Difference between revisions of "009B Sample Midterm 2, Problem 2"

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<span class="exam"> This problem has three parts:
+
<span class="exam"> Evaluate
  
::<span class="exam">a) State the Fundamental Theorem of Calculus.
+
<span class="exam">(a) &nbsp; <math style="vertical-align: -14px">\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt</math>
  
::<span class="exam">b) Compute &thinsp; <math style="vertical-align: -15px">\frac{d}{dx}\int_0^{\cos (x)}\sin (t)~dt</math>.
+
<span class="exam">(b) &nbsp; <math style="vertical-align: -14px">\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx</math>
 
 
::<span class="exam">c) Evaluate <math style="vertical-align: -14px">\int_{0}^{\pi/4}\sec^2 x~dx</math>.
 
  
  
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!Foundations: &nbsp;  
 
!Foundations: &nbsp;  
 
|-
 
|-
|'''1.''' What does Part 1 of the Fundamental Theorem of Calculus say about&nbsp; <math style="vertical-align: -15px">\frac{d}{dx}\int_0^x\sin(t)~dt?</math>
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|How would you integrate &nbsp;<math style="vertical-align: -12px">\int (2x+1)\sqrt{x^2+x}~dx?</math>
 
|-
 
|-
 
|
 
|
::Part 1 of the Fundamental Theorem of Calculus says that&nbsp; <math style="vertical-align: -15px">\frac{d}{dx}\int_0^x\sin(t)~dt=\sin(x).</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=x^2+x.</math>
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; Then, &nbsp;<math style="vertical-align: -4px">du=(2x+1)~dx.</math>
 
|-
 
|-
|'''2.''' What does Part 2 of the Fundamental Theorem of Calculus say about <math style="vertical-align: -15px">\int_a^b\sec^2x~dx,</math> where <math style="vertical-align: -5px">a,b</math> are constants?
+
|
 +
&nbsp; &nbsp; &nbsp; &nbsp; Thus,  
 
|-
 
|-
 
|
 
|
::Part 2 of the Fundamental Theorem of Calculus says that <math style="vertical-align: -15px">\int_a^b\sec^2x~dx=F(b)-F(a),</math> where <math style="vertical-align: 0px">F</math> is any antiderivative of <math style="vertical-align: 0px">\sec^2x.</math>
+
&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 +
\displaystyle{\int (2x+1)\sqrt{x^2+x}~dx} & = & \displaystyle{\int \sqrt{u}~du}\\
 +
&&\\
 +
& = & \displaystyle{\frac{2}{3}u^{3/2}+C}\\
 +
&&\\
 +
& = & \displaystyle{\frac{2}{3}(x^2+x)^{3/2}+C.}
 +
\end{array}</math>
 
|}
 
|}
 +
  
 
'''Solution:'''
 
'''Solution:'''
  
 
'''(a)'''
 
'''(a)'''
 
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
|The Fundamental Theorem of Calculus has two parts.
+
|We multiply the product inside the integral to get
 
|-
 
|-
|'''The Fundamental Theorem of Calculus, Part 1'''
+
|  
|-
+
&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
|
+
\displaystyle{\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt} & = & \displaystyle{\int_1^2 \bigg(8t^3-10+12-\frac{15}{t^3}\bigg)~dt}\\
:Let <math style="vertical-align: -4px">f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: -14px">F(x)=\int_a^x f(t)~dt.</math>
+
&&\\
|-
+
& = & \displaystyle{\int_1^2 (8t^3+2-15t^{-3})~dt.}
|
+
\end{array}</math>
:Then, <math style="vertical-align: 0px">F</math> is a differentiable function on <math style="vertical-align: -5px">(a,b),</math> and <math style="vertical-align: -5px">F'(x)=f(x).</math>
 
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
|'''The Fundamental Theorem of Calculus, Part 2'''
+
|We integrate to get
 
|-
 
|-
|
+
| &nbsp;&nbsp; &nbsp; &nbsp; <math style="vertical-align: -16px">\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt=\left. 2t^4+2t+\frac{15}{2}t^{-2}\right|_1^2.</math>
:Let <math style="vertical-align: -4px">f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: 0px">F</math> be any antiderivative of <math style="vertical-align: -4px">f.</math>
+
|-
 +
|We now evaluate to get
 
|-
 
|-
|
+
|  
:Then, <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a).</math>
+
&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 +
\displaystyle{\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt} & = & \displaystyle{2(2)^4+2(2)+\frac{15}{2(2)^2}-\bigg(2+2+\frac{15}{2}\bigg)}\\
 +
&&\\
 +
& = & \displaystyle{36+\frac{15}{8}-4-\frac{15}{2}}\\
 +
&&\\
 +
& = & \displaystyle{\frac{211}{8}.}
 +
\end{array}</math>
 
|}
 
|}
 +
 
'''(b)'''
 
'''(b)'''
 
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
|Let <math style="vertical-align: -15px">F(x)=\int_0^{\cos (x)}\sin (t)~dt.</math> The problem is asking us to find <math style="vertical-align: -5px">F'(x).</math>
+
|We use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 +
|-
 +
|Let &nbsp;<math style="vertical-align: -2px">u=x^4+2x^2+4.</math>  
 
|-
 
|-
|Let <math style="vertical-align: -5px">g(x)=\cos(x)</math> and <math style="vertical-align: -14px">G(x)=\int_0^x \sin(t)~dt.</math>
+
|Then, &nbsp;<math style="vertical-align: -5px">du=(4x^3+4x)dx</math>&nbsp; and &nbsp;<math style="vertical-align: -14px">\frac{du}{4}=(x^3+x)dx.</math>  
 
|-
 
|-
|Then, <math style="vertical-align: -5px">F(x)=G(g(x)).</math>
+
|Also, we need to change the bounds of integration.  
|}
 
