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

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&nbsp; &nbsp; &nbsp; &nbsp; You could use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.  
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&nbsp; &nbsp; &nbsp; &nbsp; You can use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.  
 
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|&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math style="vertical-align: -5px">u=\ln(x).</math>
 
|&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math style="vertical-align: -5px">u=\ln(x).</math>
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
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|We need to use &nbsp;<math style="vertical-align: 0px">u</math>-substitution. Let &nbsp;<math style="vertical-align: -2px">u=1+x^3.</math>  
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|We use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.  
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|Let &nbsp;<math style="vertical-align: -2px">u=1+x^3.</math>  
 
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|Then, &nbsp;<math style="vertical-align: 0px">du=3x^2dx</math>&nbsp; and &nbsp;<math style="vertical-align: -13px">\frac{du}{3}=x^2dx.</math>
 
|Then, &nbsp;<math style="vertical-align: 0px">du=3x^2dx</math>&nbsp; and &nbsp;<math style="vertical-align: -13px">\frac{du}{3}=x^2dx.</math>
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
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|We need to use &nbsp;<math>u</math>-substitution.  
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|We use &nbsp;<math>u</math>-substitution.  
 
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|Let &nbsp;<math style="vertical-align: -5px">u=\sin(x).</math>  
 
|Let &nbsp;<math style="vertical-align: -5px">u=\sin(x).</math>  
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|Also, we need to change the bounds of integration.
 
|Also, we need to change the bounds of integration.
 
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|Plugging in our values into the equation &nbsp;<math style="vertical-align: -5px">u=\sin(x),</math>  
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|Plugging in our values into the equation &nbsp;<math style="vertical-align: -5px">u=\sin(x),</math> we get
 
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|we get &nbsp;<math style="vertical-align: -15px">u_1=\sin\bigg(\frac{\pi}{4}\bigg)=\frac{\sqrt{2}}{2}</math>&nbsp; and &nbsp;<math style="vertical-align: -16px">u_2=\sin\bigg(\frac{\pi}{2}\bigg)=1.</math>&nbsp;
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -15px">u_1=\sin\bigg(\frac{\pi}{4}\bigg)=\frac{\sqrt{2}}{2}</math>&nbsp; and &nbsp;<math style="vertical-align: -16px">u_2=\sin\bigg(\frac{\pi}{2}\bigg)=1.</math>&nbsp;
 
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|Therefore, the integral becomes  
 
|Therefore, the integral becomes  

Revision as of 12:30, 14 March 2017

Evaluate the indefinite and definite 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 x^2\sqrt{1+x^3}~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{\pi}{4}}^{\frac{\pi}{2}} \frac{\cos(x)}{\sin^2(x)}~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 \frac{\ln x}{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=\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:  
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=1+x^3.}
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=3x^2dx}   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}{3}=x^2dx.}
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}{3}\int \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 x^2\sqrt{1+x^3}~dx} & = & \displaystyle{\frac{1}{3}\int \sqrt{u}~du}\\ &&\\ & = & \displaystyle{\frac{2}{9}u^{\frac{3}{2}}+C}\\ &&\\ & = & \displaystyle{\frac{2}{9}(1+x^3)^{\frac{3}{2}}+C.} \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=\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.}
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=\sin(x),} 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=\sin\bigg(\frac{\pi}{4}\bigg)=\frac{\sqrt{2}}{2}}   and   
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 \int_{\frac{\sqrt{2}}{2}}^1 \frac{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 _{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\cos(x)}{\sin^2(x)}~dx} & = & \displaystyle{\int_{\frac{\sqrt{2}}{2}}^1 \frac{1}{u^2}~du}\\ &&\\ & = & \displaystyle{\left.\frac{-1}{u}\right|_{\frac{\sqrt{2}}{2}}^1}\\ &&\\ & = & \displaystyle{-\frac{1}{1}-\frac{-1}{\frac{\sqrt{2}}{2}}}\\ &&\\ & = & \displaystyle{-1+\sqrt{2}.} \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{2}{9}(1+x^3)^{\frac{3}{2}}+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 -1+\sqrt{2}}

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