Difference between revisions of "009B Sample Final 2, Problem 6"

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!Step 1:    
 
!Step 1:    
 
|-
 
|-
−
|
+
|We start by using trig substitution.
 +
|-
 +
|Let &nbsp;<math>x=4\sec \theta.</math>
 
|-
 
|-
−
|
+
|Then, &nbsp;<math>dx=4\sec \theta \tan \theta d\theta.</math>
 
|-
 
|-
−
|
+
|So, the integral becomes
 
|-
 
|-
−
|
+
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 +
\displaystyle{\int \frac{1}{x^2\sqrt{x^2-16}}~dx} & = & \displaystyle{\int \frac{4\sec \theta \tan \theta}{16\sec^2\theta \sqrt{16\sec^2 \theta -16}}~d\theta}\\
 +
&&\\
 +
& = & \displaystyle{\int \frac{4\sec \theta \tan \theta}{16\sec^2\theta (4\tan \theta)} ~d\theta}\\
 +
&&\\
 +
& = & \displaystyle{\int \frac{1}{16\sec \theta} ~d\theta.}
 +
\end{array}</math>
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
−
|
+
|Now, we integrate to get
−
|-
 
−
|
 
 
|-
 
|-
−
|
+
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
−
|-
+
\displaystyle{\int \frac{1}{x^2\sqrt{x^2-16}}~dx} & = & \displaystyle{\frac{1}{16}\cos\theta~d\theta}\\
−
|
+
&&\\
 +
& = & \displaystyle{\frac{1}{16}\sin \theta +C}\\
 +
&&\\
 +
& = & \displaystyle{\frac{1}{16}\bigg(\frac{\sqrt{x^2-16}}{x}\bigg)+C.}
 +
\end{array}</math>
 
|}
 
|}
  
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
−
|
+
|First, we write
 
|-
 
|-
−
|
+
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 +
\displaystyle{\int_{-\pi}^\pi \sin^3x\cos^3x~dx} & = & \displaystyle{\int_{-\pi}^{\pi} \sin^3x \cos^2x \cos x~dx}\\
 +
&&\\
 +
& = & \displaystyle{\int_{-\pi}^{\pi} \sin^3x (1-\sin^2x)\cos x~dx.}
 +
\end{array}</math>
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
−
|
+
|Now, we use &nbsp;<math>u</math>-substitution.
 +
|-
 +
|Let &nbsp;<math>u=\sin x.</math>&nbsp; Then, &nbsp;<math>du=\cos x dx.</math>
 +
|-
 +
|Since this is a definite integral, we need to change the bounds of integration.
 +
|-
 +
|Then, we have
 +
|-
 +
|&nbsp;<math>u_1=\sin(-\pi)=0</math>&nbsp; and &nbsp;<math>u_2=\sin(\pi)=0.</math>
 +
|-
 +
|So, we have
 
|-
 
|-
−
|
+
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 +
\displaystyle{\int_{-\pi}^\pi \sin^3x\cos^3x~dx} & = & \displaystyle{\int_0^0 u^3(1-u^2)~du}\\
 +
&&\\
 +
& = & \displaystyle{0.}
 +
\end{array}</math>
 
|}
 
|}
  
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
−
|
+
|First, we write
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp;<math>\int_0^1 \frac{x-3}{x^2+6x+5}~dx=\int_0^1 \frac{x-3}{(x+1)(x+5)}~dx.</math>
 +
|-
 +
|Now, we use partial fraction decomposition. Wet set
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{x-3}{(x+1)(x+5)}=\frac{A}{x+1}+\frac{B}{x+5}.</math>
 +
|-
 +
|If we multiply both sides of this equation by &nbsp;<math>(x+1)(x+5),</math> we get
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp;<math>x-3=A(x+5)+B(x+1).</math>
 +
|-
 +
|If we  let &nbsp;<math>x=-1,</math> we get <math>A=-1.</math>
 +
|-
 +
|If we  let &nbsp;<math>x=-5,</math> we get <math>B=2.</math>
 +
|-
 +
|So, we have
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{x-3}{(x+1)(x+5)}=\frac{-1}{x+1}+\frac{2}{x+5}.</math>
 
|-
 
|-
 
|
 
|
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 2: &nbsp;
 
!Step 2: &nbsp;
 +
|-
 +
|Now, we have
 
|-
 
|-
 
|
 
|
 +
&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 +
\displaystyle{\int_0^1 \frac{x-3}{(x+1)(x+5)}~dx} & = & \displaystyle{\int_0^1 \frac{-1}{x+1}+\frac{2}{x+5}~dx}\\
 +
&&\\
 +
& = & \displaystyle{\int_0^1 \frac{-1}{x+1}~dx+\int_0^1 \frac{2}{x+5}~dx.}
 +
\end{array}</math>
 
|-
 
|-
−
|
+
|Now, we use &nbsp;<math>u</math>-substitution for both of these integrals.
 +
|-
 +
|Let &nbsp;<math>u=x+1.</math>&nbsp; Then, &nbsp;<math>du=dx.</math>
 +
|-
 +
|Let &nbsp;<math>t=x+5.</math>&nbsp; Then, &nbsp;<math>dt=dx.</math>
 +
|-
 +
|Since these are definite integrals, we need to change the bounds of integration.
 +
|-
 +
|We have &nbsp;<math>u_1=0+1=1</math>&nbsp; and &nbsp;<math>u_2=1+1=2.</math>
 +
|-
 +
|Also, &nbsp;<math>t_1=0+5=5</math>&nbsp; and &nbsp;<math>u_2=1+5=6.</math> 
 +
|-
 +
|Therefore, we get
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 +
\displaystyle{\int_0^1 \frac{x-3}{(x+1)(x+5)}~dx} & = & \displaystyle{\int_1^2 \frac{-1}{u}~du+\int_5^6 \frac{2}{t}~dt}\\
 +
&&\\
 +
& = & \displaystyle{-\ln|u|\bigg|_1^2+2\ln|t|\bigg|_5^6}\\
 +
&&\\
 +
& = & \displaystyle{-\ln(2)+2\ln(6)-2\ln(5).}
 +
\end{array}</math>
 