 
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 2: &nbsp;
 
 
|-
 
|-
|If we take the derivative of both sides of the last equation, we get <math style="vertical-align: -5px">F'(x)=G'(g(x))g'(x)</math> by the Chain Rule.
+
|Plugging in our values into the equation &nbsp;<math style="vertical-align: -4px">u=x^4+2x^2+4,</math>&nbsp; we get
|}
 
 
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 3: &nbsp;
 
 
|-
 
|-
|Now, <math style="vertical-align: -5px">g'(x)=-\sin(x)</math> and <math style="vertical-align: -5px">G'(x)=\sin(x)</math> by the '''Fundamental Theorem of Calculus, Part 1'''.
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">u_1=0^4+2(0)^2+4=4</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">u_2=2^4+2(2)^2+4=28.</math>
 
|-
 
|-
|Since <math style="vertical-align: -6px">G'(g(x))=\sin(g(x))=\sin(\cos(x)),</math> we have
+
|Therefore, the integral becomes
 
|-
 
|-
|
+
|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -14px">\frac{1}{4}\int_4^{28}\sqrt{u}~du.</math>
::<math style="vertical-align: -5px">F'(x)=G'(g(x))\cdot g'(x)=\sin(\cos(x))\cdot(-\sin(x)).</math>
 
 
|}
 
|}
 
'''(c)'''
 
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1: &nbsp;  
+
!Step 2: &nbsp;
 
|-
 
|-
| Using the '''Fundamental Theorem of Calculus, Part 2''', we have
+
|We now have
 
|-
 
|-
|
+
|  
::<math>\int_{0}^{\frac{\pi}{4}}\sec^2 x~dx=\tan(x)\biggr|_{0}^{\pi/4}.</math>
+
&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
|}
+
\displaystyle{\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx} & = & \displaystyle{\frac{1}{4}\int_4^{28}\sqrt{u}~du}\\
 
+
&&\\
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+
& = & \displaystyle{\left.\frac{1}{6}u^{\frac{3}{2}}\right|_4^{28}}\\
!Step 2: &nbsp;
+
&&\\
 +
& = & \displaystyle{\frac{1}{6}(28^{\frac{3}{2}}-4^{\frac{3}{2}})}\\
 +
&&\\
 +
& = & \displaystyle{\frac{1}{6}((\sqrt{28})^3-(\sqrt{4})^3)}\\
 +
&&\\
 +
& = & \displaystyle{\frac{1}{6}((2\sqrt{7})^3-2^3).}
 +
\end{array}</math>
 
|-
 
|-
|So, we get
+
|Therefore,  
 
|-
 
|-
|
+
| &nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -16px">\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx=\frac{28\sqrt{7}-4}{3}.</math>
::<math style="vertical-align: -16px">\int_{0}^{\frac{\pi}{4}}\sec^2 x~dx=\tan \bigg(\frac{\pi}{4}\bigg)-\tan (0)=1.</math>
 
 
|}
 
|}
 +
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
|&nbsp;&nbsp; '''(a)'''
+
|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>\frac{211}{8}</math>
|-
 
|&nbsp;&nbsp; '''The Fundamental Theorem of Calculus, Part 1'''
 
|-
 
|&nbsp;&nbsp; Let <math style="vertical-align: -4px">f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: -14px">F(x)=\int_a^x f(t)~dt.</math>
 
|-
 
|&nbsp;&nbsp; Then, <math style="vertical-align: 0px">F</math> is a differentiable function on <math style="vertical-align: -5px">(a,b),</math> and <math style="vertical-align: -5px">F'(x)=f(x).</math>
 
|-
 
|&nbsp;&nbsp; '''The Fundamental Theorem of Calculus, Part 2'''
 
|-
 
|&nbsp;&nbsp; Let <math style="vertical-align: -4px">f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: 0px">F</math> be any antiderivative of <math style="vertical-align: -4px">f.</math>
 
|-
 
|&nbsp;&nbsp; Then, <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a).</math>
 
|-
 
|&nbsp;&nbsp; '''(b)''' &nbsp; <math style="vertical-align: -15px">\frac{d}{dx}\int_0^{\cos (x)}\sin (t)~dt\,=\,\sin(\cos(x))\cdot(-\sin(x))</math>
 
 
|-
 
|-
|&nbsp;&nbsp; '''(c)''' <math style="vertical-align: -14px">\int_{0}^{\pi/4}\sec^2 x~dx\,=\,1</math>
+
|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>\frac{28\sqrt{7}-4}{3}</math>
 
|}
 
|}
 
[[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

Revision as of 12:15, 9 April 2017

Evaluate

(a)  

(b)  


Foundations:  
How would you integrate  

        You can use  -substitution.

        Let  
        Then,  

        Thus,

       


Solution:

(a)

Step 1:  
We multiply the product inside the integral to get

       

Step 2:  
We integrate to get
      
We now evaluate to get

       

(b)

Step 1:  
We use  -substitution.
Let  
Then,    and  
Also, we need to change the bounds of integration.
Plugging in our values into the equation    we get
         and  
Therefore, the integral becomes
       
Step 2:  
We now have

       

Therefore,
       


Final Answer:  
    (a)    
    (b)    

Return to Sample Exam