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

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<span class="exam"> This problem has three parts:
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<span class="exam"> Evaluate
  
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::<span class="exam">a) State the Fundamental Theorem of Calculus.
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<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>.
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<span class="exam">(b) &nbsp; <math style="vertical-align: -14px">\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx</math>
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::<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;  
 
|-
 
|-
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|'''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>
 
|-
 
|-
 
|
 
|
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::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>
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&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>
 
|-
 
|-
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|'''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,  
 
|-
 
|-
 
|
 
|
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::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>
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&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;  
 
|-
 
|-
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|The Fundamental Theorem of Calculus has two parts.
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|We multiply the product inside the integral to get
 
|-
 
|-
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|'''The Fundamental Theorem of Calculus, Part 1'''
+
|  
−
|-
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&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
−
|
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\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}\\
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: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>
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&&\\
−
|-
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& = & \displaystyle{\int_1^2 (8t^3+2-15t^{-3})~dt.}
−
|
+
\end{array}</math>
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: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;
 
|-
 
|-
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|'''The Fundamental Theorem of Calculus, Part 2'''
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|We integrate to get
 
|-
 
|-
−
|
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| &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>
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: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;  
 
|-
 
|-
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|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>
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|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>  
 
|-
 
|-
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|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>  
 
|-
 
|-
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|Then, <math style="vertical-align: -5px">F(x)=G(g(x)).</math>
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|Also, we need to change the bounds of integration.  
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|}
 
−
 
 
−
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
−
!Step 2: &nbsp;
 
 
|-
 
|-
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|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.
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|Plugging in our values into the equation &nbsp;<math style="vertical-align: -4px">u=x^4+2x^2+4,</math>&nbsp; we get
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|}
 
−
 
 
−
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
−
!Step 3: &nbsp;
 
 
|-
 
|-
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|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'''.
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|&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>
 
|-
 
|-
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|Since <math style="vertical-align: -6px">G'(g(x))=\sin(g(x))=\sin(\cos(x)),</math> we have
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|Therefore, the integral becomes
 
|-
 
|-
−
|
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|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -14px">\frac{1}{4}\int_4^{28}\sqrt{u}~du.</math>
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::<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)   Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt}

(b)   Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx}


Foundations:  
How would you integrate  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int (2x+1)\sqrt{x^2+x}~dx?}

        You can use  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u} -substitution.

        Let  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=x^2+x.}
        Then,  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle du=(2x+1)~dx.}

        Thus,

        Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}}


Solution:

(a)

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

        Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}\\ &&\\ & = & \displaystyle{\int_1^2 (8t^3+2-15t^{-3})~dt.} \end{array}}

Step 2:  
We integrate to get
       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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.}
We now evaluate to get

        Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}}

(b)

Step 1:  
We use  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u} -substitution.
Let  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=x^4+2x^2+4.}
Then,    and  
Also, we need to change the bounds of integration.
Plugging in our values into the equation  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=x^4+2x^2+4,}   we get
       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_1=0^4+2(0)^2+4=4}   and  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_2=2^4+2(2)^2+4=28.}
Therefore, the integral becomes
        Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{4}\int_4^{28}\sqrt{u}~du.}
Step 2:  
We now have

        Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}\\ &&\\ & = & \displaystyle{\left.\frac{1}{6}u^{\frac{3}{2}}\right|_4^{28}}\\ &&\\ & = & \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}}

Therefore,
        Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx=\frac{28\sqrt{7}-4}{3}.}


Final Answer:  
    (a)     Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{211}{8}}
    (b)     Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{28\sqrt{7}-4}{3}}

Return to Sample Exam

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