|}
 
|}
  
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
−
|'''(a)'''  
+
|&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp; <math>\frac{1}{16}\bigg(\frac{\sqrt{x^2-16}}{x}\bigg)+C</math>
 
|-
 
|-
−
|'''(b)'''  
+
|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp; <math>0</math>
 
|-
 
|-
−
|'''(c)'''  
+
|&nbsp; &nbsp;'''(c)'''&nbsp; &nbsp; <math>-\ln(2)+2\ln(6)-2\ln(5)</math>
 
|}
 
|}
 
[[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

Revision as of 13:08, 4 March 2017

Evaluate the following integrals:

(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 \frac{dx}{x^2\sqrt{x^2-16}}}

(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_{-\pi}^\pi \sin^3x\cos^3x~dx}

(c)  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_0^1 \frac{x-3}{x^2+6x+5}~dx}

Foundations:  


Solution:

(a)

Step 1:  
We start by using trig 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 x=4\sec \theta.}
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 dx=4\sec \theta \tan \theta d\theta.}
So, 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 \begin{array}{rcl} \displaystyle{\int \frac{1}{x^2\sqrt{x^2-16}}~dx} & = & \displaystyle{\int \frac{4\sec \theta \tan \theta}{16\sec^2\theta \sqrt{16\sec^2 \theta -16}}~d\theta}\\ &&\\ & = & \displaystyle{\int \frac{4\sec \theta \tan \theta}{16\sec^2\theta (4\tan \theta)} ~d\theta}\\ &&\\ & = & \displaystyle{\int \frac{1}{16\sec \theta} ~d\theta.} \end{array}}
Step 2:  
Now, 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 \begin{array}{rcl} \displaystyle{\int \frac{1}{x^2\sqrt{x^2-16}}~dx} & = & \displaystyle{\frac{1}{16}\cos\theta~d\theta}\\ &&\\ & = & \displaystyle{\frac{1}{16}\sin \theta +C}\\ &&\\ & = & \displaystyle{\frac{1}{16}\bigg(\frac{\sqrt{x^2-16}}{x}\bigg)+C.} \end{array}}

(b)

Step 1:  
First, we write
        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_{-\pi}^\pi \sin^3x\cos^3x~dx} & = & \displaystyle{\int_{-\pi}^{\pi} \sin^3x \cos^2x \cos x~dx}\\ &&\\ & = & \displaystyle{\int_{-\pi}^{\pi} \sin^3x (1-\sin^2x)\cos x~dx.} \end{array}}
Step 2:  
Now, 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=\sin 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=\cos x dx.}
Since this is a definite integral, we need to change the bounds of integration.
Then, we 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 u_1=\sin(-\pi)=0}   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=\sin(\pi)=0.}
So, we 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_{-\pi}^\pi \sin^3x\cos^3x~dx} & = & \displaystyle{\int_0^0 u^3(1-u^2)~du}\\ &&\\ & = & \displaystyle{0.} \end{array}}

(c)

Step 1:  
First, we write
       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_0^1 \frac{x-3}{x^2+6x+5}~dx=\int_0^1 \frac{x-3}{(x+1)(x+5)}~dx.}
Now, we use partial fraction decomposition. Wet set
       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{x-3}{(x+1)(x+5)}=\frac{A}{x+1}+\frac{B}{x+5}.}
If we multiply both sides of this equation by  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 (x+1)(x+5),} 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 x-3=A(x+5)+B(x+1).}
If we 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 x=-1,} 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 A=-1.}
If we 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 x=-5,} 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 B=2.}
So, we 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 \frac{x-3}{(x+1)(x+5)}=\frac{-1}{x+1}+\frac{2}{x+5}.}
Step 2:  
Now, we 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_0^1 \frac{x-3}{(x+1)(x+5)}~dx} & = & \displaystyle{\int_0^1 \frac{-1}{x+1}+\frac{2}{x+5}~dx}\\ &&\\ & = & \displaystyle{\int_0^1 \frac{-1}{x+1}~dx+\int_0^1 \frac{2}{x+5}~dx.} \end{array}}

Now, 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 for both of these integrals.
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+1.}   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=dx.}
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 t=x+5.}   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 dt=dx.}
Since these are definite integrals, we need to change the bounds of integration.
We have    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=1+1=2.}
Also,  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 t_1=0+5=5}   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=1+5=6.}
Therefore, 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 \begin{array}{rcl} \displaystyle{\int_0^1 \frac{x-3}{(x+1)(x+5)}~dx} & = & \displaystyle{\int_1^2 \frac{-1}{u}~du+\int_5^6 \frac{2}{t}~dt}\\ &&\\ & = & \displaystyle{-\ln|u|\bigg|_1^2+2\ln|t|\bigg|_5^6}\\ &&\\ & = & \displaystyle{-\ln(2)+2\ln(6)-2\ln(5).} \end{array}}


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{1}{16}\bigg(\frac{\sqrt{x^2-16}}{x}\bigg)+C}
   (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 0}
   (c)    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 -\ln(2)+2\ln(6)-2\ln(5)}

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