Difference between revisions of "009B Sample Midterm 3, Problem 3"
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Kayla Murray (talk | contribs) |
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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: |
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u substitution |
Solution:
(a)
Step 1: |
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We proceed using u substitution. Let . Then, . |
Therefore, we have |
Step 2: |
---|
We integrate to get |
(b)
Step 1: |
---|
Step 2: |
---|
Final Answer: |
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(a) |
(b) |