009B Sample Final 1, Problem 4 - Revision history

Kayla Murray at 21:01, 20 May 2017

2017-05-20T21:01:14Z

Kayla Murray at 19:43, 18 March 2017

2017-03-18T19:43:44Z

Kayla Murray at 19:02, 27 February 2017

2017-02-27T19:02:36Z

Kayla Murray at 18:54, 27 February 2017

2017-02-27T18:54:57Z

Kayla Murray at 18:49, 27 February 2017

2017-02-27T18:49:12Z

Kayla Murray at 18:44, 27 February 2017

2017-02-27T18:44:16Z

Kayla Murray at 02:11, 19 February 2017

2017-02-19T02:11:38Z

Grad at 06:58, 26 February 2016

2016-02-26T06:58:20Z

Grad at 06:57, 26 February 2016

2016-02-26T06:57:44Z

Grad: /* 4 */

2016-02-26T06:57:27Z

@@ Line 18: / Line 18: @@
 |'''2.''' Recall the Pythagorean identity
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">\sin^2(x)=1-\cos^2(x).</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">\sin^2(x)+\cos^2(x)=1.</math>
 |}
@@ Line 88: / Line 88: @@
 & = & \displaystyle{\int \frac{2x^2+x}{2x^2+x}~dx+\int\frac{1-x}{2x^2+x}~dx}\\
 &&\\
-& = & \displaystyle{\int ~dx+\int\frac{1-x}{2x^2+x}~dx}.\\
+& = & \displaystyle{\int 1~dx+\int\frac{1-x}{2x^2+x}~dx}.\\
 \end{array}</math>
 |}
@@ Line 123: / Line 123: @@
+|
 &nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
-\displaystyle{\int \frac{2x^2+1}{2x^2+x}~dx} & = & \displaystyle{\int ~dx+\int\frac{1}{x}~dx+\int\frac{-3}{2x+1}~dx}\\
+\displaystyle{\int \frac{2x^2+1}{2x^2+x}~dx} & = & \displaystyle{\int 1~dx+\int\frac{1}{x}~dx+\int\frac{-3}{2x+1}~dx}\\
 &&\\
 & = & \displaystyle{x+\ln x+ \int\frac{-3}{2x+1}~dx.}\\
@@ Line 152: / Line 152: @@
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;<math>\int \frac{2x^2+1}{2x^2+x}~dx\,=\,x+\ln x-\frac{3}{2}\ln (2x+1) +C.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 |}

@@ Line 16: / Line 16: @@
 |&nbsp; &nbsp; &nbsp; &nbsp;for some constants <math style="vertical-align: -4px">A,B.</math>
 |-
-|'''2.''' We have the Pythagorean identity
+|'''2.''' Recall the Pythagorean identity
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">\sin^2(x)=1-\cos^2(x).</math>
@@ Line 36: / Line 36: @@
 |Now, we proceed by &nbsp;<math>u</math>-substitution.
 |-
-|Let &nbsp;<math style="vertical-align: 0px">u=t^3.</math> &nbsp; Then, &nbsp; <math style="vertical-align: 0px">du=3t^2dt</math>&nbsp; and &nbsp;<math style="vertical-align: -14px">\frac{du}{3}=t^2dt.</math>
+|Let &nbsp;<math style="vertical-align: 0px">u=t^3.</math> &nbsp;
 |-
 |So, we have
@@ Line 132: / Line 134: @@
 |For the final remaining integral, we use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 |-
-|Let &nbsp;<math style="vertical-align: -2px">u=2x+1.</math>&nbsp; Then, &nbsp;<math style="vertical-align: 0px">du=2\,dx</math>&nbsp; and &nbsp;<math style="vertical-align: -14px">\frac{du}{2}=dx.</math>
+|Let &nbsp;<math style="vertical-align: -2px">u=2x+1.</math>&nbsp;
 |-
-|Thus, our final integral becomes
+|Thus, our integral becomes
 |-
+|

