009B Sample Final 2, Problem 6 - Revision history

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Kayla Murray: Created page with "Consider the region bounded by the following two functions: :::::::: $y=2(-x^2+9)$ and
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Created page with "Consider the region bounded by the following two functions: :::::::: <math style="vertical-align: -5px">y=2(-x^2+9)</math> and <math styl..."

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Consider the region bounded by the following two functions:
:::::::: <math style="vertical-align: -5px">y=2(-x^2+9)</math> and <math style="vertical-align: -4px">y=0</math>.

a) Using the lower sum with three rectangles having equal width, approximate the area.

b) Using the upper sum with three rectangles having equal width, approximate the area.

c) Find the actual area of the region.

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'''Solution:'''

'''(a)'''

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'''(b)'''

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'''(c)'''

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[[009B_Sample_Final_2|'''Return to Sample Exam''']]

@@ Line 14: / Line 14: @@
 |&nbsp; &nbsp; &nbsp; &nbsp;The correct substitution is &nbsp;<math style="vertical-align: -1px">x=4\sec^2\theta.</math>
 |-
-|'''2.''' We have the Pythagorean identity
+|'''2.''' Recall the Pythagorean identity
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">\cos^2(x)=1-\sin^2(x).</math>
@@ Line 56: / Line 56: @@
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\int \frac{1}{x^2\sqrt{x^2-16}}~dx} & = & \displaystyle{\frac{1}{16}\cos\theta~d\theta}\\
+\displaystyle{\int \frac{1}{x^2\sqrt{x^2-16}}~dx} & = & \displaystyle{\int \frac{1}{16}\cos\theta~d\theta}\\
 &&\\
 & = & \displaystyle{\frac{1}{16}\sin \theta +C}\\
@@ Line 89: / Line 89: @@
 |Then, we have
 |-
-|&nbsp;<math style="vertical-align: -5px">u_1=\sin(-\pi)=0</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">u_2=\sin(\pi)=0.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">u_1=\sin(-\pi)=0</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">u_2=\sin(\pi)=0.</math>
 |-
 |So, we have
@@ Line 148: / Line 148: @@
 |Since these are definite integrals, we need to change the bounds of integration.
 |-
-|We have &nbsp;<math style="vertical-align: -4px">u_1=0+1=1</math>&nbsp; and &nbsp;<math style="vertical-align: -4px">u_2=1+1=2.</math>
+|We have
 |-
-|Also, &nbsp;<math style="vertical-align: -4px">t_1=0+5=5</math>&nbsp; and &nbsp;<math style="vertical-align: -4px">t_2=1+5=6.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -4px">u_1=0+1=1</math>&nbsp; and &nbsp;<math style="vertical-align: -4px">u_2=1+1=2.</math>
 |-
 |Therefore, we get

