009B Sample Final 1, Problem 6 - Revision history

Kayla Murray: Replaced content with " Evaluate the improper integrals: (a) $\int_0^{\infty} xe^{-x}~dx$ (b) $\int_1^4 \..."$

2017-12-03T00:19:37Z

Replaced content with "<span class="exam"> Evaluate the improper integrals: <span class="exam">(a) <math>\int_0^{\infty} xe^{-x}~dx</math> <span class="exam">(b) <math>\int_1^4 \..."

Kayla Murray at 21:02, 20 May 2017

2017-05-20T21:02:54Z

Kayla Murray at 19:47, 18 March 2017

2017-03-18T19:47:10Z

Kayla Murray at 17:33, 27 February 2017

2017-02-27T17:33:37Z

Show changes

Kayla Murray at 02:12, 19 February 2017

2017-02-19T02:12:46Z

Grad at 06:10, 26 February 2016

2016-02-26T06:10:20Z

Grad: /* 4 */

2016-02-26T06:10:06Z

Grad: /* 3 */

2016-02-26T06:09:35Z

Grad at 06:06, 26 February 2016

2016-02-26T06:06:25Z

Grad: /* 2 */

2016-02-26T06:06:02Z

@@ Line 5: / Line 5: @@
 <span class="exam">(b) &nbsp; <math>\int_1^4 \frac{dx}{\sqrt{4-x}}</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009B Sample Final 1, Problem 6 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009B Sample Final 1, Problem 6 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 110: / Line 110: @@
 !Step 1: &nbsp;
 |-
-|First, we write &nbsp;<math>\int_1^4 \frac{dx}{\sqrt{4-x}}=\lim_{a\rightarrow 4} \int_1^a\frac{dx}{\sqrt{4-x}}.</math>
+|First, we write &nbsp;<math>\int_1^4 \frac{dx}{\sqrt{4-x}}=\lim_{a\rightarrow 4^-} \int_1^a\frac{dx}{\sqrt{4-x}}.</math>
 |-
 |Now, we proceed by &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
@@ Line 125: / Line 125: @@
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;<math>\int_1^4 \frac{dx}{\sqrt{4-x}}\,=\,\lim_{a\rightarrow 4} \int_3^{4-a}\frac{-1}{\sqrt{u}}~du.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\int_1^4 \frac{dx}{\sqrt{4-x}}\,=\,\lim_{a\rightarrow 4^-} \int_3^{4-a}\frac{-1}{\sqrt{u}}~du.</math>
 |}
@@ Line 135: / Line 135: @@
+|
 &nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
-\displaystyle{\int_1^4 \frac{dx}{\sqrt{4-x}}} & = & \displaystyle{\lim_{a\rightarrow 4} -2u^{\frac{1}{2}}\bigg|_{3}^{4-a}}\\
+\displaystyle{\int_1^4 \frac{dx}{\sqrt{4-x}}} & = & \displaystyle{\lim_{a\rightarrow 4^-} -2u^{\frac{1}{2}}\bigg|_{3}^{4-a}}\\
 &&\\
-& = & \displaystyle{\lim_{a\rightarrow 4}-2\sqrt{4-a}+2\sqrt{3}}\\
+& = & \displaystyle{\lim_{a\rightarrow 4^-}-2\sqrt{4-a}+2\sqrt{3}}\\
 &&\\
 & = & \displaystyle{2\sqrt{3}}.\\

@@ Line 63: / Line 63: @@
 |Plugging in our values into the equation &nbsp;<math style="vertical-align: -4px">u=-x,</math>&nbsp; we get
 |-
-|<math style="vertical-align: -5px">u_1=0</math>&nbsp; and &nbsp;<math style="vertical-align: -3px">u_2=-a.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">u_1=0</math>&nbsp; and &nbsp;<math style="vertical-align: -3px">u_2=-a.</math>
 |-
 |Thus, the integral becomes
@@ Line 120: / Line 120: @@
 |Plugging in our values into the equation &nbsp;<math style="vertical-align: -4px">u=4-x,</math>&nbsp; we get
 |-
-|<math style="vertical-align: -5px">u_1=4-1=3</math>&nbsp; and &nbsp;<math style="vertical-align: -3px">u_2=4-a.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">u_1=4-1=3</math>&nbsp; and &nbsp;<math style="vertical-align: -3px">u_2=4-a.</math>
 |-
 |Thus, the integral becomes

