Difference between revisions of "009B Sample Midterm 1, Problem 2"

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::<math>f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)~dx</math>.
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::<math>f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)~dx.</math>
 
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::<math style="vertical-align: 0px">f_{\text{avg}}=\frac{1}{2-0}\int_0^2 2x^3(1+x^2)^4~dx=\int_0^2 x^3(1+x^2)^4~dx.</math>  
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::<math>\begin{array}{rcl}
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\displaystyle{f_{\text{avg}}} & = & \displaystyle{\frac{1}{2-0}\int_0^2 2x^3(1+x^2)^4~dx}\\
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&&\\
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& = & \displaystyle{\int_0^2 x^3(1+x^2)^4~dx.}\\
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\end{array}</math>
 
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
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|Now, we use <math>u</math>-substitution. Let <math style="vertical-align: -2px">u=1+x^2</math>. Then, <math style="vertical-align: 0px">du=2x dx</math> and <math style="vertical-align: -13px">\frac{du}{2}=xdx</math>. Also, <math style="vertical-align: 0px">x^2=u-1</math>.
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|Now, we use <math>u</math>-substitution. Let <math style="vertical-align: -2px">u=1+x^2.</math> Then, <math style="vertical-align: 0px">du=2x dx</math> and <math style="vertical-align: -13px">\frac{du}{2}=xdx.</math> Also, <math style="vertical-align: 0px">x^2=u-1.</math>
 
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|We need to change the bounds on the integral. We have <math style="vertical-align: -4px">u_1=1+0^2=1</math> and <math style="vertical-align: -3px">u_2=1+2^2=5</math>.
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|We need to change the bounds on the integral. We have <math style="vertical-align: -4px">u_1=1+0^2=1</math> and <math style="vertical-align: -3px">u_2=1+2^2=5.</math>
 
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|So, the integral becomes <math style="vertical-align: -19px">f_{\text{avg}}=\int_0^2 x\cdot x^2 (1+x^2)^4~dx=\frac{1}{2}\int_1^5(u-1)u^4~du=\frac{1}{2}\int_1^5(u^5-u^4)~du</math>.
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|So, the integral becomes  
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::<math>\begin{array}{rcl}
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\displaystyle{f_{\text{avg}}} & = & \displaystyle{\int_0^2 x\cdot x^2 (1+x^2)^4~dx}\\
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&&\\
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& = & \displaystyle{\frac{1}{2}\int_1^5(u-1)u^4~du}\\
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&&\\
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& = & \displaystyle{\frac{1}{2}\int_1^5(u^5-u^4)~du.}\\
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\end{array}</math>
 
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|We integrate to get
 
|We integrate to get
 
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| &nbsp; &nbsp; <math>f_{\text{avg}}=\left.\frac{u^6}{12}-\frac{u^5}{10}\right|_{1}^5=\left.u^5\bigg(\frac{u}{12}-\frac{1}{10}\bigg)\right|_{1}^5.</math>
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::<math>\begin{array}{rcl}
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\displaystyle{f_{\text{avg}}} & = & \displaystyle{\left.\frac{u^6}{12}-\frac{u^5}{10}\right|_{1}^5}\\
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&&\\
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& = & \displaystyle{\left.u^5\bigg(\frac{u}{12}-\frac{1}{10}\bigg)\right|_{1}^5.}\\
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\end{array}</math>
 
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|We evaluate to get
 
|We evaluate to get
 
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| &nbsp; &nbsp; <math style="vertical-align: -20px">f_{\text{avg}}=5^5\bigg(\frac{5}{12}-\frac{1}{10}\bigg)-1^5\bigg(\frac{1}{12}-\frac{1}{10}\bigg)=3125\bigg(\frac{19}{60}\bigg)-\frac{-1}{60}=\frac{59376}{60}=\frac{4948}{5}</math>.
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::<math>\begin{array}{rcl}
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\displaystyle{f_{\text{avg}}} & = & \displaystyle{5^5\bigg(\frac{5}{12}-\frac{1}{10}\bigg)-1^5\bigg(\frac{1}{12}-\frac{1}{10}\bigg)}\\
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&&\\
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& = & \displaystyle{3125\bigg(\frac{19}{60}\bigg)-\frac{-1}{60}}\\
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&&\\
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& = & \displaystyle{\frac{59376}{60}}\\
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&&\\
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& = & \displaystyle{\frac{4948}{5}.}\\
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\end{array}</math>
 
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Revision as of 12:58, 18 April 2016

Find the average value of the function on the given interval.

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 f(x)=2x^3(1+x^2)^4,~~~[0,2]}


Foundations:  
The average value of a function 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 f(x)} on an interval 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,b]} is given 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 f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)~dx.}

Solution:

Step 1:  
Using the formula given in Foundations, 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{f_{\text{avg}}} & = & \displaystyle{\frac{1}{2-0}\int_0^2 2x^3(1+x^2)^4~dx}\\ &&\\ & = & \displaystyle{\int_0^2 x^3(1+x^2)^4~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=1+x^2.} 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 dx} 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}{2}=xdx.} 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 x^2=u-1.}
We need to change the bounds on the integral. We have Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u_{1}=1+0^{2}=1} 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+2^2=5.}
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{f_{\text{avg}}} & = & \displaystyle{\int_0^2 x\cdot x^2 (1+x^2)^4~dx}\\ &&\\ & = & \displaystyle{\frac{1}{2}\int_1^5(u-1)u^4~du}\\ &&\\ & = & \displaystyle{\frac{1}{2}\int_1^5(u^5-u^4)~du.}\\ \end{array}}
Step 3:  
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{f_{\text{avg}}} & = & \displaystyle{\left.\frac{u^6}{12}-\frac{u^5}{10}\right|_{1}^5}\\ &&\\ & = & \displaystyle{\left.u^5\bigg(\frac{u}{12}-\frac{1}{10}\bigg)\right|_{1}^5.}\\ \end{array}}
Step 4:  
We 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{f_{\text{avg}}} & = & \displaystyle{5^5\bigg(\frac{5}{12}-\frac{1}{10}\bigg)-1^5\bigg(\frac{1}{12}-\frac{1}{10}\bigg)}\\ &&\\ & = & \displaystyle{3125\bigg(\frac{19}{60}\bigg)-\frac{-1}{60}}\\ &&\\ & = & \displaystyle{\frac{59376}{60}}\\ &&\\ & = & \displaystyle{\frac{4948}{5}.}\\ \end{array}}
Final Answer:  
    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{4948}{5}}

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