009B Sample Midterm 2, Problem 2 - Revision history

Kayla Murray at 01:00, 13 November 2017

2017-11-13T01:00:44Z

Kayla Murray at 01:31, 5 November 2017

2017-11-05T01:31:14Z

Kayla Murray at 20:37, 14 March 2017

2017-03-14T20:37:42Z

Kayla Murray at 01:23, 27 February 2017

2017-02-27T01:23:43Z

Kayla Murray at 00:12, 19 February 2017

2017-02-19T00:12:17Z

Kayla Murray at 01:58, 8 February 2017

2017-02-08T01:58:52Z

Kayla Murray at 16:43, 7 February 2017

2017-02-07T16:43:47Z

Kayla Murray at 16:29, 6 February 2017

2017-02-06T16:29:02Z

Kayla Murray at 16:15, 6 February 2017

2017-02-06T16:15:52Z

Kayla Murray at 17:17, 5 February 2017

2017-02-05T17:17:09Z

Show changes

@@ Line 6: / Line 6: @@
 <hr>
-(insert picture of handwritten solution)
+[[009B Sample Midterm 2, Problem 2 Solution|'''<u>Solution</u>''']]
 [[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 5: / Line 5: @@
 <span class="exam">(b) &nbsp; <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;"
+[[009B Sample Midterm 2, Problem 2 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']]
 [[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 12: / Line 12: @@
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; You could use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
+&nbsp; &nbsp; &nbsp; &nbsp; You can use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math style="vertical-align: -2px">u=x^2+x.</math>
@@ Line 79: / Line 79: @@
 |Also, we need to change the bounds of integration.
 |-
-|Plugging in our values into the equation &nbsp;<math style="vertical-align: -4px">u=x^4+2x^2+4,</math>
+|Plugging in our values into the equation &nbsp;<math style="vertical-align: -4px">u=x^4+2x^2+4,</math>&nbsp; we get
 |-
-|we get &nbsp;<math style="vertical-align: -5px">u_1=0^4+2(0)^2+4=4</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">u_2=2^4+2(2)^2+4=28.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">u_1=0^4+2(0)^2+4=4</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">u_2=2^4+2(2)^2+4=28.</math>
 |-
 |Therefore, the integral becomes

@@ Line 9: / Line 9: @@
 !Foundations: &nbsp;
 |-
-|How would you integrate <math style="vertical-align: -12px">\int (2x+1)\sqrt{x^2+x}~dx?</math>
+|How would you integrate &nbsp;<math style="vertical-align: -12px">\int (2x+1)\sqrt{x^2+x}~dx?</math>
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; You could use <math style="vertical-align: 0px">u</math>-substitution.
+&nbsp; &nbsp; &nbsp; &nbsp; You could use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; Let <math style="vertical-align: -2px">u=x^2+x.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math style="vertical-align: -2px">u=x^2+x.</math>
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; Then, <math style="vertical-align: -4px">du=(2x+1)~dx.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; Then, &nbsp;<math style="vertical-align: -4px">du=(2x+1)~dx.</math>
 |-
+|
@@ Line 71: / Line 71: @@
 !Step 1: &nbsp;
 |-
-|We use <math style="vertical-align: 0px">u</math>-substitution.
+|We use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 |-
-|Let <math style="vertical-align: -2px">u=x^4+2x^2+4.</math>
+|Let &nbsp;<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>
+|Then, &nbsp;<math style="vertical-align: -5px">du=(4x^3+4x)dx</math>&nbsp; and &nbsp;<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: -4px">u=x^4+2x^2+4,</math>
+|Plugging in our values into the equation &nbsp;<math style="vertical-align: -4px">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: -5px">u_2=2^4+2(2)^2+4=28.</math>
+|we get &nbsp;<math style="vertical-align: -5px">u_1=0^4+2(0)^2+4=4</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">u_2=2^4+2(2)^2+4=28.</math>
 |-
 |Therefore, the integral becomes

