MathAdmin: Created page with "(a) State '''both parts''' of the Fundamental Theorem of Calculus. (b) Evaluate the integral :: $\int_0^1 \frac{d}{dx} \bigg(e^{\ta..."$

2017-04-10T16:47:37Z

Created page with "(a) State '''both parts''' of the Fundamental Theorem of Calculus. (b) Evaluate the integral ::<math>\int_0^1 \frac{d}{dx} \bigg(e^{\ta..."

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(a) State '''both parts''' of the Fundamental Theorem of Calculus.

(b) Evaluate the integral

::<math>\int_0^1 \frac{d}{dx} \bigg(e^{\tan^{-1}(x)}\bigg)dx</math>

(c) Compute

::<math>\frac{d}{dx}\int_1^{\frac{1}{x}} \sin t~dt</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' What does Part 2 of the Fundamental Theorem of Calculus say about  <math style="vertical-align: -15px">\int_a^b\sec^2x~dx</math>  where  <math style="vertical-align: -5px">a,b</math>  are constants?
|-
|
        Part 2 of the Fundamental Theorem of Calculus says that
|-
|        <math style="vertical-align: -15px">\int_a^b\sec^2x~dx=F(b)-F(a)</math>  where  <math style="vertical-align: 0px">F</math>  is any antiderivative of  <math style="vertical-align: 0px">\sec^2x.</math>
|-
|'''2.''' What does Part 1 of the Fundamental Theorem of Calculus say about  <math style="vertical-align: -15px">\frac{d}{dx}\int_0^x\sin(t)~dt?</math>
|-
|
        Part 1 of the Fundamental Theorem of Calculus says that
|-
|        <math style="vertical-align: -15px">\frac{d}{dx}\int_0^x\sin(t)~dt=\sin(x).</math>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|The Fundamental Theorem of Calculus has two parts.
|-
|'''The Fundamental Theorem of Calculus, Part 1'''
|-
|       Let  <math>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: 0px">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>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|'''The Fundamental Theorem of Calculus, Part 2'''
|-
|       Let  <math>f</math>  be continuous on  <math>[a,b]</math>  and let  <math style="vertical-align: 0px">F</math>  be any antiderivative of  <math>f.</math>
|-
|       Then,  <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a).</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|The Fundamental Theorem of Calculus Part 2 says that
|-
|        <math>\int_0^1 \frac{d}{dx}\bigg(e^{\arctan(x)}\bigg)~dx=F(1)-F(0)</math>
|-
|where  <math style="vertical-align: -5px">F(x)</math>  is any antiderivative of  <math style="vertical-align: -15px">\frac{d}{dx}\bigg(e^{\arctan(x)}\bigg).</math>
|-
|Thus, we can take
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|       <math>F(x)=e^{\arctan(x)}</math>
|-
|since then <math style="vertical-align: -15px">F'(x)=\frac{d}{dx}\bigg(e^{\arctan(x)}\bigg).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\int_0^1 \frac{d}{dx}\bigg(e^{\arctan(x)}\bigg)~dx} & = & \displaystyle{F(1)-F(0)}\\
&&\\
& = & \displaystyle{e^{\arctan(1)}-e^{\arctan(0)}}\\
&&\\
& = & \displaystyle{e^{\frac{\pi}{4}}-e^0}\\
&&\\
& = & \displaystyle{e^{\frac{\pi}{4}}-1.}
\end{array}</math>
|}

'''(c)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Using the Fundamental Theorem of Calculus Part 1 and the Chain Rule, we have
|-
|       <math>\frac{d}{dx}\int_1^{\frac{1}{x}} \sin t~dt=\sin\bigg(\frac{1}{x}\bigg)\frac{d}{dx}\bigg(\frac{1}{x}\bigg).</math>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Hence, we have
|-
|       <math>\frac{d}{dx}\int_1^{\frac{1}{x}} \sin t~dt=\sin\bigg(\frac{1}{x}\bigg)\bigg(-\frac{1}{x^2}\bigg).</math>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|   '''(a)'''    See above
|-
|   '''(b)'''    <math>e^{\frac{\pi}{4}}-1</math>
|-
|   '''(c)'''    <math>\sin\bigg(\frac{1}{x}\bigg)\bigg(-\frac{1}{x^2}\bigg)</math>
|}
[[009B_Sample_Final_2|'''Return to Sample Exam''']]

009B Sample Final 2, Problem 1 - Revision history

MathAdmin: Created page with "(a) State '''both parts''' of the Fundamental Theorem of Calculus. (b) Evaluate the integral ::\int_0^1 \frac{d}{dx} \bigg(e^{\ta..."

MathAdmin: Created page with "(a) State '''both parts''' of the Fundamental Theorem of Calculus. (b) Evaluate the integral :: $\int_0^1 \frac{d}{dx} \bigg(e^{\ta..."$