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| − | <span class="exam">Find the average value of the function on the given interval. | + | <span class="exam">Evaluate the indefinite and definite integrals. |
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| − | ::<math>f(x)=2x^3(1+x^2)^4,~~~[0,2]</math>
| + | <span class="exam">(a) <math>\int x^2\sqrt{1+x^3}~dx</math> |
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| | + | <span class="exam">(b) <math>\int _{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\cos(x)}{\sin^2(x)}~dx</math> |
| | + | <hr> |
| | + | [[009B Sample Midterm 1, Problem 2 Solution|'''<u>Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Foundations:
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| − | |-
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| − | |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
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| − | ::<math>f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)~dx.</math>
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| − | |}
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| − | '''Solution:'''
| + | [[009B Sample Midterm 1, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |Using the formula given in Foundations, we have:
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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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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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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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| − | |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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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 3:
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| − | |-
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| − | |We integrate to get
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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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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 4:
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| − | |We evaluate to get
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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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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | | <math>\frac{4948}{5}</math>
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| − | |}
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| | [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] | | [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] |
