Difference between revisions of "009B Sample Midterm 3, Problem 2"

Latest revision as of 17:34, 12 November 2017

State the fundamental theorem of calculus, and use this theorem to find the derivative of

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(x)=\int_{\cos (x)}^5 \frac{1}{1+u^{10}}~du.}

Solution

Detailed Solution

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@@ Line 3: / Line 3: @@
 ::<math>F(x)=\int_{\cos (x)}^5 \frac{1}{1+u^{10}}~du.</math>
+<hr>
+[[009B Sample Midterm 3, Problem 2 Solution|'''<u>Solution</u>''']]
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Foundations: &nbsp;
-|-
-|What does Part 1 of the Fundamental Theorem of Calculus say is the derivative of <math style="vertical-align: -16px">G(x)=\int_x^5 \frac{1}{1+u^{10}}~du?</math>
-|-
-|
-::First, we need to switch the bounds of integration.
-|-
-|
-::So, we have <math style="vertical-align: -16px">G(x)=-\int_5^x \frac{1}{1+u^{10}}~du.</math>
-|-
-|
-::By Part 1 of the Fundamental Theorem of Calculus, <math style="vertical-align: -16px">G'(x)=-\frac{1}{1+x^{10}}.</math>
-|}
-'''Solution:'''
+[[009B Sample Midterm 3, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']]
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|'''The Fundamental Theorem of Calculus, Part 1'''
-|-
-|
-: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>
-|-
-|
-: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>
-|-
-|'''The Fundamental Theorem of Calculus, Part 2'''
-|-
-|
-: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, <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a).</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|First, we have
-|-
-|
-::<math style="vertical-align: -15px">F(x)=-\int_5^{\cos (x)} \frac{1}{1+u^{10}}~du.</math>
-|-
-|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>
-|-
-|So,
-|-
-|
-::<math style="vertical-align: -5px">F(x)=-G(g(x)).</math>
-|-
-|Hence, <math style="vertical-align: -5px">F'(x)=-G'(g(x))g'(x)</math> by the Chain Rule.
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 3: &nbsp;
-|-
-|Now, <math style="vertical-align: -5px">g'(x)=-\sin(x).</math>
-|-
-|By the Fundamental Theorem of Calculus,
-|-
-|
-::<math style="vertical-align: -15px">G'(x)=\frac{1}{1+x^{10}}.</math>
-|-
-|Hence,
-|-
-|
-::<math>\begin{array}{rcl}
-\displaystyle{F'(x)} & = & \displaystyle{-\frac{1}{1+\cos^{10}x}(-\sin(x))}\\
-&&\\
-& = & \displaystyle{\frac{\sin(x)}{1+\cos^{10}x}.}\\
-\end{array}</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Final Answer: &nbsp;
-|-
-|&nbsp;&nbsp; '''The Fundamental Theorem of Calculus, Part 1'''
-|-
-|
-:&nbsp;&nbsp; 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>
-|-
-|
-:&nbsp;&nbsp; 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>
-|-
-|&nbsp;&nbsp; '''The Fundamental Theorem of Calculus, Part 2'''
-|-
-|
-:&nbsp;&nbsp; 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>
-|-
-|
-:&nbsp;&nbsp; Then, <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a).</math>
-|-
-|&nbsp;&nbsp; <math>F'(x)=\frac{\sin(x)}{1+\cos^{10}x}</math>
-|}
 [[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

Difference between revisions of "009B Sample Midterm 3, Problem 2"

Latest revision as of 17:34, 12 November 2017

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