@@ Line 47: / Line 47: @@
 !Step 2: &nbsp;
 |-
-|Now, for the one remaining integral, we use <math style="vertical-align: 0px">u</math>-substitution.
+|Now, we need to use trig substitution.
 |-
-|Let <math style="vertical-align: 0px">u=e^x</math>. Then, <math style="vertical-align: 0px">du=e^xdx</math>.
+|Let &nbsp;<math style="vertical-align: -1px">u=\sin \theta.</math>&nbsp; Then, &nbsp;<math style="vertical-align: 0px">du=\cos \theta d\theta.</math>
 |-
 |So, we have
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\int e^x(x+\sin(e^x))~dx} & = & \displaystyle{xe^x-e^x+\int \sin(u)~du}\\
+\displaystyle{\int \frac{t^2}{\sqrt{1-t^6}}~dt} & = & \displaystyle{\int \frac{\cos \theta}{3\sqrt{1-\sin^2\theta}}~d\theta}\\
 &&\\
-& = & \displaystyle{xe^x-e^x-\cos(u)+C}\\
+& = & \displaystyle{\frac{1}{3}\arcsin(u)+C}\\
 &&\\
-& = & \displaystyle{xe^x-e^x-\cos(e^x)+C}.\\
+& = & \displaystyle{\frac{1}{3}\arcsin(t^3)+C.}
 \end{array}</math>
 |}
@@ Line 188: / Line 196: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp;<math>xe^x-e^x-\cos(e^x)+C</math>
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp;<math>\frac{1}{3}\arcsin(t^3)+C</math>
 |-
 |&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp;<math style="vertical-align: -14px">x+\ln x-\frac{3}{2}\ln (2x+1) +C</math>

@@ Line 29: / Line 29: @@
 !Step 1: &nbsp;
 |-
-|We first distribute to get
+|We first note that
 |-
+|
-::<math>\int e^x(x+\sin(e^x))~dx\,=\,\int e^xx~dx+\int e^x\sin(e^x)~dx.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\int \frac{t^2}{\sqrt{1-t^6}}~dt=\int \frac{t^2}{\sqrt{1-(t^3)^2}}~dt.</math>
 |-
-|Now, for the first integral on the right hand side of the last equation, we use integration by parts.
+|Now, we proceed by &nbsp;<math>u</math>-substitution.
 |-
-|Let <math style="vertical-align: 0px">u=x</math> and <math style="vertical-align: 0px">dv=e^xdx</math>. Then, <math style="vertical-align: 0px">du=dx</math> and <math style="vertical-align: 0px">v=e^x</math>.
+|Let &nbsp;<math style="vertical-align: 0px">u=t^3.</math> &nbsp; Then, &nbsp; <math style="vertical-align: 0px">du=3t^2dt</math>&nbsp; and &nbsp;<math style="vertical-align: -14px">\frac{du}{3}=t^2dt.</math>
 |-
 |So, we have
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\int \frac{t^2}{\sqrt{1-t^6}}~dt=\int \frac{1}{3\sqrt{1-u^2}}~du.</math>
 |}

