Difference between revisions of "009B Sample Midterm 1, Problem 5"
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|Thus, the left-hand Riemann sum is | |Thus, the left-hand Riemann sum is | ||
|- | |- | ||
− | | <math | + | | |
+ | <math>\begin{array}{rcl} | ||
+ | \displaystyle{1(f(0)+f(1)+f(2))} & = & \displaystyle{1+0+-3}\\ | ||
+ | &&\\ | ||
+ | & = & \displaystyle{-2.} | ||
+ | \end{array}</math> | ||
|} | |} | ||
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|Thus, the right-hand Riemann sum is | |Thus, the right-hand Riemann sum is | ||
|- | |- | ||
− | | <math | + | | |
+ | <math>\begin{array}{rcl} | ||
+ | \displaystyle{1(f(1)+f(2)+f(3))} & = & \displaystyle{0+-3+-8}\\ | ||
+ | &&\\ | ||
+ | & = & \displaystyle{-11.} | ||
+ | \end{array}</math> | ||
|} | |} | ||
Revision as of 17:16, 7 February 2017
Let .
- a) Compute the left-hand Riemann sum approximation of with boxes.
- b) Compute the right-hand Riemann sum approximation of with boxes.
- c) Express as a limit of right-hand Riemann sums (as in the definition of the definite integral). Do not evaluate the limit.
Foundations: |
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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 Riemann sums (insert link) for more information. |
Solution:
(a)
Step 1: |
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Since our interval is and we are using 3 rectangles, each rectangle has width 1. |
So, the left-hand Riemann sum is |
Step 2: |
---|
Thus, the left-hand Riemann sum is |
|
(b)
Step 1: |
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Since our interval is and we are using 3 rectangles, each rectangle has width 1. |
So, the right-hand Riemann sum is |
Step 2: |
---|
Thus, the right-hand Riemann sum is |
|
(c)
Step 1: |
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Let be the number of rectangles used in the right-hand Riemann sum for |
The width of each rectangle is |
Step 2: |
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So, the right-hand Riemann sum is |
Finally, we let go to infinity to get a limit. |
Thus, is equal to |
Final Answer: |
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(a) |
(b) |
(c) |