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

Revision as of 14:08, 31 January 2016

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

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

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

Solution:

Step 1:
Using the formula given in the Foundations sections, we have:
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 $u=1+x^{2}$ . Then, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle du=2xdx} and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {du}{2}}=xdx} . Also, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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}}

@@ Line 7: / Line 7: @@
 !Foundations: &nbsp;
 |-
-|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>.
 |}
@@ Line 16: / Line 16: @@
 |Using the formula given in the Foundations sections, we have:
 |-
-|<math>f_{\text{avg}}=\frac{1}{2-0}\int_0^2 2x^3(1+x^2)^4dx=\int_0^2 x^3(1+x^2)^4dx</math>
+|<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>
 |}
@@ Line 26: / Line 26: @@
 |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>.
 |-
-|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^4du=\frac{1}{2}\int_1^5(u^5-u^4)du</math>.
+|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>.
 |}