@@ Line 14: / Line 14: @@
 |&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -18px">\frac{1}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}</math>
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;for some constants <math style="vertical-align: -4px">A,B</math>.
+|&nbsp; &nbsp; &nbsp; &nbsp;for some constants <math style="vertical-align: -4px">A,B.</math>
 |-
 |'''2.''' We have the Pythagorean identity
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">\sin^2(x)=1-\cos^2(x)</math>.
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">\sin^2(x)=1-\cos^2(x).</math>
 |}
@@ Line 91: / Line 91: @@
 |Now, we need to use partial fraction decomposition for the second integral.
 |-
-|Since &nbsp;<math style="vertical-align: -5px">2x^2+x=x(2x+1),</math>&nbsp; we let &nbsp;<math>\frac{1-x}{2x^2+x}=\frac{A}{x}+\frac{B}{2x+1}.</math>
+|Since &nbsp;<math style="vertical-align: -5px">2x^2+x=x(2x+1),</math>&nbsp; we let
 |-
 |Multiplying both sides of the last equation by &nbsp;<math style="vertical-align: -5px">x(2x+1),</math>
 |-
-|we get &nbsp;<math style="vertical-align: -5px">1-x=A(2x+1)+Bx.</math>
+|we get
 |-
-|If we let &nbsp;<math style="vertical-align: 0px">x=0,</math> the last equation becomes &nbsp;<math style="vertical-align: -1px">1=A.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">1-x=A(2x+1)+Bx.</math>
 |-
 |If we let &nbsp;<math style="vertical-align: -14px">x=-\frac{1}{2},</math>&nbsp; then we get &nbsp;<math style="vertical-align: -13px">\frac{3}{2}=-\frac{1}{2}\,B.</math>&nbsp; Thus, &nbsp;<math style="vertical-align: 0px">B=-3.</math>
 |-
-|So, in summation, we have &nbsp;<math>\frac{1-x}{2x^2+x}=\frac{1}{x}+\frac{-3}{2x+1}.</math>
+|So, in summation, we have
 |}
@@ Line 146: / Line 152: @@
 !Step 1: &nbsp;
 |-
-|First, we write &nbsp;<math style="vertical-align: -13px">\int\sin^3x~dx=\int \sin^2 x \sin x~dx</math>.
+|First, we write
 |-
-|Using the identity &nbsp;<math style="vertical-align: -2px">\sin^2x+\cos^2x=1,</math>&nbsp; we get &nbsp;<math style="vertical-align: -1px">\sin^2x=1-\cos^2x.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -1px">\sin^2x=1-\cos^2x.</math>
 |-
 |If we use this identity, we have
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -13px">\int\sin^3x~dx=\int (1-\cos^2x)\sin x~dx</math>.
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -13px">\int\sin^3x~dx=\int (1-\cos^2x)\sin x~dx.</math>
 |-
+|

