Difference between revisions of "009B Sample Final 3, Problem 2"
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!Foundations: | !Foundations: | ||
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| − | |<math>\int \frac{1}{1+x^2}~dx=\arctan(x)+C</math> | + | |'''1.''' |
| + | |- | ||
| + | | <math>\int \frac{1}{1+x^2}~dx=\arctan(x)+C</math> | ||
| + | |- | ||
| + | |'''2.''' How would you integrate <math style="vertical-align: -12px">\int \frac{\ln x}{x}~dx?</math> | ||
|- | |- | ||
| | | | ||
| + | You could use <math style="vertical-align: 0px">u</math>-substitution. | ||
| + | |- | ||
| + | | Let <math style="vertical-align: -5px">u=\ln(x).</math> | ||
| + | |- | ||
| + | | Then, <math style="vertical-align: -13px">du=\frac{1}{x}dx.</math> | ||
|- | |- | ||
| | | | ||
| + | Thus, | ||
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| | | | ||
| + | <math>\begin{array}{rcl} | ||
| + | \displaystyle{\int \frac{\ln x}{x}~dx} & = & \displaystyle{\int u~du}\\ | ||
| + | &&\\ | ||
| + | & = & \displaystyle{\frac{u^2}{2}+C}\\ | ||
| + | &&\\ | ||
| + | & = & \displaystyle{\frac{(\ln x)^2}{2}+C.} | ||
| + | \end{array}</math> | ||
| + | |- | ||
|} | |} | ||
Revision as of 16:07, 1 March 2017
Evaluate the following integrals.
(a)
(b)
(c)
| Foundations: |
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| 1. |
| 2. How would you integrate |
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You could use -substitution. |
| Let |
| Then, |
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Thus, |
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Solution:
(a)
| Step 1: |
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| Step 2: |
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(b)
| Step 1: |
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| Step 2: |
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(c)
| Step 1: |
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| Step 2: |
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| Final Answer: |
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| (a) |
| (b) |
| (c) |