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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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| − | |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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| − | '''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 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)^4dx=\int_0^2 x^3(1+x^2)^4dx</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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| − | |Since <math>(1+x^2)^2=1+2x^2+x^4</math>, we have
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| − | |<math>(1+x^2)^4=(1+x^2)^2(1+x^2)^2=(1+2x^2+x^4)(1+2x^2+x^4)=1+4x^2+6x^4+4x^6+x^8</math>.
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| − | |So, the integral becomes <math>f_{\text{avg}}=\int_0^2 x^3(1+4x^2+6x^4+4x^6+x^8)dx</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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| − | |We distribute to get
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| − | |<math>f_{\text{avg}}=\int_0^2 x^3(1+4x^2+6x^4+4x^6+x^8)dx=\int_0^2 (x^3+4x^5+6x^7+4x^9+x^{11})dx</math>.
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| − | |Now, we integrate to get
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| − | |<math>f_{\text{avg}}=\left.\frac{x^4}{4}+4\frac{2x^6}{3}+\frac{3x^8}{4}+\frac{2x^{10}}{5}+\frac{x^{12}}{12}\right|_{0}^{2}</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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| − | |}
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| | [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] | | [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] |