@@ Line 1: / Line 1: @@
 <span class="exam"> Evaluate
-::<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">(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) &nbsp; <math style="vertical-align: -14px">\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx</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>

@@ Line 12: / Line 12: @@
 |-
+|
-&nbsp; &nbsp; You could use <math style="vertical-align: 0px">u</math>-substitution.
+&nbsp; &nbsp; &nbsp; &nbsp; You could use <math style="vertical-align: 0px">u</math>-substitution.
 |-
-|&nbsp; &nbsp; Let <math style="vertical-align: -2px">u=x^2+x.</math> Then, <math style="vertical-align: -4px">du=(2x+1)~dx.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; Let <math style="vertical-align: -2px">u=x^2+x.</math>
 |-
+|
-&nbsp; &nbsp; Thus,
+&nbsp; &nbsp; &nbsp; &nbsp; Thus,
 |-
+|
-&nbsp; &nbsp; <math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 \displaystyle{\int (2x+1)\sqrt{x^2+x}~dx} & = & \displaystyle{\int \sqrt{u}~du}\\
 &&\\
@@ Line 39: / Line 41: @@
 |-
+|
-&nbsp; &nbsp; <math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 \displaystyle{\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt} & = & \displaystyle{\int_1^2 \bigg(8t^3-10+12-\frac{15}{t^3}\bigg)~dt}\\
 &&\\
@@ Line 51: / Line 53: @@
 |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>
+| &nbsp;&nbsp; &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>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 \displaystyle{\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt} & = & \displaystyle{2(2)^4+2(2)+\frac{15}{2(2)^2}-\bigg(2+2+\frac{15}{2}\bigg)}\\
 &&\\
@@ Line 71: / Line 73: @@
 |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>
+|Let <math style="vertical-align: -2px">u=x^4+2x^2+4.</math>
 |-
 |Also, we need to change the bounds of integration.
@@ Line 81: / Line 85: @@
 |Therefore, the integral becomes
 |-
-|&nbsp; &nbsp; <math style="vertical-align: -14px">\frac{1}{4}\int_4^{28}\sqrt{u}~du.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -14px">\frac{1}{4}\int_4^{28}\sqrt{u}~du.</math>
 |}
@@ Line 90: / Line 94: @@
 |-
+|
-&nbsp; &nbsp; <math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 \displaystyle{\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx} & = & \displaystyle{\frac{1}{4}\int_4^{28}\sqrt{u}~du}\\
 &&\\
@@ Line 104: / Line 108: @@
 |Therefore,
 |-
-| &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>
+| &nbsp; &nbsp; &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>
 |}
@@ Line 111: / Line 115: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)''' &nbsp; <math>\frac{211}{8}</math>
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>\frac{211}{8}</math>
 |-
-|'''(b)''' &nbsp; <math>\frac{28\sqrt{7}-4}{3}</math>
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>\frac{28\sqrt{7}-4}{3}</math>
 |}
 [[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ 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>
+&nbsp; &nbsp; You could use <math style="vertical-align: 0px">u</math>-substitution.
 |-
+|
-::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>
+&nbsp; &nbsp; <math>\begin{array}{rcl}
 |}
@@ Line 27: / Line 38: @@
 |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 39: / Line 55: @@
 |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 46: / Line 69: @@
 !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.
 |-
-|Plugging in our values into the equation <math style="vertical-align: -4px">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: -5px">u_2=2^4+2(2)^2+4=28.</math>
+|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>
 |-
-|Therefore, the integral becomes&nbsp; <math style="vertical-align: -14px">\frac{1}{4}\int_4^{28}\sqrt{u}~du.</math>
+|Also, we need to change the bounds of integration.
 |-
+|Plugging in our values into the equation <math style="vertical-align: -4px">u=x^4+2x^2+4,</math>
 |}
@@ Line 58: / Line 87: @@
 !Step 2: &nbsp;
 |-
-|We now have:
+|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
+|Therefore,
 |-
 | &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>