@@ Line 12: / Line 12: @@
 |'''1.''' For &nbsp;<math>\int \frac{dx}{x^2\sqrt{x^2-16}},</math>&nbsp; what would be the correct trig substitution?
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;The correct substitution is &nbsp;<math>x=4\sec^2\theta.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;The correct substitution is &nbsp;<math style="vertical-align: -1px">x=4\sec^2\theta.</math>
 |-
 |'''2.''' We have the Pythagorean identity
@@ Line 35: / Line 35: @@
 |We start by using trig substitution.
 |-
-|Let &nbsp;<math>x=4\sec \theta.</math>
+|Let &nbsp;<math style="vertical-align: -1px">x=4\sec \theta.</math>
 |-
-|Then, &nbsp;<math>dx=4\sec \theta \tan \theta ~d\theta.</math>
+|Then, &nbsp;<math style="vertical-align: -2px">dx=4\sec \theta \tan \theta ~d\theta.</math>
 |-
 |So, the integral becomes
@@ Line 81: / Line 81: @@
 !Step 2: &nbsp;
 |-
-|Now, we use &nbsp;<math>u</math>-substitution.
+|Now, we use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 |-
-|Let &nbsp;<math>u=\sin x.</math>&nbsp; Then, &nbsp;<math>du=\cos x ~dx.</math>
+|Let &nbsp;<math style="vertical-align: -1px">u=\sin x.</math>&nbsp; Then, &nbsp;<math style="vertical-align: -1px">du=\cos x ~dx.</math>
 |-
 |Since this is a definite integral, we need to change the bounds of integration.
@@ Line 89: / Line 89: @@
 |Then, we have
 |-
-|&nbsp;<math>u_1=\sin(-\pi)=0</math>&nbsp; and &nbsp;<math>u_2=\sin(\pi)=0.</math>
+|&nbsp;<math style="vertical-align: -5px">u_1=\sin(-\pi)=0</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">u_2=\sin(\pi)=0.</math>
 |-
 |So, we have
@@ Line 113: / Line 113: @@
 |&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{x-3}{(x+1)(x+5)}=\frac{A}{x+1}+\frac{B}{x+5}.</math>
 |-
-|If we multiply both sides of this equation by &nbsp;<math>(x+1)(x+5),</math>&nbsp; we get
+|If we multiply both sides of this equation by &nbsp;<math style="vertical-align: -5px">(x+1)(x+5),</math>&nbsp; we get
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp;<math>x-3=A(x+5)+B(x+1).</math>
 |-
-|If we  let &nbsp;<math>x=-1,</math>&nbsp; we get &nbsp;<math>A=-1.</math>
+|If we  let &nbsp;<math style="vertical-align: -4px">x=-1,</math>&nbsp; we get &nbsp;<math style="vertical-align: -1px">A=-1.</math>
 |-
-|If we  let &nbsp;<math>x=-5,</math>&nbsp; we get &nbsp;<math>B=2.</math>
+|If we  let &nbsp;<math style="vertical-align: -4px">x=-5,</math>&nbsp; we get &nbsp;<math style="vertical-align: -1px">B=2.</math>
 |-
 |So, we have
@@ Line 140: / Line 140: @@
 \end{array}</math>
 |-
-|Now, we use &nbsp;<math>u</math>-substitution for both of these integrals.
+|Now, we use &nbsp;<math style="vertical-align: 0px">u</math>-substitution for both of these integrals.
 |-
-|Let &nbsp;<math>u=x+1.</math>&nbsp; Then, &nbsp;<math>du=dx.</math>
+|Let &nbsp;<math style="vertical-align: -2px">u=x+1.</math>&nbsp; Then, &nbsp;<math style="vertical-align: -1px">du=dx.</math>
 |-
-|Let &nbsp;<math>t=x+5.</math>&nbsp; Then, &nbsp;<math>dt=dx.</math>
+|Let &nbsp;<math style="vertical-align: -3px">t=x+5.</math>&nbsp; Then, &nbsp;<math style="vertical-align: -1px">dt=dx.</math>
 |-
 |Since these are definite integrals, we need to change the bounds of integration.
 |-
-|We have &nbsp;<math>u_1=0+1=1</math>&nbsp; and &nbsp;<math>u_2=1+1=2.</math>
+|We have &nbsp;<math style="vertical-align: -4px">u_1=0+1=1</math>&nbsp; and &nbsp;<math style="vertical-align: -4px">u_2=1+1=2.</math>
 |-
-|Also, &nbsp;<math>t_1=0+5=5</math>&nbsp; and &nbsp;<math>u_2=1+5=6.</math>
+|Also, &nbsp;<math style="vertical-align: -4px">t_1=0+5=5</math>&nbsp; and &nbsp;<math style="vertical-align: -4px">t_2=1+5=6.</math>
 |-
 |Therefore, we get

