Difference between revisions of "009B Sample Final 3, Problem 2"

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!Step 1:    
 
!Step 1:    
 
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|First, we notice
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|&nbsp; &nbsp; &nbsp; &nbsp;<math>\int_0^{\frac{\sqrt{3}}{4}} \frac{1}{1+16x^2}~dx=\int_0^{\frac{\sqrt{3}}{4}} \frac{1}{1+(4x)^2}~dx.</math>
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|Now, we use &nbsp;<math>u</math>-substitution.
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|Let &nbsp;<math style="vertical-align: -5px">u=4x.</math>
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|Then, &nbsp;<math style="vertical-align: -5px">du=4dx</math>&nbsp; and &nbsp;<math>\frac{du}{4}=dx.</math>
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|-
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|Also, we need to change the bounds of integration.
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|-
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|Plugging in our values into the equation &nbsp;<math style="vertical-align: -5px">u=4x,</math>
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|-
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|we get &nbsp;<math style="vertical-align: -15px">u_1=4(0)=0</math>&nbsp; and &nbsp;<math style="vertical-align: -16px">u_2=4\bigg(\frac{\sqrt{3}}{4}\bigg)=\sqrt{3}.</math>
 
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|Therefore, the integral becomes
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -19px">\frac{1}{4}\int_0^{\sqrt{3}} \frac{1}{1+u^2}~du.</math>
 
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
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|We now have
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&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
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\displaystyle{\int_0^{\frac{\sqrt{3}}{4}} \frac{1}{1+16x^2}~dx} & = & \displaystyle{\frac{1}{4}\int_0^{\sqrt{3}} \frac{1}{1+u^2}~du}\\
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&&\\
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& = & \displaystyle{\frac{1}{4}\arctan(u)\bigg|_0^{\sqrt{3}}}\\
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&&\\
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& = & \displaystyle{\frac{1}{4}\arctan(\sqrt{3})-\frac{1}{4}\arctan(0)}\\
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&&\\
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& = & \displaystyle{\frac{1}{4}\bigg(\frac{\pi}{3}\bigg)-0}\\
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&&\\
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& = & \displaystyle{\frac{\pi}{12}.}
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\end{array}</math>
 
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
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|'''(a)'''  
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|&nbsp; &nbsp;'''(a)''' &nbsp; &nbsp;<math>\frac{\pi}{12}</math>
 
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|'''(b)'''  
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|&nbsp; &nbsp;'''(b)''' &nbsp; &nbsp;
 
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|'''(c)'''  
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|&nbsp; &nbsp;'''(c)''' &nbsp; &nbsp;
 
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[[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

Revision as of 15:18, 1 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_0^{\frac{\sqrt{3}}{4}} \frac{1}{1+16x^2}~dx}

(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 \frac{x^2}{(1+x^3)^2}}

(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_1^e \frac{\cos(\ln(x))}{x}~dx}

Foundations:  
1.
       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{1}{1+x^2}~dx=\arctan(x)+C}
2. 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 \frac{\ln x}{x}~dx?}

        You could 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=\ln(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=\frac{1}{x}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 \frac{\ln x}{x}~dx} & = & \displaystyle{\int u~du}\\ &&\\ & = & \displaystyle{\frac{u^2}{2}+C}\\ &&\\ & = & \displaystyle{\frac{(\ln x)^2}{2}+C.} \end{array}}


Solution:

(a)

Step 1:  
First, we notice
       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^{\frac{\sqrt{3}}{4}} \frac{1}{1+16x^2}~dx=\int_0^{\frac{\sqrt{3}}{4}} \frac{1}{1+(4x)^2}~dx.}
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=4x.}
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=4dx}   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 \frac{du}{4}=dx.}
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=4x,}
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=4(0)=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=4\bigg(\frac{\sqrt{3}}{4}\bigg)=\sqrt{3}.}
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_0^{\sqrt{3}} \frac{1}{1+u^2}~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^{\frac{\sqrt{3}}{4}} \frac{1}{1+16x^2}~dx} & = & \displaystyle{\frac{1}{4}\int_0^{\sqrt{3}} \frac{1}{1+u^2}~du}\\ &&\\ & = & \displaystyle{\frac{1}{4}\arctan(u)\bigg|_0^{\sqrt{3}}}\\ &&\\ & = & \displaystyle{\frac{1}{4}\arctan(\sqrt{3})-\frac{1}{4}\arctan(0)}\\ &&\\ & = & \displaystyle{\frac{1}{4}\bigg(\frac{\pi}{3}\bigg)-0}\\ &&\\ & = & \displaystyle{\frac{\pi}{12}.} \end{array}}

(b)

Step 1:  
Step 2:  

(c)

Step 1:  
Step 2:  


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{\pi}{12}}
   (b)    
   (c)    

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