009B Sample Midterm 1, Problem 1 - Revision history

Kayla Murray at 22:42, 12 November 2017

2017-11-12T22:42:42Z

Kayla Murray at 01:48, 5 November 2017

2017-11-05T01:48:53Z

Kayla Murray at 20:38, 14 March 2017

2017-03-14T20:38:28Z

Kayla Murray at 20:30, 14 March 2017

2017-03-14T20:30:22Z

Kayla Murray at 23:14, 1 March 2017

2017-03-01T23:14:29Z

Kayla Murray at 00:53, 27 February 2017

2017-02-27T00:53:16Z

Kayla Murray at 00:08, 19 February 2017

2017-02-19T00:08:25Z

Kayla Murray at 23:31, 7 February 2017

2017-02-07T23:31:50Z

Kayla Murray at 16:01, 7 February 2017

2017-02-07T16:01:12Z

Kayla Murray at 15:59, 7 February 2017

2017-02-07T15:59:38Z

@@ Line 8: / Line 8: @@
 <hr>
-(insert picture of handwritten solution)
+[[009B Sample Midterm 1, Problem 1 Solution|'''<u>Solution</u>''']]
 [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
-<span class="exam">Evaluate the indefinite and definite integrals.
+<span class="exam"> Let &nbsp;<math style="vertical-align: -5px">f(x)=1-x^2</math>.
-<span class="exam">(a) &nbsp; <math>\int x^2\sqrt{1+x^3}~dx</math>
+<span class="exam">(a) Compute the left-hand Riemann sum approximation of &nbsp;<math style="vertical-align: -14px">\int_0^3 f(x)~dx</math>&nbsp; with &nbsp;<math style="vertical-align: 0px">n=3</math>&nbsp; boxes.
-<span class="exam">(b) &nbsp; <math>\int _{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\cos(x)}{\sin^2(x)}~dx</math>
+<span class="exam">(b) Compute the right-hand Riemann sum approximation of &nbsp;<math style="vertical-align: -14px">\int_0^3 f(x)~dx</math>&nbsp; with &nbsp;<math style="vertical-align: 0px">n=3</math>&nbsp; boxes.
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+(insert picture of handwritten solution)
 [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 75: / Line 75: @@
 |Also, we need to change the bounds of integration.
 |-
-|Plugging in our values into the equation &nbsp;<math style="vertical-align: -5px">u=\sin(x),</math> we get
+|Plugging in our values into the equation &nbsp;<math style="vertical-align: -5px">u=\sin(x),</math>&nbsp; we get
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -15px">u_1=\sin\bigg(\frac{\pi}{4}\bigg)=\frac{\sqrt{2}}{2}</math>&nbsp; and &nbsp;<math style="vertical-align: -16px">u_2=\sin\bigg(\frac{\pi}{2}\bigg)=1.</math>&nbsp;

@@ 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: -5px">u=\ln(x).</math>
@@ Line 38: / Line 38: @@
 !Step 1: &nbsp;
 |-
-|We need to use &nbsp;<math style="vertical-align: 0px">u</math>-substitution. Let &nbsp;<math style="vertical-align: -2px">u=1+x^3.</math>
+|We use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 |-
 |Then, &nbsp;<math style="vertical-align: 0px">du=3x^2dx</math>&nbsp; and &nbsp;<math style="vertical-align: -13px">\frac{du}{3}=x^2dx.</math>
@@ Line 65: / Line 67: @@
 !Step 1: &nbsp;
 |-
-|We need to use &nbsp;<math>u</math>-substitution.
+|We use &nbsp;<math>u</math>-substitution.
 |-
 |Let &nbsp;<math style="vertical-align: -5px">u=\sin(x).</math>
@@ Line 73: / Line 75: @@
 |Also, we need to change the bounds of integration.
 |-
-|Plugging in our values into the equation &nbsp;<math style="vertical-align: -5px">u=\sin(x),</math>
+|Plugging in our values into the equation &nbsp;<math style="vertical-align: -5px">u=\sin(x),</math> we get
 |-
-|we get &nbsp;<math style="vertical-align: -15px">u_1=\sin\bigg(\frac{\pi}{4}\bigg)=\frac{\sqrt{2}}{2}</math>&nbsp; and &nbsp;<math style="vertical-align: -16px">u_2=\sin\bigg(\frac{\pi}{2}\bigg)=1.</math>&nbsp;
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -15px">u_1=\sin\bigg(\frac{\pi}{4}\bigg)=\frac{\sqrt{2}}{2}</math>&nbsp; and &nbsp;<math style="vertical-align: -16px">u_2=\sin\bigg(\frac{\pi}{2}\bigg)=1.</math>&nbsp;
 |-
 |Therefore, the integral becomes

@@ Line 85: / Line 85: @@
 !Step 2: &nbsp;
 |-
-|We now have:
+|We now have
 |-
+|

