009B Sample Midterm 3, Problem 2 - Revision history

MathAdmin at 01:34, 13 November 2017

2017-11-13T01:34:08Z

MathAdmin at 19:19, 9 April 2017

2017-04-09T19:19:18Z

MathAdmin: Created page with "State the fundamental theorem of calculus, and use this theorem to find the derivative of :: $F(x)=\int_{\cos (x)}^5 \frac{1}{1+u^{10}}~du.$ {..."

2016-04-19T00:49:27Z

Created page with "<span class="exam">State the fundamental theorem of calculus, and use this theorem to find the derivative of ::<math>F(x)=\int_{\cos (x)}^5 \frac{1}{1+u^{10}}~du.</math> {..."

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<span class="exam">State the fundamental theorem of calculus, and use this theorem to find the derivative of

::<math>F(x)=\int_{\cos (x)}^5 \frac{1}{1+u^{10}}~du.</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|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:'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|'''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:  
|-
|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:  
|-
|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:  
|-
|   '''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>
|-
|   <math>F'(x)=\frac{\sin(x)}{1+\cos^{10}x}</math>
|}
[[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

← Older revision		Revision as of 01:34, 13 November 2017
Line 3:		Line 3:
	::<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>

		+	<hr>
		+	[[009B Sample Midterm 3, Problem 2 Solution\|'''<u>Solution</u>''']]

−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Foundations:  ~~
−	\|-
−	~~\|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>~~
−	\|}

		+	[[009B Sample Midterm 3, Problem 2 Detailed Solution\|'''<u>Detailed Solution</u>''']]

−	~~'''Solution:'''~~
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 1:  ~~
−	\|-
−	~~\|'''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:  ~~
−	\|-
−	~~\|First,~~
−	\|-
−	~~\|        <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>~~
−	\|-
−	~~\|Therefore,~~
−	\|-
−	\|
−	~~        <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:  ~~
−	\|-
−	~~\|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:  ~~
−	\|-
−	~~\|        See Step 1 above~~
−	\|-
−	~~\|       <math>F'(x)=\frac{\sin(x)}{1+\cos^{10}x}</math>~~
−	\|}
	[[009B_Sample_Midterm_3\|'''<u>Return to Sample Exam</u>''']]		[[009B_Sample_Midterm_3\|'''<u>Return to Sample Exam</u>''']]

@@ Line 7: / Line 7: @@
 !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>
+|What does Part 1 of the Fundamental Theorem of Calculus
 |-
+|
-::First, we need to switch the bounds of integration.
+&nbsp;&nbsp;&nbsp;&nbsp; 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>
+&nbsp;&nbsp;&nbsp;&nbsp; So, we have &nbsp;<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>
+&nbsp;&nbsp;&nbsp;&nbsp; By Part 1 of the Fundamental Theorem of Calculus, &nbsp;<math style="vertical-align: -16px">G'(x)=-\frac{1}{1+x^{10}}.</math>
 |}
 '''Solution:'''
@@ Line 26: / Line 29: @@
 |-
+|
-: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>
+Let &nbsp;<math style="vertical-align: -5px">f</math>&nbsp; be continuous on &nbsp;<math style="vertical-align: -5px">[a,b]</math>&nbsp; and let &nbsp;<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>
+Then, &nbsp;<math style="vertical-align: -1px">F</math>&nbsp; is a differentiable function on &nbsp;<math style="vertical-align: -5px">(a,b)</math>&nbsp; and &nbsp;<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>
+Let &nbsp;<math style="vertical-align: -5px">f</math>&nbsp; be continuous on &nbsp;<math style="vertical-align: -5px">[a,b]</math>&nbsp; and let &nbsp;<math style="vertical-align: -1px">F</math>&nbsp; be any antiderivative of &nbsp;<math style="vertical-align: -5px">f.</math> Then,
 |-
+|
-:Then, <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a).</math>
+&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a).</math>
 |}
@@ Line 43: / Line 46: @@
 !Step 2: &nbsp;
 |-
-|First, we have
+|First,
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <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>
+|Now, let &nbsp;<math style="vertical-align: -5px">g(x)=\cos(x)</math>&nbsp; and &nbsp;<math style="vertical-align: -15px">G(x)=\int_5^x \frac{1}{1+u^{10}}~du.</math>
 |-
-|So,
+|Therefore,
 |-
+|
-::<math style="vertical-align: -5px">F(x)=-G(g(x)).</math>
+&nbsp; &nbsp; &nbsp; &nbsp; <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.
+|by the Chain Rule.
 |}
@@ Line 61: / Line 67: @@
 !Step 3: &nbsp;
 |-
-|Now, <math style="vertical-align: -5px">g'(x)=-\sin(x).</math>
+|Now,
 |-
 |By the Fundamental Theorem of Calculus,
 |-
+|
-::<math style="vertical-align: -15px">G'(x)=\frac{1}{1+x^{10}}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -15px">G'(x)=\frac{1}{1+x^{10}}.</math>
 |-
 |Hence,
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 \displaystyle{F'(x)} & = & \displaystyle{-\frac{1}{1+\cos^{10}x}(-\sin(x))}\\
 &&\\
@@ Line 77: / Line 85: @@
 \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; &nbsp; &nbsp; See Step 1 above
 |-
-|&nbsp;&nbsp; <math>F'(x)=\frac{\sin(x)}{1+\cos^{10}x}</math>
+|&nbsp;&nbsp; &nbsp; &nbsp; <math>F'(x)=\frac{\sin(x)}{1+\cos^{10}x}</math>
 |}
 [[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

009B Sample Midterm 3, Problem 2 - Revision history

MathAdmin at 01:34, 13 November 2017

MathAdmin at 19:19, 9 April 2017

MathAdmin: Created page with "State the fundamental theorem of calculus, and use this theorem to find the derivative of ::F(x)=\int_{\cos (x)}^5 \frac{1}{1+u^{10}}~du. {..."

MathAdmin: Created page with "State the fundamental theorem of calculus, and use this theorem to find the derivative of :: $F(x)=\int_{\cos (x)}^5 \frac{1}{1+u^{10}}~du.$ {..."