009B Sample Midterm 2, Problem 3 - Revision history

MathAdmin at 19:15, 9 April 2017

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MathAdmin at 22:04, 18 April 2016

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MathAdmin: Created page with " Evaluate ::a) $\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt$ ::
2016-02-03T06:57:56Z

Created page with " Evaluate ::a) <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..."

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 Evaluate

::a) <math style="vertical-align: -14px">\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt</math>

::b) <math style="vertical-align: -14px">\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Review <math style="vertical-align: 0px">u</math>-substitution
|}

'''Solution:'''

'''(a)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We multiply the product inside the integral to get
|-
|    <math style="vertical-align: -16px">\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt=\int_1^2 \bigg(8t^3-10+12-\frac{15}{t^3}\bigg)~dt=\int_1^2 (8t^3+2-15t^{-3})~dt</math>.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|We integrate to get
|-
|    <math style="vertical-align: -16px">\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt=\left. 2t^4+2t+\frac{15}{2}t^{-2}\right|_1^2</math>.
|-
|We now evaluate to get
|-
|    <math style="vertical-align: -16px">\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt=2(2)^4+2(2)+\frac{15}{2(2)^2}-\bigg(2+2+\frac{15}{2}\bigg)=36+\frac{15}{8}-4-\frac{15}{2}=\frac{211}{8}</math>.
|}

'''(b)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -2px">u=x^4+2x^2+4</math>. Then, <math style="vertical-align: -5px">du=(4x^3+4x)dx</math> and <math style="vertical-align: -14px">\frac{du}{4}=(x^3+x)dx</math>. Also, we need to change the bounds of integration.
|-
|Plugging in our values into the equation <math style="vertical-align: -2px">u=x^4+2x^2+4</math>, we get <math style="vertical-align: -5px">u_1=0^4+2(0)^2+4=4</math> and <math style="vertical-align: -4px">u_2=2^4+2(2)^2+4=28</math>.
|-
|Therefore, the integral becomes  <math style="vertical-align: -14px">\frac{1}{4}\int_4^{28}\sqrt{u}~du</math>.
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|We now have:
|-
|    <math style="vertical-align: -16px">\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx=\frac{1}{4}\int_4^{28}\sqrt{u}~du=\left.\frac{1}{6}u^{\frac{3}{2}}\right|_4^{28}=\frac{1}{6}(28^{\frac{3}{2}}-4^{\frac{3}{2}})=\frac{1}{6}((\sqrt{28})^3-(\sqrt{4})^3)=\frac{1}{6}((2\sqrt{7})^3-2^3)</math>.
|-
|So, we have
|-
|    <math style="vertical-align: -16px">\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx=\frac{28\sqrt{7}-4}{3}</math>.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)'''   <math>\frac{211}{8}</math>
|-
|'''(b)'''   <math>\frac{28\sqrt{7}-4}{3}</math>
|}
[[009B_Sample_Midterm_2|'''Return to Sample Exam''']]

@@ Line 12: / Line 12: @@
 |-
+|
-::You could use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -2px">u=x^2+x.</math> Then, <math style="vertical-align: -4px">du=(2x+1)~dx.</math>
+::You could use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -2px">u=x^2+x.</math> Then, <math style="vertical-align: -5px">du=(2x+1)~dx.</math> Thus,
 |-
+|
-::Thus, <math style="vertical-align: -12px">\int (2x+1)\sqrt{x^2+x}~dx\,=\,\int \sqrt{u}\,=\,\frac{2}{3}u^{3/2}+C\,=\,\frac{2}{3}(x^2+x)^{3/2}+C.</math>
+::<math>\begin{array}{rcl}
 |}
 '''Solution:'''
@@ Line 25: / Line 31: @@
 |We multiply the product inside the integral to get
 |-
-| &nbsp;&nbsp; <math style="vertical-align: -16px">\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt=\int_1^2 \bigg(8t^3-10+12-\frac{15}{t^3}\bigg)~dt=\int_1^2 (8t^3+2-15t^{-3})~dt</math>.
 |}
@@ Line 33: / Line 44: @@
 |We integrate to get
 |-
-| &nbsp;&nbsp; <math style="vertical-align: -16px">\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt=\left. 2t^4+2t+\frac{15}{2}t^{-2}\right|_1^2</math>.
 |-
 |We now evaluate to get
 |-
-| &nbsp;&nbsp; <math style="vertical-align: -16px">\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt=2(2)^4+2(2)+\frac{15}{2(2)^2}-\bigg(2+2+\frac{15}{2}\bigg)=36+\frac{15}{8}-4-\frac{15}{2}=\frac{211}{8}</math>.
 |}
@@ Line 44: / Line 63: @@
 !Step 1: &nbsp;
 |-
-|We use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -2px">u=x^4+2x^2+4</math>. Then, <math style="vertical-align: -5px">du=(4x^3+4x)dx</math> and <math style="vertical-align: -14px">\frac{du}{4}=(x^3+x)dx</math>. Also, we need to change the bounds of integration.
+|We use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -2px">u=x^4+2x^2+4.</math> Then, <math style="vertical-align: -5px">du=(4x^3+4x)dx</math> and <math style="vertical-align: -14px">\frac{du}{4}=(x^3+x)dx.</math> Also, we need to change the bounds of integration.
 |-
-|Plugging in our values into the equation <math style="vertical-align: -2px">u=x^4+2x^2+4</math>, we get <math style="vertical-align: -5px">u_1=0^4+2(0)^2+4=4</math> and <math style="vertical-align: -4px">u_2=2^4+2(2)^2+4=28</math>.
+|Therefore, the integral becomes
 |-
-|Therefore, the integral becomes&nbsp; <math style="vertical-align: -14px">\frac{1}{4}\int_4^{28}\sqrt{u}~du</math>.
 |-
+|
@@ Line 58: / Line 80: @@
 |We now have:
 |-
-| &nbsp;&nbsp; <math style="vertical-align: -16px">\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx=\frac{1}{4}\int_4^{28}\sqrt{u}~du=\left.\frac{1}{6}u^{\frac{3}{2}}\right|_4^{28}=\frac{1}{6}(28^{\frac{3}{2}}-4^{\frac{3}{2}})=\frac{1}{6}((\sqrt{28})^3-(\sqrt{4})^3)=\frac{1}{6}((2\sqrt{7})^3-2^3)</math>.
 |-
 |So, we have
 |-
-| &nbsp;&nbsp; <math style="vertical-align: -16px">\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx=\frac{28\sqrt{7}-4}{3}</math>.
 |}

← Older revision		Revision as of 16:17, 3 February 2016
Line 1:		Line 1:
−	<span class="exam"> Evaluate	+	<span class="exam"> Evaluate:

	::<span class="exam">a) <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">a) <math style="vertical-align: -14px">\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt</math>

@@ Line 102: / Line 102: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)''' &nbsp; <math>\frac{211}{8}</math>
+|&nbsp;&nbsp; '''(a)''' &nbsp; <math>\frac{211}{8}</math>
 |-
-|'''(b)''' &nbsp; <math>\frac{28\sqrt{7}-4}{3}</math>
+|&nbsp;&nbsp; '''(b)''' &nbsp; <math>\frac{28\sqrt{7}-4}{3}</math>
 |}
 [[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 9: / Line 9: @@
 !Foundations: &nbsp;
 |-
-|Review <math style="vertical-align: 0px">u</math>-substitution
+|How would you integrate <math style="vertical-align: -12px">\int (2x+1)\sqrt{x^2+x}~dx?</math>
 |}
 '''Solution:'''