Difference between revisions of "009B Sample Midterm 2, Problem 1"
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!Foundations: | !Foundations: | ||
|- | |- | ||
− | |See the page on [[Riemann_Sums|'''Riemann Sums''']]. | + | |Recall: |
+ | |- | ||
+ | |'''1.''' The height of each rectangle in the left-hand Riemann sum is given by choosing the left endpoint of the interval. | ||
+ | |- | ||
+ | |'''2.''' The height of each rectangle in the right-hand Riemann sum is given by choosing the right endpoint of the interval. | ||
+ | |- | ||
+ | |'''3.''' See the page on [[Riemann_Sums|'''Riemann Sums''']] for more information. | ||
|} | |} | ||
Revision as of 14:11, 8 April 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: |
---|
Recall: |
1. The height of each rectangle in the left-hand Riemann sum is given by choosing the left endpoint of the interval. |
2. The height of each rectangle in the right-hand Riemann sum is given by choosing the right endpoint of the interval. |
3. See the page on Riemann Sums for more information. |
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 |
. |
Step 2: |
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Thus, the left-endpoint Riemann sum is |
. |
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) The left-endpoint Riemann sum is , which overestimates the area of . |
(b) Using left-endpoint Riemann sums: |