@@ Line 1: / Line 1: @@
 <span class="exam"> Evaluate the improper integrals:
-::<span class="exam">a) <math>\int_0^{\infty} xe^{-x}~dx</math>
+<span class="exam">(a) <math>\int_0^{\infty} xe^{-x}~dx</math>
-::<span class="exam">b) <math>\int_1^4 \frac{dx}{\sqrt{4-x}}</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 28: / Line 29: @@
 ::Let <math style="vertical-align: 0px">u=x</math> and <math style="vertical-align: 0px">dv=e^xdx</math>.
 |}
 '''Solution:'''
@@ Line 128: / Line 130: @@
 \end{array}</math>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 95: / Line 95: @@
 |}
 '''(b)'''
@@ Line 130: / Line 129: @@
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Final Answer: &nbsp;

@@ Line 134: / Line 134: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)''' <math style="vertical-align: -3px">1</math>
+|'''(a)''' &nbsp;<math style="vertical-align: -3px">1</math>
 |-
-|'''(b)''' <math style="vertical-align: -4px">2\sqrt{3}</math>
+|'''(b)''' &nbsp;<math style="vertical-align: -4px">2\sqrt{3}</math>
 |}
 [[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 107: / Line 107: @@
 |Since the integral is a definite integral, we need to change the bounds of integration.
 |-
-|Plugging in our values into the equation <math style="vertical-align: -1px">u=4-x</math>, we get <math style="vertical-align: -5px">u_1=4-1=3</math> and <math style="vertical-align: -3px">u_2=4-a</math>.
+|Plugging in our values into the equation <math style="vertical-align: -1px">u=4-x</math>, we get <math style="vertical-align: -5px">u_1=4-1=3</math>&thinsp; and <math style="vertical-align: -3px">u_2=4-a</math>.
 |-
 |Thus, the integral becomes
 |-
+|
-::<math>\int_1^4 \frac{dx}{\sqrt{4-x}}=\lim_{a\rightarrow 4} \int_3^{4-a}\frac{-1}{\sqrt{u}}~du</math>.
+::<math>\int_1^4 \frac{dx}{\sqrt{4-x}}\,=\,\lim_{a\rightarrow 4} \int_3^{4-a}\frac{-1}{\sqrt{u}}~du.</math>
 |}
@@ Line 126: / Line 126: @@
 & = & \displaystyle{\lim_{a\rightarrow 4}-2\sqrt{4-a}+2\sqrt{3}}\\
 &&\\
-& = & \displaystyle{2\sqrt{3}}\\
+& = & \displaystyle{2\sqrt{3}}.\\
 \end{array}</math>
 |}
 == 4 ==
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

← Older revision		Revision as of 06:06, 26 February 2016
Line 31:		Line 31:
	'''Solution:'''		'''Solution:'''

−	~~== 2 ==~~
	'''(a)'''		'''(a)'''

@@ Line 44: / Line 44: @@
 |-
+|
-::<math>\int_0^{\infty} xe^{-x}~dx=\lim_{a\rightarrow \infty} \left.-xe^{-x}\right|_0^a-\int_0^a-e^{-x}dx</math>
+::<math>\int_0^{\infty} xe^{-x}~dx=\lim_{a\rightarrow \infty} \left.-xe^{-x}\right|_0^a-\int_0^a-e^{-x}\,dx.</math>
 |}
@@ Line 64: / Line 64: @@
 & = & \displaystyle{\lim_{a\rightarrow \infty} -xe^{-x}\bigg|_0^a-e^{u}\bigg|_0^{-a}}\\
 &&\\
-& = & \displaystyle{\lim_{a\rightarrow \infty} -ae^{-a}-(e^{-a}-1)}\\
+& = & \displaystyle{\lim_{a\rightarrow \infty} -ae^{-a}-(e^{-a}-1)}.\\
 \end{array}</math>
 |}
@@ Line 79: / Line 79: @@
 & = & \displaystyle{\lim_{a\rightarrow \infty} \frac{-a}{e^a}-\frac{1}{e^a}+1}\\
 &&\\
-& = & \displaystyle{\lim_{a\rightarrow \infty} \frac{-a-1}{e^a}+1}\\
+& = & \displaystyle{\lim_{a\rightarrow \infty} \frac{-a-1}{e^a}+1}.\\
 \end{array}</math>
 |-
-|Using L'Hopital's Rule, we get
+|Using L'Hôpital's Rule, we get
 |-
+|
@@ Line 90: / Line 90: @@
 & = & \displaystyle{0+1}\\
 &&\\
-& = & \displaystyle{1}\\
+& = & \displaystyle{1}.\\
 \end{array}</math>
 |-
+|
 |}
 == 3 ==
 '''(b)'''