@@ Line 9: / Line 9: @@
 !Foundations: &nbsp;
 |-
-| How would you integrate <math style="vertical-align: -12px">\int \frac{\ln x}{x}~dx?</math>
+| How would you integrate &nbsp; <math style="vertical-align: -12px">\int \frac{\ln x}{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: -5px">u=\ln(x).</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math style="vertical-align: -5px">u=\ln(x).</math>
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; Then, <math style="vertical-align: -13px">du=\frac{1}{x}dx.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; Then, &nbsp;<math style="vertical-align: -13px">du=\frac{1}{x}dx.</math>
 |-
+|
@@ Line 38: / Line 38: @@
 !Step 1: &nbsp;
 |-
-|We need to use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -2px">u=1+x^3.</math>
+|We need to use &nbsp;<math style="vertical-align: 0px">u</math>-substitution. Let &nbsp;<math style="vertical-align: -2px">u=1+x^3.</math>
 |-
-|Then, <math style="vertical-align: 0px">du=3x^2dx</math> and&thinsp; <math style="vertical-align: -13px">\frac{du}{3}=x^2dx.</math>
+|Then, &nbsp;<math style="vertical-align: 0px">du=3x^2dx</math>&nbsp; and &nbsp;<math style="vertical-align: -13px">\frac{du}{3}=x^2dx.</math>
 |-
 |Therefore, the integral becomes
@@ Line 65: / Line 65: @@
 !Step 1: &nbsp;
 |-
-|We need to use <math>u</math>-substitution.
+|We need to use &nbsp;<math>u</math>-substitution.
 |-
-|Let <math style="vertical-align: -5px">u=\sin(x).</math>
+|Let &nbsp;<math style="vertical-align: -5px">u=\sin(x).</math>
 |-
-|Then, <math style="vertical-align: -5px">du=\cos(x)dx.</math>
+|Then, &nbsp;<math style="vertical-align: -5px">du=\cos(x)dx.</math>
 |-
 |Also, we need to change the bounds of integration.
 |-
-|Plugging in our values into the equation <math style="vertical-align: -5px">u=\sin(x),</math>
+|Plugging in our values into the equation &nbsp;<math style="vertical-align: -5px">u=\sin(x),</math>
 |-
-|we get <math style="vertical-align: -15px">u_1=\sin\bigg(\frac{\pi}{4}\bigg)=\frac{\sqrt{2}}{2}</math> and <math style="vertical-align: -16px">u_2=\sin\bigg(\frac{\pi}{2}\bigg)=1.</math>
+|we get &nbsp;<math style="vertical-align: -15px">u_1=\sin\bigg(\frac{\pi}{4}\bigg)=\frac{\sqrt{2}}{2}</math>&nbsp; and &nbsp;<math style="vertical-align: -16px">u_2=\sin\bigg(\frac{\pi}{2}\bigg)=1.</math>&nbsp;
 |-
 |Therefore, the integral becomes

@@ Line 11: / Line 11: @@
 |-
+|
-&nbsp; &nbsp; You could use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -5px">u=\ln(x).</math> Then, <math style="vertical-align: -13px">du=\frac{1}{x}dx.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; You could use <math style="vertical-align: 0px">u</math>-substitution.
 |-
+|
-&nbsp; &nbsp; Thus, <math style="vertical-align: -12px">\int \frac{\ln x}{x}~dx=\int u~du=\frac{u^2}{2}+C=\frac{(\ln x)^2}{2}+C.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; Thus,
 |}
@@ Line 30: / Line 43: @@
 |Therefore, the integral becomes
 |-
-|&nbsp; &nbsp; <math style="vertical-align: -13px">\frac{1}{3}\int \sqrt{u}~du.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -13px">\frac{1}{3}\int \sqrt{u}~du.</math>
 |}
@@ Line 36: / Line 49: @@
 !Step 2: &nbsp;
 |-
-|We now have:
+|We now have
 |-
-|&nbsp; &nbsp; <math>\begin{array}{rcl}
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 \displaystyle{\int x^2\sqrt{1+x^3}~dx} & = & \displaystyle{\frac{1}{3}\int \sqrt{u}~du}\\
 &&\\
@@ Line 51: / Line 64: @@
 !Step 1: &nbsp;
 |-
-|Again, we need to use <math>u</math>-substitution. Let <math style="vertical-align: -5px">u=\sin(x).</math> Then, <math style="vertical-align: -5px">du=\cos(x)dx.</math>
+|We need to use <math>u</math>-substitution.
 |-
 |Also, we need to change the bounds of integration.
@@ Line 61: / Line 78: @@
 |Therefore, the integral becomes
 |-
-|&nbsp; &nbsp; <math style="vertical-align: -19px">\int_{\frac{\sqrt{2}}{2}}^1 \frac{1}{u^2}~du.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -19px">\int_{\frac{\sqrt{2}}{2}}^1 \frac{1}{u^2}~du.</math>
 |}
@@ Line 70: / Line 87: @@
 |-
+|
-&nbsp; &nbsp; <math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 \displaystyle{\int _{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\cos(x)}{\sin^2(x)}~dx} & = & \displaystyle{\int_{\frac{\sqrt{2}}{2}}^1 \frac{1}{u^2}~du}\\
 &&\\
@@ Line 85: / Line 102: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)''' &nbsp; <math>\frac{2}{9}(1+x^3)^{\frac{3}{2}}+C</math>
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>\frac{2}{9}(1+x^3)^{\frac{3}{2}}+C</math>
 |-
-|'''(b)''' &nbsp; <math>-1+\sqrt{2}</math>
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>-1+\sqrt{2}</math>
 |}
 [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 36: / Line 36: @@
 |We now have:
 |-
-|&nbsp; &nbsp; <math style="vertical-align: -13px">\int x^2\sqrt{1+x^3}~dx=\frac{1}{3}\int \sqrt{u}~du=\frac{2}{9}u^{\frac{3}{2}}+C=\frac{2}{9}(1+x^3)^{\frac{3}{2}}+C.</math>
+|&nbsp; &nbsp; <math>\begin{array}{rcl}
 |}
@@ Line 59: / Line 65: @@
 |We now have:
 |-
 |&nbsp; &nbsp; <math>\int _{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\cos(x)}{\sin^2(x)}~dx=\int_{\frac{\sqrt{2}}{2}}^1 \frac{1}{u^2}~du=\left.\frac{-1}{u}\right|_{\frac{\sqrt{2}}{2}}^1=-\frac{1}{1}-\frac{-1}{\frac{\sqrt{2}}{2}}=-1+\sqrt{2}.</math>
 |}