009B Sample Midterm 1, Problem 2 - Revision history

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MathAdmin: Created page with "Find the average value of the function on the given interval. :: $f(x)=2x^3(1+x^2)^4,~~~[0,2]$ {| class="mw-collapsible mw-collapsed" style = "..."

2016-02-01T22:36:32Z

Created page with "<span class="exam">Find the average value of the function on the given interval. ::<math>f(x)=2x^3(1+x^2)^4,~~~[0,2]</math> {| class="mw-collapsible mw-collapsed" style = "..."

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<span class="exam">Find the average value of the function on the given interval.

::<math>f(x)=2x^3(1+x^2)^4,~~~[0,2]</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|The average value of a function <math style="vertical-align: -5px">f(x)</math> on an interval <math style="vertical-align: -5px">[a,b]</math> is given by <math style="vertical-align: -18px">f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)~dx</math>.
|}

'''Solution:'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Using the formula given in Foundations, we have:
|-
|    <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>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|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>.
|-
|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>.
|-
|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>.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|We integrate to get
|-
|     <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>
|-
|
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 4:  
|-
|We evaluate to get
|-
|     <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>.
|-
|
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|     <math>\frac{4948}{5}</math>
|-
|
|}
[[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
-<span class="exam"> Otis Taylor plots the price per share of a stock that he owns as a function of time
+<span class="exam">Evaluate the indefinite and definite integrals.
-<span class="exam">and finds that it can be approximated by the function
+<span class="exam">(a) &nbsp; <math>\int x^2\sqrt{1+x^3}~dx</math>
-::<math>s(t)=t(25-5t)+18</math>
+<span class="exam">(b) &nbsp; <math>\int _{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\cos(x)}{\sin^2(x)}~dx</math>
-<span class="exam">Find the average price of the stock over the first five years.
+[[009B Sample Midterm 1, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
-<span class="exam">Find the average value of the function on the given interval.
+<span class="exam"> Otis Taylor plots the price per share of a stock that he owns as a function of time
-::<math>f(x)=2x^3(1+x^2)^4,~~~[0,2]</math>
+<span class="exam">and finds that it can be approximated by the function
@@ Line 7: / Line 13: @@
 !Foundations: &nbsp;
 |-
-|The average value of a function <math style="vertical-align: -5px">f(x)</math> on an interval <math style="vertical-align: -5px">[a,b]</math> is given by
+|The average value of a function &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; on an interval &nbsp;<math style="vertical-align: -5px">[a,b]</math>&nbsp; is given by
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -18px">f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)~dx.</math>
 |}
 '''Solution:'''
@@ Line 17: / Line 23: @@
 !Step 1: &nbsp;
 |-
-|Using the formula given in Foundations, we have:
+|This problem wants us to find the average value of &nbsp;<math style="vertical-align: -5px">s(t)</math>&nbsp; over the interval &nbsp;<math style="vertical-align: -5px">[0,5].</math>
 |-
+| &nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: 0px">s_{\text{avg}}=\frac{1}{5-0} \int_0^5 t(25-5t)+18~dt.</math>
 |}
@@ Line 30: / Line 33: @@
 !Step 2: &nbsp;
 |-
-|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>
+|First, we distribute to get
 |-
-|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>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>s_{\text{avg}}=\frac{1}{5} \int_0^5 25t-5t^2+18~dt.</math>
 |-
-|So, the integral becomes
+|Then, we integrate to get
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>s_{\text{avg}}=\left. \frac{1}{5}\bigg[\frac{25t^2}{2}-\frac{5t^3}{3}+18t\bigg]\right|_0^5.</math>
 |}
@@ Line 49: / Line 45: @@
 !Step 3: &nbsp;
 |-
-|We integrate to get
+|We now evaluate to get
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-::<math>\begin{array}{rcl}
+\displaystyle{s_{\text{avg}}} & = & \displaystyle{\frac{1}{5}\bigg[\frac{25(5)^2}{2}-\frac{5(5)^3}{3}+18(5)\bigg]-0}\\
-\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.}\\
+& \approx & \displaystyle{$38.83.}
 \end{array}</math>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; <math>\frac{4948}{5}</math>
+| &nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{233}{6}\approx $38.83</math>
 |-
+|
 |}
 [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 10: / Line 10: @@
 |-
+|
-::<math>f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)~dx</math>.
+::<math>f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)~dx.</math>
 |}
@@ Line 20: / Line 20: @@
 |-
+|
-::<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>
+::<math>\begin{array}{rcl}
 |}
@@ Line 26: / Line 30: @@
 !Step 2: &nbsp;
 |-
-|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>.
+|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>
 |-
-|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>.
+|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>
 |-
-|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>.
+|So, the integral becomes
 |}
@@ Line 38: / Line 51: @@
 |We integrate to get
 |-
-| &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>
-|-
+::<math>\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.}\\
 |}
@@ Line 50: / Line 64: @@
 |We evaluate to get
 |-
-| &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>.
-|-
+::<math>\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}}\\
 |}

@@ Line 7: / Line 7: @@
 !Foundations: &nbsp;
 |-
-|The average value of a function <math style="vertical-align: -5px">f(x)</math> on an interval <math style="vertical-align: -5px">[a,b]</math> is given by <math style="vertical-align: -18px">f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)~dx</math>.
+|The average value of a function <math style="vertical-align: -5px">f(x)</math> on an interval <math style="vertical-align: -5px">[a,b]</math> is given by
 |}
@@ Line 16: / Line 19: @@
 |Using the formula given in Foundations, we have:
 |-
-| &nbsp; &nbsp;<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>
 |}