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

From Math Wiki
Jump to navigation Jump to search
 
(One intermediate revision by the same user not shown)
Line 1: Line 1:
−
<span class="exam"> This problem has three parts:
+
<span class="exam"> Evaluate
  
−
::<span class="exam">a) State the Fundamental Theorem of Calculus.
+
<span class="exam">(a) &nbsp; <math style="vertical-align: -14px">\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt</math>
  
−
::<span class="exam">b) Compute &thinsp; <math style="vertical-align: -15px">\frac{d}{dx}\int_0^{\cos (x)}\sin (t)~dt</math>.
+
<span class="exam">(b) &nbsp; <math style="vertical-align: -14px">\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx</math>
  
−
::<span class="exam">c) Evaluate <math style="vertical-align: -14px">\int_{0}^{\pi/4}\sec^2 x~dx</math>.
+
<hr>
 +
[[009B Sample Midterm 2, Problem 2 Solution|'''<u>Solution</u>''']]
  
  
−
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+
[[009B Sample Midterm 2, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']]
−
!Foundations: &nbsp;
 
−
|-
 
−
|'''1.''' What does Part 1 of the Fundamental Theorem of Calculus say about&nbsp; <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&nbsp; <math style="vertical-align: -15px">\frac{d}{dx}\int_0^x\sin(t)~dt=\sin(x).</math>
 
−
|-
 
−
|'''2.''' 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>
 
−
|}
 
  
−
'''Solution:'''
 
  
−
'''(a)'''
 
−
 
−
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
−
!Step 1: &nbsp;
 
−
|-
 
−
|The Fundamental Theorem of Calculus has two parts.
 
−
|-
 
−
|'''The Fundamental Theorem of Calculus, Part 1'''
 
−
|-
 
−
|
 
−
:Let <math style="vertical-align: -4px">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: &nbsp;
 
−
|-
 
−
|'''The Fundamental Theorem of Calculus, Part 2'''
 
−
|-
 
−
|
 
−
:Let <math style="vertical-align: -4px">f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: 0px">F</math> be any antiderivative of <math style="vertical-align: -4px">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: &nbsp;
 
−
|-
 
−
|Let <math style="vertical-align: -15px">F(x)=\int_0^{\cos (x)}\sin (t)~dt.</math> The problem is asking us to find <math style="vertical-align: -5px">F'(x).</math>
 
−
|-
 
−
|Let <math style="vertical-align: -5px">g(x)=\cos(x)</math> and <math style="vertical-align: -14px">G(x)=\int_0^x \sin(t)~dt.</math>
 
−
|-
 
−
|Then, <math style="vertical-align: -5px">F(x)=G(g(x)).</math>
 
−
|}
 
−
 
−
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
−
!Step 2: &nbsp;
 
−
|-
 
−
|If we take the derivative of both sides of the last equation, we get <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> and <math style="vertical-align: -5px">G'(x)=\sin(x)</math> by the '''Fundamental Theorem of Calculus, Part 1'''.
 
−
|-
 
−
|Since <math style="vertical-align: -6px">G'(g(x))=\sin(g(x))=\sin(\cos(x)),</math> we have
 
−
|-
 
−
|
 
−
::<math style="vertical-align: -5px">F'(x)=G'(g(x))\cdot g'(x)=\sin(\cos(x))\cdot(-\sin(x)).</math>
 
−
|}
 
−
 
−
'''(c)'''
 
−
 
−
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
−
!Step 1: &nbsp;
 
−
|-
 
−
| Using the '''Fundamental Theorem of Calculus, Part 2''', we have
 
−
|-
 
−
|
 
−
::<math>\int_{0}^{\frac{\pi}{4}}\sec^2 x~dx=\tan(x)\biggr|_{0}^{\pi/4}.</math>
 
−
|}
 
−
 
−
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
−
!Step 2: &nbsp;
 
−
|-
 
−
|So, we get
 
−
|-
 
−
|
 
−
::<math style="vertical-align: -16px">\int_{0}^{\frac{\pi}{4}}\sec^2 x~dx=\tan \bigg(\frac{\pi}{4}\bigg)-\tan (0)=1.</math>
 
−
|}
 
−
 
−
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
−
!Final Answer: &nbsp;
 
−
|-
 
−
|&nbsp;&nbsp; '''(a)'''
 
−
|-
 
−
|&nbsp;&nbsp; '''The Fundamental Theorem of Calculus, Part 1'''
 
−
|-
 
−
|&nbsp;&nbsp; Let <math style="vertical-align: -4px">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: 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>
 
−
|-
 
−
|&nbsp;&nbsp; '''The Fundamental Theorem of Calculus, Part 2'''
 
−
|-
 
−
|&nbsp;&nbsp; Let <math style="vertical-align: -4px">f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: 0px">F</math> be any antiderivative of <math style="vertical-align: -4px">f.</math>
 
−
|-
 
−
|&nbsp;&nbsp; Then, <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a).</math>
 
−
|-
 
−
|&nbsp;&nbsp; '''(b)''' &nbsp; <math style="vertical-align: -15px">\frac{d}{dx}\int_0^{\cos (x)}\sin (t)~dt\,=\,\sin(\cos(x))\cdot(-\sin(x))</math>
 
−
|-
 
−
|&nbsp;&nbsp; '''(c)''' <math style="vertical-align: -14px">\int_{0}^{\pi/4}\sec^2 x~dx\,=\,1</math>
 
−
|}
 
 
[[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 17:09, 12 November 2017

Evaluate

(a)  

(b)  

Solution


Detailed Solution


Return to Sample Exam

Retrieved from "https://wiki.math.ucr.edu/index.php?title=009B_Sample_Midterm_2,_Problem_2&oldid=1897"