Difference between revisions of "009B Sample Midterm 1, Problem 4"

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<span class="exam">Evaluate the integral:
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<span class="exam"> Evaluate the indefinite and definite integrals.
  
::<math>\int \sin^3x \cos^2x~dx</math>
+
<span class="exam">(a) &nbsp; <math>\int x^2 e^x~dx</math>
  
 +
<span class="exam">(b) &nbsp; <math>\int_{1}^{e} x^3\ln x~dx</math>
  
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Foundations: &nbsp;
 
|-
 
|'''1.''' Recall the trig identity
 
|-
 
| &nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -2px">\sin^2x+\cos^2x=1</math>
 
|-
 
|'''2.''' How would you integrate &nbsp;<math style="vertical-align: -14px">\int \sin^2x\cos x~dx?</math>
 
|-
 
|
 
&nbsp; &nbsp; &nbsp; &nbsp; You can use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math style="vertical-align: -2px">u=\sin x.</math>
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp; Then, &nbsp;<math style="vertical-align: -1px">du=\cos x~dx.</math>
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp; Thus,
 
|-
 
|
 
&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
\displaystyle{\int \sin^2x\cos x~dx} & = & \displaystyle{\int u^2~du}\\
 
&&\\
 
& = & \displaystyle{\frac{u^3}{3}+C}\\
 
&&\\
 
& = & \displaystyle{\frac{\sin^3x}{3}+C.}
 
\end{array}</math>
 
|}
 
  
 +
<hr>
 +
[[009B Sample Midterm 1, Problem 4 Solution|'''<u>Solution</u>''']]
  
'''Solution:'''
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 1: &nbsp;
 
|-
 
|First, we write
 
|-
 
| &nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -13px">\int\sin^3x\cos^2x~dx=\int (\sin x) \sin^2x\cos^2x~dx.</math>
 
|-
 
|Using the identity &nbsp;<math style="vertical-align: -4px">\sin^2x+\cos^2x=1,</math>
 
|-
 
|we get
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -1px">\sin^2x=1-\cos^2x.</math>
 
|-
 
|If we use this identity, we have
 
|-
 
|
 
&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
\displaystyle{\int\sin^3x\cos^2x~dx} & = & \displaystyle{\int (\sin x) (1-\cos^2x)\cos^2x~dx}\\
 
&&\\
 
& = & \displaystyle{\int (\cos^2x-\cos^4x)\sin(x)~dx.}
 
\end{array}</math>
 
|}
 
  
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+
[[009B Sample Midterm 1, Problem 4 Detailed Solution|'''<u>Detailed Solution</u>''']]
!Step 2: &nbsp;
 
|-
 
|Now, we use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 
|-
 
|Let &nbsp;<math style="vertical-align: -5px">u=\cos(x).</math>
 
|-
 
|Then, &nbsp;<math style="vertical-align: -5px">du=-\sin(x)dx.</math>
 
|-
 
|Therefore,
 
|-
 
|
 
&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 
\displaystyle{\int\sin^3x\cos^2x~dx} & = & \displaystyle{\int -(u^2-u^4)~du}\\
 
&&\\
 
& = & \displaystyle{\frac{-u^3}{3}+\frac{u^5}{5}+C}\\
 
&&\\
 
& = & \displaystyle{\frac{\cos^5x}{5}-\frac{\cos^3x}{3}+C.}
 
\end{array}</math>
 
|}
 
  
  
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Final Answer: &nbsp;
 
|-
 
| &nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{\cos^5x}{5}-\frac{\cos^3x}{3}+C</math>
 
|}
 
 
[[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 16:05, 12 November 2017

Evaluate the indefinite and definite integrals.

(a)  

(b)  



Solution


Detailed Solution


Return to Sample Exam