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  
 
|-
 
|-
|&nbsp; &nbsp; <math style="vertical-align: -12px">\int u~dv=uv-\int v~du.</math>
+
|&nbsp; &nbsp; &nbsp; &nbsp; <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>
 
|-
 
|-
 
|
 
|
&nbsp; &nbsp; You could use integration by parts.
+
&nbsp; &nbsp; &nbsp; &nbsp; You could use integration by parts.
 
|-
 
|-
 
|
 
|
&nbsp; &nbsp; 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>
+
&nbsp; &nbsp; &nbsp; &nbsp; Let <math style="vertical-align: -1px">u=\ln x</math> and <math style="vertical-align: 0px">dv=x~dx.</math>  
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; Then, <math style="vertical-align: -13px">du=\frac{1}{x}dx</math> and <math style="vertical-align: -12px">v=\frac{x^2}{2}.</math>
 
|-
 
|-
 
|
 
|
&nbsp; &nbsp; 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>
+
&nbsp; &nbsp; &nbsp; &nbsp; <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
 
|-
 
|-
| &nbsp;&nbsp; <math style="vertical-align: -12px">\int x^2 e^x~dx=x^2e^x-\int 2xe^x~dx.</math>
+
| &nbsp; &nbsp; &nbsp; &nbsp; <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
 
|-
 
|-
| &nbsp; &nbsp; <math>\begin{array}{rcl}
+
| &nbsp; &nbsp; &nbsp; &nbsp; <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
 
|-
 
|-
 
|  
 
|  
&nbsp; &nbsp; <math>\begin{array}{rcl}
+
&nbsp; &nbsp; &nbsp; &nbsp; <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  
 
|-
 
|-
| &nbsp; &nbsp; <math>\begin{array}{rcl}
+
| &nbsp; &nbsp; &nbsp; &nbsp; <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: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
|'''(a)''' &nbsp; <math>x^2e^x-2xe^x+2e^x+C</math>
+
|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>x^2e^x-2xe^x+2e^x+C</math>
 
|-
 
|-
|'''(b)''' &nbsp; <math>\frac{3e^4+1}{16}</math>
+
|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <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:  
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:  
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)    

Return to Sample Exam