@@ Line 37: / Line 37: @@
 |Let &nbsp;<math>x=4\sec \theta.</math>
 |-
-|Then, &nbsp;<math>dx=4\sec \theta \tan \theta d\theta.</math>
+|Then, &nbsp;<math>dx=4\sec \theta \tan \theta ~d\theta.</math>
 |-
 |So, the integral becomes
@@ Line 83: / Line 83: @@
 |Now, we use &nbsp;<math>u</math>-substitution.
 |-
-|Let &nbsp;<math>u=\sin x.</math>&nbsp; Then, &nbsp;<math>du=\cos x dx.</math>
+|Let &nbsp;<math>u=\sin x.</math>&nbsp; Then, &nbsp;<math>du=\cos x ~dx.</math>
 |-
 |Since this is a definite integral, we need to change the bounds of integration.
@@ Line 113: / Line 113: @@
 |&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{x-3}{(x+1)(x+5)}=\frac{A}{x+1}+\frac{B}{x+5}.</math>
 |-
-|If we multiply both sides of this equation by &nbsp;<math>(x+1)(x+5),</math> we get
+|If we multiply both sides of this equation by &nbsp;<math>(x+1)(x+5),</math>&nbsp; we get
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp;<math>x-3=A(x+5)+B(x+1).</math>
 |-
-|If we  let &nbsp;<math>x=-1,</math> we get <math>A=-1.</math>
+|If we  let &nbsp;<math>x=-1,</math>&nbsp; we get &nbsp;<math>A=-1.</math>
 |-
-|If we  let &nbsp;<math>x=-5,</math> we get <math>B=2.</math>
+|If we  let &nbsp;<math>x=-5,</math>&nbsp; we get &nbsp;<math>B=2.</math>
 |-
 |So, we have

@@ Line 27: / Line 27: @@
 !Step 1: &nbsp;
 |-
+|We start by using trig substitution.
 |-
+|Then, &nbsp;<math>dx=4\sec \theta \tan \theta d\theta.</math>
 |-
+|So, the integral becomes
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |}
@@ Line 39: / Line 47: @@
 !Step 2: &nbsp;
 |-
+|Now, we integrate to get
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-|-
+\displaystyle{\int \frac{1}{x^2\sqrt{x^2-16}}~dx} & = & \displaystyle{\frac{1}{16}\cos\theta~d\theta}\\
+&&\\
 |}
@@ Line 53: / Line 63: @@
 !Step 1: &nbsp;
 |-
+|First, we write
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |}
@@ Line 61: / Line 75: @@
 !Step 2: &nbsp;
 |-
+|Now, we use &nbsp;<math>u</math>-substitution.
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |}
@@ Line 71: / Line 99: @@
 !Step 1: &nbsp;
 |-
+|First, we write
 |-
+|
@@ Line 78: / Line 124: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Step 2: &nbsp;
 |-
+|
 |-
+|Now, we use &nbsp;<math>u</math>-substitution for both of these integrals.
 |}
@@ Line 88: / Line 161: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)'''
+|&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp; <math>\frac{1}{16}\bigg(\frac{\sqrt{x^2-16}}{x}\bigg)+C</math>
 |-
-|'''(b)'''
+|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp; <math>0</math>
 |-
-|'''(c)'''
+|&nbsp; &nbsp;'''(c)'''&nbsp; &nbsp; <math>-\ln(2)+2\ln(6)-2\ln(5)</math>
 |}
 [[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
 <span class="exam"> Evaluate the following integrals:
-<span class="exam">(a) <math>\int \frac{dx}{x^2\sqrt{x^2-16}}</math>
+<span class="exam">(a) &nbsp;<math>\int \frac{dx}{x^2\sqrt{x^2-16}}</math>
-<span class="exam">(b) <math>\int_{-\pi}^\pi \sin^3x\cos^3x~dx</math>
+<span class="exam">(b) &nbsp;<math>\int_{-\pi}^\pi \sin^3x\cos^3x~dx</math>
-<span class="exam">(c) <math>\int_0^1 \frac{x-3}{x^2+6x+5}~dx</math>
+<span class="exam">(c) &nbsp;<math>\int_0^1 \frac{x-3}{x^2+6x+5}~dx</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 1: / Line 1: @@
-<span class="exam">Consider the region bounded by the following two functions:
+<span class="exam"> Evaluate the following integrals:
-<span class="exam">a) Using the lower sum with three rectangles having equal width, approximate the area.
+::<span class="exam">a) <math>\int \frac{dx}{x^2\sqrt{x^2-16}}</math>
-<span class="exam">b) Using the upper sum with three rectangles having equal width, approximate the area.
+::<span class="exam">b) <math>\int_{-\pi}^\pi \sin^3x\cos^3x~dx</math>
-<span class="exam">c) Find the actual area of the region.
+::<span class="exam">c) <math>\int_0^1 \frac{x-3}{x^2+6x+5}~dx</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"