Difference between revisions of "009B Sample Midterm 1, Problem 5"
Jump to navigation
Jump to search
Kayla Murray (talk | contribs) |
Kayla Murray (talk | contribs) |
||
| Line 1: | Line 1: | ||
<span class="exam">Let <math>f(x)=1-x^2</math>. | <span class="exam">Let <math>f(x)=1-x^2</math>. | ||
| − | + | <span class="exam">(a) Compute the left-hand Riemann sum approximation of <math style="vertical-align: -14px">\int_0^3 f(x)~dx</math> with <math style="vertical-align: 0px">n=3</math> boxes. | |
| − | + | ||
| − | + | <span class="exam">(b) Compute the right-hand Riemann sum approximation of <math style="vertical-align: -14px">\int_0^3 f(x)~dx</math> with <math style="vertical-align: 0px">n=3</math> boxes. | |
| + | |||
| + | <span class="exam">(c) Express <math style="vertical-align: -14px">\int_0^3 f(x)~dx</math> as a limit of right-hand Riemann sums (as in the definition of the definite integral). Do not evaluate the limit. | ||
Revision as of 17:10, 18 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: |
|---|
| 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: |
|---|
| 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: |
|---|
| 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: |
|---|
| Let be the number of rectangles used in the right-hand Riemann sum for |
| The width of each rectangle is |
| Step 2: |
|---|
| So, the right-hand Riemann sum is |
| Finally, we let go to infinity to get a limit. |
| Thus, is equal to |
| Final Answer: |
|---|
| (a) |
| (b) |
| (c) |