Difference between revisions of "009B Sample Midterm 1, Problem 5"
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!Foundations: | !Foundations: | ||
|- | |- | ||
| − | |Link to Riemann sums | + | |Link to Riemann sums page |
|} | |} | ||
'''Solution:''' | '''Solution:''' | ||
| + | |||
| + | '''(a)''' | ||
{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
!Step 1: | !Step 1: | ||
| + | |- | ||
| + | |Since our interval is <math>[0,3]</math> and we are using 3 rectangles, each rectangle has width 1. So, the left-hand Riemann sum is | ||
| + | |- | ||
| + | |<math>1(f(0)+f(1)+f(2))</math>. | ||
|- | |- | ||
| | | | ||
| + | |} | ||
| + | |||
| + | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
| + | !Step 2: | ||
| + | |- | ||
| + | |Thus, the left-hand Riemann sum is | ||
| + | |- | ||
| + | |<math>1(f(0)+f(1)+f(2))=1+0+-3=-2</math>. | ||
| + | |} | ||
| + | |||
| + | '''(b)''' | ||
| + | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
| + | !Step 1: | ||
|- | |- | ||
| − | | | + | |Since our interval is <math>[0,3]</math> and we are using 3 rectangles, each rectangle has width 1. So, the right-hand Riemann sum is |
|- | |- | ||
| − | | | + | |<math>1(f(1)+f(2)+f(3))</math>. |
|- | |- | ||
| | | | ||
| Line 27: | Line 46: | ||
{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
!Step 2: | !Step 2: | ||
| + | |- | ||
| + | |Thus, the right-hand Riemann sum is | ||
| + | |- | ||
| + | |<math>1(f(1)+f(2)+f(3))=0+-3+-8=-11</math>. | ||
| + | |} | ||
| + | |||
| + | '''(c)''' | ||
| + | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
| + | !Step 1: | ||
| + | |- | ||
| + | |Let <math>n</math> be the number of rectangles used in the right-hand Riemann sum for <math>f(x)=1-x^2</math>. | ||
| + | |- | ||
| + | |The width of each rectangle is <math>\Delta x=\frac{3-0}{n}=\frac{3}{n}</math>. | ||
|- | |- | ||
| | | | ||
|- | |- | ||
| | | | ||
| + | |} | ||
| + | |||
| + | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
| + | !Step 2: | ||
| + | |- | ||
| + | |So, the right-hand Riemann sum is | ||
| + | |- | ||
| + | |<math>\Delta x \bigg(f\bigg(\frac{3}{n}\bigg)+f\bigg(2\frac{3}{n}\bigg)+f\bigg(3\frac{3}{n}\bigg)+\ldots +f(3)\bigg)</math>. | ||
|- | |- | ||
| − | | | + | |Now, we let <math>n</math> go to infinity to get a limit. |
|- | |- | ||
| − | | | + | |So, the area of <math>S</math> is equal to <math>\lim_{n\to\infty} \frac{3}{n}\sum_{i=1}^{n}f\bigg(i\frac{3}{n}\bigg)</math>. |
|} | |} | ||
| Line 40: | Line 80: | ||
!Final Answer: | !Final Answer: | ||
|- | |- | ||
| − | | | + | |'''(a)''' <math>-2</math> |
| + | |- | ||
| + | |'''(b)''' <math>-11</math> | ||
|- | |- | ||
| − | | | + | |'''(c)''' <math>\lim_{n\to\infty} \frac{3}{n}\sum_{i=1}^{n}f\bigg(i\frac{3}{n}\bigg)</math> |
|} | |} | ||
[[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] | [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] | ||
Revision as of 14:47, 31 January 2016
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: |
|---|
| Link to Riemann sums page |
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 |
| . |
| Now, we let go to infinity to get a limit. |
| So, the area of is equal to . |
| Final Answer: |
|---|
| (a) |
| (b) |
| (c) |