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

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&nbsp; &nbsp; You could use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -5px">u=\ln(x).</math> Then, <math style="vertical-align: -13px">du=\frac{1}{x}dx.</math>
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&nbsp; &nbsp; &nbsp; &nbsp; You could use <math style="vertical-align: 0px">u</math>-substitution.  
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|&nbsp; &nbsp; &nbsp; &nbsp; Let <math style="vertical-align: -5px">u=\ln(x).</math>
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|&nbsp; &nbsp; &nbsp; &nbsp; Then, <math style="vertical-align: -13px">du=\frac{1}{x}dx.</math>
 
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&nbsp; &nbsp; Thus, <math style="vertical-align: -12px">\int \frac{\ln x}{x}~dx=\int u~du=\frac{u^2}{2}+C=\frac{(\ln x)^2}{2}+C.</math>
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&nbsp; &nbsp; &nbsp; &nbsp; Thus,  
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&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
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\displaystyle{\int \frac{\ln x}{x}~dx} & = & \displaystyle{\int u~du}\\
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&&\\
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& = & \displaystyle{\frac{u^2}{2}+C}\\
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&&\\
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& = & \displaystyle{\frac{(\ln x)^2}{2}+C.}
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\end{array}</math>
 
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|Therefore, the integral becomes  
 
|Therefore, the integral becomes  
 
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|&nbsp; &nbsp; <math style="vertical-align: -13px">\frac{1}{3}\int \sqrt{u}~du.</math>
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|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -13px">\frac{1}{3}\int \sqrt{u}~du.</math>
 
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
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|We now have:
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|We now have
 
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|&nbsp; &nbsp; <math>\begin{array}{rcl}
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
\displaystyle{\int x^2\sqrt{1+x^3}~dx} & = & \displaystyle{\frac{1}{3}\int \sqrt{u}~du}\\
 
\displaystyle{\int x^2\sqrt{1+x^3}~dx} & = & \displaystyle{\frac{1}{3}\int \sqrt{u}~du}\\
 
&&\\
 
&&\\
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
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|Again, we need to use <math>u</math>-substitution. Let <math style="vertical-align: -5px">u=\sin(x).</math> Then, <math style="vertical-align: -5px">du=\cos(x)dx.</math>  
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|We need to use <math>u</math>-substitution.  
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|-
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|Let <math style="vertical-align: -5px">u=\sin(x).</math>  
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|Then, <math style="vertical-align: -5px">du=\cos(x)dx.</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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|Therefore, the integral becomes  
 
|Therefore, the integral becomes  
 
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|&nbsp; &nbsp; <math style="vertical-align: -19px">\int_{\frac{\sqrt{2}}{2}}^1 \frac{1}{u^2}~du.</math>
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|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -19px">\int_{\frac{\sqrt{2}}{2}}^1 \frac{1}{u^2}~du.</math>
 
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&nbsp; &nbsp; <math>\begin{array}{rcl}
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&nbsp; &nbsp; &nbsp; &nbsp; <math>\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{\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}\\
 
&&\\
 
&&\\
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
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|'''(a)''' &nbsp; <math>\frac{2}{9}(1+x^3)^{\frac{3}{2}}+C</math>
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|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>\frac{2}{9}(1+x^3)^{\frac{3}{2}}+C</math>
 
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|'''(b)''' &nbsp; <math>-1+\sqrt{2}</math>
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|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>-1+\sqrt{2}</math>
 
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[[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

Revision as of 15:31, 7 February 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 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:  
We need to 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 need to 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 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\bigg(\frac{\pi}{2}\bigg)=1.}
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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