Difference between revisions of "009B Sample Midterm 3, Problem 3"
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| Line 8: | Line 8: | ||
!Foundations: | !Foundations: | ||
|- | |- | ||
| − | | | + | | u substitution |
|} | |} | ||
| Line 17: | Line 17: | ||
!Step 1: | !Step 1: | ||
|- | |- | ||
| − | | | + | |We proceed using u substitution. Let <math>u=x^3</math>. Then, <math>du=3x^2dx</math>. |
| + | |- | ||
| + | |Therefore, we have | ||
|- | |- | ||
| − | | | + | |<math>\int x^2\sin (x^3) dx=\int \frac{\sin(u)}{3}du</math> |
|} | |} | ||
| Line 25: | Line 27: | ||
!Step 2: | !Step 2: | ||
|- | |- | ||
| − | | | + | |We integrate to get |
|- | |- | ||
| − | | | + | |<math>\int x^2\sin (x^3) dx=\frac{-1}{3}\cos(u)+C=\frac{-1}{3}\cos(x^3)+C</math> |
|} | |} | ||
| Line 58: | Line 60: | ||
!Final Answer: | !Final Answer: | ||
|- | |- | ||
| − | |'''(a)''' | + | |'''(a)''' <math>\frac{-1}{3}\cos(x^3)+C</math> |
|- | |- | ||
|'''(b)''' | |'''(b)''' | ||
|} | |} | ||
[[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']] | [[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']] | ||
Revision as of 13:38, 31 January 2016
Compute the following integrals:
- a)
- b)
| Foundations: |
|---|
| u substitution |
Solution:
(a)
| Step 1: |
|---|
| We proceed using u substitution. Let . Then, . |
| Therefore, we have |
| Step 2: |
|---|
| We integrate to get |
(b)
| Step 1: |
|---|
| Step 2: |
|---|
| Final Answer: |
|---|
| (a) |
| (b) |