@@ Line 1: / Line 1: @@
 <span class="exam"> Compute the following integrals.
-<span class="exam">(a) <math>\int e^x(x+\sin(e^x))~dx</math>
+<span class="exam">(a) &nbsp;<math>\int \frac{t^2}{\sqrt{1-t^6}}~dt</math>
-<span class="exam">(b) <math>\int \frac{2x^2+1}{2x^2+x}~dx</math>
+<span class="exam">(b) &nbsp;<math>\int \frac{2x^2+1}{2x^2+x}~dx</math>
-<span class="exam">(c) <math>\int \sin^3x~dx</math>
+<span class="exam">(c) &nbsp;<math>\int \sin^3x~dx</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Foundations: &nbsp;
 |-
-|Recall:
+|'''1.''' Through partial fraction decomposition, we can write the fraction
 |-
-|'''1.''' Integration by parts tells us that <math style="vertical-align: -12px">\int u~dv=uv-\int v~du</math>.
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -18px">\frac{1}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}</math>
 |-
-|'''2.''' Through partial fraction decomposition, we can write the fraction &nbsp;<math style="vertical-align: -18px">\frac{1}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}</math> &nbsp;for some constants <math style="vertical-align: -4px">A,B</math>.
+|&nbsp; &nbsp; &nbsp; &nbsp;for some constants <math style="vertical-align: -4px">A,B</math>.
 |-
-|'''3.''' We have the Pythagorean identity <math style="vertical-align: -5px">\sin^2(x)=1-\cos^2(x)</math>.
+|'''2.''' We have the Pythagorean identity
 |}
@@ Line 70: / Line 72: @@
 !Step 1: &nbsp;
 |-
-|First, we add and subtract <math style="vertical-align: 0px">x</math> from the numerator.
+|First, we add and subtract &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; from the numerator.
 |-
 |So, we have
@@ Line 89: / Line 91: @@
 |Now, we need to use partial fraction decomposition for the second integral.
 |-
-|Since <math style="vertical-align: -5px">2x^2+x=x(2x+1)</math>, we let <math>\frac{1-x}{2x^2+x}=\frac{A}{x}+\frac{B}{2x+1}</math>.
+|Since &nbsp;<math style="vertical-align: -5px">2x^2+x=x(2x+1),</math>&nbsp; we let &nbsp;<math>\frac{1-x}{2x^2+x}=\frac{A}{x}+\frac{B}{2x+1}.</math>
 |-
-|Multiplying both sides of the last equation by <math style="vertical-align: -5px">x(2x+1)</math>,
+|Multiplying both sides of the last equation by &nbsp;<math style="vertical-align: -5px">x(2x+1),</math>
 |-
-|we get <math style="vertical-align: -5px">1-x=A(2x+1)+Bx</math>.
+|we get &nbsp;<math style="vertical-align: -5px">1-x=A(2x+1)+Bx.</math>
 |-
-|If we let <math style="vertical-align: 0px">x=0</math>, the last equation becomes <math style="vertical-align: -1px">1=A</math>.
+|If we let &nbsp;<math style="vertical-align: 0px">x=0,</math> the last equation becomes &nbsp;<math style="vertical-align: -1px">1=A.</math>
 |-
-|If we let <math style="vertical-align: -14px">x=-\frac{1}{2}</math>, then we get &thinsp;<math style="vertical-align: -13px">\frac{3}{2}=-\frac{1}{2}\,B</math>. Thus, <math style="vertical-align: 0px">B=-3</math>.
+|If we let &nbsp;<math style="vertical-align: -14px">x=-\frac{1}{2},</math>&nbsp; then we get &nbsp;<math style="vertical-align: -13px">\frac{3}{2}=-\frac{1}{2}\,B.</math>&nbsp; Thus, &nbsp;<math style="vertical-align: 0px">B=-3.</math>
 |-
-|So, in summation, we have&thinsp; <math>\frac{1-x}{2x^2+x}=\frac{1}{x}+\frac{-3}{2x+1}</math>.
+|So, in summation, we have &nbsp;<math>\frac{1-x}{2x^2+x}=\frac{1}{x}+\frac{-3}{2x+1}.</math>
 |}
@@ Line 108: / Line 110: @@
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 \displaystyle{\int \frac{2x^2+1}{2x^2+x}~dx} & = & \displaystyle{\int ~dx+\int\frac{1}{x}~dx+\int\frac{-3}{2x+1}~dx}\\
 &&\\
-& = & \displaystyle{x+\ln x+ \int\frac{-3}{2x+1}~dx}.\\
+& = & \displaystyle{x+\ln x+ \int\frac{-3}{2x+1}~dx.}\\
 \end{array}</math>
 |}
