Difference between revisions of "009B Sample Midterm 1, Problem 3"
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|'''1.''' Integration by parts tells us that | |'''1.''' Integration by parts tells us that | ||
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− | | <math style="vertical-align: -12px">\int u~dv=uv-\int v~du.</math> | + | | <math style="vertical-align: -12px">\int u~dv=uv-\int v~du.</math> |
|- | |- | ||
|'''2.''' How would you integrate <math style="vertical-align: -12px">\int x\ln x~dx?</math> | |'''2.''' How would you integrate <math style="vertical-align: -12px">\int x\ln x~dx?</math> | ||
|- | |- | ||
| | | | ||
− | You could use integration by parts. | + | You could use integration by parts. |
|- | |- | ||
| | | | ||
− | 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> | ||
|- | |- | ||
| | | | ||
− | | + | <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> | ||
|} | |} | ||
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|We proceed using integration by parts. | |We proceed using integration by parts. | ||
|- | |- | ||
− | |Let <math style="vertical-align: 0px">u=x^2</math> and <math style="vertical-align: 0px">dv=e^xdx.</math> Then, <math style="vertical-align: 0px">du=2xdx</math> and <math style="vertical-align: 0px">v=e^x.</math> | + | |Let <math style="vertical-align: 0px">u=x^2</math> and <math style="vertical-align: 0px">dv=e^xdx.</math> |
+ | |- | ||
+ | |Then, <math style="vertical-align: 0px">du=2xdx</math> and <math style="vertical-align: 0px">v=e^x.</math> | ||
|- | |- | ||
|Therefore, we have | |Therefore, we have | ||
|- | |- | ||
− | | <math style="vertical-align: -12px">\int x^2 e^x~dx=x^2e^x-\int 2xe^x~dx.</math> | + | | <math style="vertical-align: -12px">\int x^2 e^x~dx=x^2e^x-\int 2xe^x~dx.</math> |
|} | |} | ||
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|Now, we need to use integration by parts again. | |Now, we need to use integration by parts again. | ||
|- | |- | ||
− | |Let <math style="vertical-align: 0px">u=2x</math> and <math style="vertical-align: 0px">dv=e^xdx.</math> Then, <math style="vertical-align: 0px">du=2dx</math> and <math style="vertical-align: 0px">v=e^x.</math> | + | |Let <math style="vertical-align: 0px">u=2x</math> and <math style="vertical-align: 0px">dv=e^xdx.</math> |
+ | |- | ||
+ | |Then, <math style="vertical-align: 0px">du=2dx</math> and <math style="vertical-align: 0px">v=e^x.</math> | ||
|- | |- | ||
|Building on the previous step, we have | |Building on the previous step, we have | ||
|- | |- | ||
− | | <math>\begin{array}{rcl} | + | | <math>\begin{array}{rcl} |
\displaystyle{\int x^2 e^x~dx} & = & \displaystyle{x^2e^x-\bigg(2xe^x-\int 2e^x~dx\bigg)}\\ | \displaystyle{\int x^2 e^x~dx} & = & \displaystyle{x^2e^x-\bigg(2xe^x-\int 2e^x~dx\bigg)}\\ | ||
&&\\ | &&\\ | ||
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|We proceed using integration by parts. | |We proceed using integration by parts. | ||
|- | |- | ||
− | |Let <math style="vertical-align: -1px">u=\ln x</math> and <math style="vertical-align: 0px">dv=x^3dx.</math> Then, <math style="vertical-align: -13px">du=\frac{1}{x}dx</math> and <math style="vertical-align: -14px">v=\frac{x^4}{4}.</math> | + | |Let <math style="vertical-align: -1px">u=\ln x</math> and <math style="vertical-align: 0px">dv=x^3dx.</math> |
+ | |- | ||
+ | |Then, <math style="vertical-align: -13px">du=\frac{1}{x}dx</math> and <math style="vertical-align: -14px">v=\frac{x^4}{4}.</math> | ||
|- | |- | ||
|Therefore, we have | |Therefore, we have | ||
|- | |- | ||
| | | | ||
− | <math>\begin{array}{rcl} | + | <math>\begin{array}{rcl} |
\displaystyle{\int_{1}^{e} x^3\ln x~dx} & = & \displaystyle{\left.\ln x \bigg(\frac{x^4}{4}\bigg)\right|_{1}^{e}-\int_1^e \frac{x^3}{4}~dx}\\ | \displaystyle{\int_{1}^{e} x^3\ln x~dx} & = & \displaystyle{\left.\ln x \bigg(\frac{x^4}{4}\bigg)\right|_{1}^{e}-\int_1^e \frac{x^3}{4}~dx}\\ | ||
&&\\ | &&\\ | ||
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|Now, we evaluate to get | |Now, we evaluate to get | ||
|- | |- | ||
− | | <math>\begin{array}{rcl} | + | | <math>\begin{array}{rcl} |
\displaystyle{\int_{1}^{e} x^3\ln x~dx} & = & \displaystyle{\bigg((\ln e) \frac{e^4}{4}-\frac{e^4}{16}\bigg)-\bigg((\ln 1) \frac{1^4}{4}-\frac{1^4}{16}\bigg)}\\ | \displaystyle{\int_{1}^{e} x^3\ln x~dx} & = & \displaystyle{\bigg((\ln e) \frac{e^4}{4}-\frac{e^4}{16}\bigg)-\bigg((\ln 1) \frac{1^4}{4}-\frac{1^4}{16}\bigg)}\\ | ||
&&\\ | &&\\ | ||
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!Final Answer: | !Final Answer: | ||
|- | |- | ||
− | |'''(a)''' <math>x^2e^x-2xe^x+2e^x+C</math> | + | | '''(a)''' <math>x^2e^x-2xe^x+2e^x+C</math> |
|- | |- | ||
− | |'''(b)''' <math>\frac{3e^4+1}{16}</math> | + | | '''(b)''' <math>\frac{3e^4+1}{16}</math> |
|} | |} | ||
[[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] | [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] |
Revision as of 16:48, 7 February 2017
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 |
You could use integration by parts. |
Let and |
Then, and |
|
Solution:
(a)
Step 1: |
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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: |
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(a) |
(b) |