Difference between revisions of "009B Sample Midterm 1, Problem 3"
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− | ::Let <math style="vertical-align: -1px">u=\ln x</math> and <math style="vertical-align: 0px">dv=x~dx.</math> Then, <math style="vertical-align: -13px">du=\frac{1}{x}dx</math> and <math style="vertical-align: -12px">v=\frac{x^2}{2}.</math> | + | ::Let <math style="vertical-align: -1px">u=\ln x</math> and <math style="vertical-align: 0px">dv=x~dx.</math> Then, <math style="vertical-align: -13px">du=\frac{1}{x}dx</math> and <math style="vertical-align: -12px">v=\frac{x^2}{2}.</math> Thus, |
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− | :: | + | ::<math>\begin{array}{rcl} |
+ | \displaystyle{\int x\ln x~dx} & = & \displaystyle{\frac{x^2\ln x}{2}-\int \frac{x}{2}~dx}\\ | ||
+ | &&\\ | ||
+ | & = & \displaystyle{\frac{x^2\ln x}{2}-\frac{x^2}{4}+C.}\\ | ||
+ | \end{array}</math> | ||
|} | |} | ||
Revision as of 13:09, 18 April 2016
Evaluate the indefinite and definite integrals.
- a)
- b)
Foundations: |
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1. Integration by parts tells us that |
2. How would you integrate |
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Solution:
(a)
Step 1: |
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We proceed using integration by parts. Let and Then, and |
Therefore, we have |
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Step 2: |
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Now, we need to use integration by parts again. Let and Then, and |
Building on the previous step, we have |
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(b)
Step 1: |
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We proceed using integration by parts. Let and Then, and |
Therefore, we have |
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Step 2: |
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Now, we evaluate to get |
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Final Answer: |
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(a) |
(b) |