Difference between revisions of "009B Sample Midterm 1, Problem 1"
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!Foundations: | !Foundations: | ||
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'''Solution:''' | '''Solution:''' | ||
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+ | '''(a)''' | ||
+ | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
+ | !Step 1: | ||
+ | |- | ||
+ | |We need to use substitution. Let <math>u=1+x^3</math>. Then, <math>du=3x^2dx</math> and <math>\frac{du}{3}=x^2dx</math>. | ||
+ | |- | ||
+ | |Therefore, the integral becomes <math>\frac{1}{3}\int \sqrt{u}du</math>. | ||
+ | |} | ||
+ | |||
+ | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
+ | !Step 2: | ||
+ | |- | ||
+ | |We now have: | ||
+ | |- | ||
+ | |<math>\int x^2\sqrt{1+x^3}dx=\frac{1}{3}\int \sqrt{u}du=\frac{2}{9}u^{\frac{3}{2}}+C=\frac{2}{9}(1+x^3)^{\frac{3}{2}}+C</math>. | ||
+ | |} | ||
+ | |||
+ | '''(b)''' | ||
{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
!Step 1: | !Step 1: | ||
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!Final Answer: | !Final Answer: | ||
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− | | | + | |(a) <math>\frac{2}{9}(1+x^3)^{\frac{3}{2}}+C</math> |
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− | | | + | |(b) <math></math> |
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[[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] | [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] |
Revision as of 14:56, 26 January 2016
Evaluate the indefinite and definite integrals.
- a)
- b)
Foundations: |
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Solution:
(a)
Step 1: |
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We need to use substitution. Let . Then, and . |
Therefore, the integral becomes . |
Step 2: |
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We now have: |
. |
(b)
Step 1: |
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Step 2: |
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Final Answer: |
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(a) |
(b) |