@@ Line 118: / Line 120: @@
 !Step 4: &nbsp;
 |-
-|For the final remaining integral, we use <math style="vertical-align: 0px">u</math>-substitution.
+|For the final remaining integral, we use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 |-
-|Let <math style="vertical-align: -2px">u=2x+1</math>. Then, <math style="vertical-align: 0px">du=2\,dx</math> and&thinsp; <math style="vertical-align: -14px">\frac{du}{2}=dx</math>.
+|Let &nbsp;<math style="vertical-align: -2px">u=2x+1.</math>&nbsp; Then, &nbsp;<math style="vertical-align: 0px">du=2\,dx</math>&nbsp; and &nbsp;<math style="vertical-align: -14px">\frac{du}{2}=dx.</math>
 |-
 |Thus, our final integral becomes
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 \displaystyle{\int \frac{2x^2+1}{2x^2+x}~dx} & = & \displaystyle{x+\ln x+ \int\frac{-3}{2x+1}~dx}\\
 &&\\
 & = & \displaystyle{x+\ln x+\int\frac{-3}{2u}~du}\\
 &&\\
-& = & \displaystyle{x+\ln x-\frac{3}{2}\ln u +C}.\\
+& = & \displaystyle{x+\ln x-\frac{3}{2}\ln u +C.}\\
 \end{array}</math>
 |-
@@ Line 136: / Line 138: @@
 |-
+|
-::<math>\int \frac{2x^2+1}{2x^2+x}~dx\,=\,x+\ln x-\frac{3}{2}\ln (2x+1) +C.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\int \frac{2x^2+1}{2x^2+x}~dx\,=\,x+\ln x-\frac{3}{2}\ln (2x+1) +C.</math>
 |}
@@ Line 144: / Line 146: @@
 !Step 1: &nbsp;
 |-
-|First, we write <math style="vertical-align: -13px">\int\sin^3x~dx=\int \sin^2 x \sin x~dx</math>.
+|First, we write &nbsp;<math style="vertical-align: -13px">\int\sin^3x~dx=\int \sin^2 x \sin x~dx</math>.
 |-
-|Using the identity <math style="vertical-align: -2px">\sin^2x+\cos^2x=1</math>, we get <math style="vertical-align: -1px">\sin^2x=1-\cos^2x</math>.
+|Using the identity &nbsp;<math style="vertical-align: -2px">\sin^2x+\cos^2x=1,</math>&nbsp; we get &nbsp;<math style="vertical-align: -1px">\sin^2x=1-\cos^2x.</math>
 |-
 |If we use this identity, we have
 |-
-| &nbsp; &nbsp; <math style="vertical-align: -13px">\int\sin^3x~dx=\int (1-\cos^2x)\sin x~dx</math>.
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -13px">\int\sin^3x~dx=\int (1-\cos^2x)\sin x~dx</math>.
 |-
+|
@@ Line 158: / Line 160: @@
 !Step 2: &nbsp;
 |-
-|Now, we proceed by <math>u</math>-substitution. Let <math>u=\cos x</math>. Then, <math style="vertical-align: -1px">du=-\sin x dx</math>.
+|Now, we proceed by &nbsp;<math>u</math>-substitution.
 |-
 |So we have
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 \displaystyle{\int\sin^3x~dx} & = & \displaystyle{\int -(1-u^2)~du}\\
 &&\\
@@ Line 178: / Line 182: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)''' &nbsp;<math>xe^x-e^x-\cos(e^x)+C</math>
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp;<math>xe^x-e^x-\cos(e^x)+C</math>
 |-
-|'''(b)''' &nbsp;<math style="vertical-align: -14px">x+\ln x-\frac{3}{2}\ln (2x+1) +C</math>
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp;<math style="vertical-align: -14px">x+\ln x-\frac{3}{2}\ln (2x+1) +C</math>
 |-
-|'''(c)''' &nbsp;<math style="vertical-align: -14px">-\cos x+\frac{\cos^3x}{3}+C</math>
+|&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp;<math style="vertical-align: -14px">-\cos x+\frac{\cos^3x}{3}+C</math>
 |}
 [[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

← Older revision		Revision as of 06:58, 26 February 2016
Line 64:		Line 64:
	\|}		\|}

−	~~== 3 ==~~
	'''(b)'''		'''(b)'''

← Older revision		Revision as of 06:57, 26 February 2016
Line 139:		Line 139:
	\|}		\|}

−	~~== 4 ==~~
	'''(c)'''		'''(c)'''

@@ Line 169: / Line 169: @@
 & = & \displaystyle{-u+\frac{u^3}{3}+C}\\
 &&\\
-& = & \displaystyle{-\cos x+\frac{\cos^3x}{3}+C}\\
+& = & \displaystyle{-\cos x+\frac{\cos^3x}{3}+C}.\\
 \end{array}</math>
 |-