frac{4}{n}\sum_{i=0}^{n-1}f\bigg(1+i\frac{4}{n}\bigg)</math>. | + | |So, the area of <math style="vertical-align: 0px">S</math> is equal to <math style="vertical-align: -20px">\lim_{n\to\infty} \frac{4}{n}\sum_{i=0}^{n-1}f\bigg(1+i\frac{4}{n}\bigg)\,=\,\lim_{n\to\infty} \frac{4}{n}\sum_{i=0}^{n-1}\bigg(1+i\frac{4}{n}\bigg)^{-2}</math>. |
|} | |} | ||
Line 66: | Line 66: | ||
|'''(a)''' The left-endpoint Riemann sum is <math style="vertical-align: -20px">\frac{205}{144}</math>, which overestimates the area of <math style="vertical-align: 0px">S</math>. | |'''(a)''' The left-endpoint Riemann sum is <math style="vertical-align: -20px">\frac{205}{144}</math>, which overestimates the area of <math style="vertical-align: 0px">S</math>. | ||
|- | |- | ||
− | |'''(b)''' Using left-endpoint Riemann sums: <math style="vertical-align: -20px">\lim_{n\to\infty} \frac{4}{n}\sum_{i=0}^{n-1} | + | |'''(b)''' Using left-endpoint Riemann sums: |
+ | |- | ||
+ | | <math style="vertical-align: -20px">\lim_{n\to\infty} \frac{4}{n}\sum_{i=0}^{n-1}\bigg(1+i\frac{4}{n}\bigg)^{-2}</math> | ||
|} | |} | ||
[[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] | [[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] |
Revision as of 10:49, 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.
Foundations: |
---|
See the page on Riemann Sums. |
Solution:
(a)
Step 1: |
---|
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 |
. |
Step 2: |
---|
Thus, the left-endpoint Riemann sum is |
. |
The left-endpoint Riemann sum overestimates the area of . |
(b)
Step 1: |
---|
Let be the number of rectangles used in the left-endpoint Riemann sum for . |
The width of each rectangle is . |
Step 2: |
---|
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: |
---|
(a) The left-endpoint Riemann sum is , which overestimates the area of . |
(b) Using left-endpoint Riemann sums: |