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>
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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> Thus,
 
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::Thus, <math style="vertical-align: -15px">\int x\ln x~dx\,=\,\frac{x^2\ln x}{2}-\int \frac{x}{2}~dx\,=\,\frac{x^2\ln x}{2}-\frac{x^2}{4}+C.</math>
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::<math>\begin{array}{rcl}
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\displaystyle{\int x\ln x~dx} & = & \displaystyle{\frac{x^2\ln x}{2}-\int \frac{x}{2}~dx}\\
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&&\\
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& = & \displaystyle{\frac{x^2\ln x}{2}-\frac{x^2}{4}+C.}\\
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\end{array}</math>
 
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Revision as of 13:09, 18 April 2016

Evaluate the indefinite and definite integrals.

a)
b)


Foundations:  
1. Integration by parts tells us that
2. How would you integrate
You could use integration by parts.
Let and Then, and Thus,

Solution:

(a)

Step 1:  
We proceed using integration by parts. Let and Then, and
Therefore, we have
Step 2:  
Now, we need to use integration by parts again. Let and Then, and
Building on the previous step, we have

(b)

Step 1:  
We proceed using integration by parts. Let and Then, and
Therefore, we have
Step 2:  
Now, we evaluate to get
Final Answer:  
(a)  
(b)  

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