Difference between revisions of "009B Sample Midterm 1, Problem 2"
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!Foundations: | !Foundations: | ||
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| − | |The average value of a function <math>f(x)</math> on an interval <math>[a,b]</math> is given by <math>f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)dx</math>. | + | |The average value of a function <math>f(x)</math> on an interval <math>[a,b]</math> is given by <math>f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)~dx</math>. |
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|Using the formula given in the Foundations sections, we have: | |Using the formula given in the Foundations sections, we have: | ||
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| − | |<math>f_{\text{avg}}=\frac{1}{2-0}\int_0^2 2x^3(1+x^2)^ | + | |<math>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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|We need to change the bounds on the integral. We have <math>u_1=1+0^2=1</math> and <math>u_2=1+2^2=5</math>. | |We need to change the bounds on the integral. We have <math>u_1=1+0^2=1</math> and <math>u_2=1+2^2=5</math>. | ||
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| − | |So, the integral becomes <math>f_{\text{avg}}=\int_0^2 x x^2 (1+x^2)^4 dx=\frac{1}{2}\int_1^5(u-1)u^ | + | |So, the integral becomes <math>f_{\text{avg}}=\int_0^2 x 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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Revision as of 14:08, 31 January 2016
Find the average value of the function on the given interval.
![{\displaystyle f(x)=2x^{3}(1+x^{2})^{4},~~~[0,2]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/77a161ffdc7696e36689931b0c93c832f51cbe78)
| Foundations: |
|---|
| The average value of a function 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 the Foundations sections, 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 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} |
| Step 2: |
|---|
| Now, we use 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 (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=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 f_{\text{avg}}=\int_0^2 x 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} . |
| 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 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} |
| 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 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}} . |
| 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}} |