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| | ::<math>F(x)=\int_{\cos (x)}^5 \frac{1}{1+u^{10}}~du.</math> | | ::<math>F(x)=\int_{\cos (x)}^5 \frac{1}{1+u^{10}}~du.</math> |
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| | + | [[009B Sample Midterm 3, 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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| − | |What does Part 1 of the Fundamental Theorem of Calculus
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| − | |say is the derivative of <math style="vertical-align: -16px">G(x)=\int_x^5 \frac{1}{1+u^{10}}~du?</math>
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| − | First, we need to switch the bounds of integration.
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| − | So, we have <math style="vertical-align: -16px">G(x)=-\int_5^x \frac{1}{1+u^{10}}~du.</math>
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| − | By Part 1 of the Fundamental Theorem of Calculus, <math style="vertical-align: -16px">G'(x)=-\frac{1}{1+x^{10}}.</math>
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| − | |}
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| | + | [[009B Sample Midterm 3, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | '''Solution:'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |'''The Fundamental Theorem of Calculus, Part 1'''
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| − | Let <math style="vertical-align: -5px">f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: -14px">F(x)=\int_a^x f(t)~dt.</math>
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| − | Then, <math style="vertical-align: -1px">F</math> is a differentiable function on <math style="vertical-align: -5px">(a,b)</math> and <math style="vertical-align: -5px">F'(x)=f(x).</math>
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| − | |'''The Fundamental Theorem of Calculus, Part 2'''
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| − | Let <math style="vertical-align: -5px">f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: -1px">F</math> be any antiderivative of <math style="vertical-align: -5px">f.</math> Then,
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| − | <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a).</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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| − | |First,
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| − | | <math style="vertical-align: -15px">F(x)=-\int_5^{\cos (x)} \frac{1}{1+u^{10}}~du.</math>
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| − | |Now, let <math style="vertical-align: -5px">g(x)=\cos(x)</math> and <math style="vertical-align: -15px">G(x)=\int_5^x \frac{1}{1+u^{10}}~du.</math>
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| − | |Therefore,
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| − | <math style="vertical-align: -5px">F(x)=-G(g(x)).</math>
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| − | |Hence,
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| − | | <math style="vertical-align: -5px">F'(x)=-G'(g(x))g'(x)</math>
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| − | |by the Chain Rule.
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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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| − | |Now,
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| − | | <math style="vertical-align: -5px">g'(x)=-\sin(x).</math>
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| − | |By the Fundamental Theorem of Calculus,
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| − | <math style="vertical-align: -15px">G'(x)=\frac{1}{1+x^{10}}.</math>
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| − | |Hence,
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{F'(x)} & = & \displaystyle{-\frac{1}{1+\cos^{10}x}(-\sin(x))}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{\sin(x)}{1+\cos^{10}x}.}\\
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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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| − | | See Step 1 above
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| − | | <math>F'(x)=\frac{\sin(x)}{1+\cos^{10}x}</math>
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| − | |}
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| | [[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']] | | [[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']] |
