009B Sample Midterm 1, Problem 1 - Revision history

MathAdmin at 22:47, 12 November 2017

2017-11-12T22:47:23Z

MathAdmin at 18:59, 9 April 2017

2017-04-09T18:59:22Z

MathAdmin at 22:06, 18 April 2016

2016-04-18T22:06:44Z

MathAdmin at 20:50, 18 April 2016

2016-04-18T20:50:30Z

MathAdmin at 22:02, 8 April 2016

2016-04-08T22:02:15Z

MathAdmin: Created page with "Evaluate the indefinite and definite integrals. ::a) $\int x^2\sqrt{1+x^3}~dx$ ::b) $\int _{\frac{\pi..."$

2016-02-01T22:34:51Z

Created page with "Evaluate the indefinite and definite integrals. ::a) <math>\int x^2\sqrt{1+x^3}~dx</math> ::b) <math>\int _{\frac{\pi..."

New page

Evaluate the indefinite and definite integrals.

::a) <math>\int x^2\sqrt{1+x^3}~dx</math>
::b) <math>\int _{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\cos(x)}{\sin^2(x)}~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 need to use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -2px">u=1+x^3</math>. Then, <math style="vertical-align: 0px">du=3x^2dx</math> and  <math style="vertical-align: -13px">\frac{du}{3}=x^2dx</math>.
|-
|Therefore, the integral becomes  <math style="vertical-align: -13px">\frac{1}{3}\int \sqrt{u}~du</math>.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|We now have:
|-
|    <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>.
|}

'''(b)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|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>. 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>, 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>.
|-
|Therefore, the integral becomes <math style="vertical-align: -19px">\int_{\frac{\sqrt{2}}{2}}^1 \frac{1}{u^2}~du</math>.
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|We now have:
|-
|    <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>.
|-
|
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)'''   <math>\frac{2}{9}(1+x^3)^{\frac{3}{2}}+C</math>
|-
|'''(b)'''   <math>-1+\sqrt{2}</math>
|}
[[009B_Sample_Midterm_1|'''Return to Sample Exam''']]

@@ 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;
+[[009B Sample Midterm 1, Problem 1 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009B Sample Midterm 1, Problem 1 Detailed Solution|'''<u>Detailed 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">a) <math>\int x^2\sqrt{1+x^3}~dx</math>
+<span class="exam">(a) &nbsp; <math>\int x^2\sqrt{1+x^3}~dx</math>
-::<span class="exam">b) <math>\int _{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\cos(x)}{\sin^2(x)}~dx</math>
@@ Line 8: / 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>
 |-
+|
-::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> Thus,
+&nbsp; &nbsp; &nbsp; &nbsp; Thus,
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 \displaystyle{\int \frac{\ln x}{x}~dx} & = & \displaystyle{\int u~du}\\
 &&\\
 & = & \displaystyle{\frac{u^2}{2}+C}\\
 &&\\
-& = & \displaystyle{\frac{(\ln x)^2}{2}+C.}\\
+& = & \displaystyle{\frac{(\ln x)^2}{2}+C.}
 \end{array}</math>
 |}
 '''Solution:'''
@@ Line 28: / 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> Then, <math style="vertical-align: 0px">du=3x^2dx</math> and&thinsp; <math style="vertical-align: -13px">\frac{du}{3}=x^2dx.</math>
+|We use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 |-
-|Therefore, the integral becomes&thinsp;
+|Let &nbsp;<math style="vertical-align: -2px">u=1+x^3.</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>
-::<math style="vertical-align: -13px">\frac{1}{3}\int \sqrt{u}~du.</math>
+|-
 |}
@@ Line 39: / Line 52: @@
 !Step 2: &nbsp;
 |-
-|We now have:
+|We now have
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-::<math>\begin{array}{rcl}
+\displaystyle{\int x^2\sqrt{1+x^3}~dx} & = & \displaystyle{\frac{1}{3}\int \sqrt{u}~du}\\
 &&\\
 & = & \displaystyle{\frac{2}{9}u^{\frac{3}{2}}+C}\\
 &&\\
-& = & \displaystyle{\frac{2}{9}(1+x^3)^{\frac{3}{2}}+C.}\\
+& = & \displaystyle{\frac{2}{9}(1+x^3)^{\frac{3}{2}}+C.}
 \end{array}</math>
 |}
@@ Line 57: / Line 67: @@
 !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> Also, we need to change the bounds of integration.
+|We use &nbsp;<math>u</math>-substitution.
 |-
-|Plugging in our values into the equation <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>
+|Let &nbsp;<math style="vertical-align: -5px">u=\sin(x).</math>
 |-
 |Therefore, the integral becomes
 |-
+|&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: @@
 !Step 2: &nbsp;
 |-
-|We now have:
+|We now have
 |-
+|
-::<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 80: / Line 97: @@
 & = & \displaystyle{-\frac{1}{1}-\frac{-1}{\frac{\sqrt{2}}{2}}}\\
 &&\\
-& = & \displaystyle{-1+\sqrt{2}.}\\
+& = & \displaystyle{-1+\sqrt{2}.}
 \end{array}</math>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Final Answer: &nbsp;
 |-
-|&nbsp;&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>
 |-
-|&nbsp;&nbsp; '''(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 11: / Line 11: @@
 |-
+|
-::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>
+::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> Thus,
 |-
+|
-::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>
+::<math>\begin{array}{rcl}
 |}
 '''Solution:'''
@@ Line 22: / Line 28: @@
 !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>. Then, <math style="vertical-align: 0px">du=3x^2dx</math> and&thinsp; <math style="vertical-align: -13px">\frac{du}{3}=x^2dx</math>.
+|We need to use <math style="vertical-align: 0px">u</math>-substitution. Let <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>
 |-
-|Therefore, the integral becomes&thinsp; <math style="vertical-align: -13px">\frac{1}{3}\int \sqrt{u}~du</math>.
+|Therefore, the integral becomes&thinsp;
 |}
@@ Line 32: / Line 41: @@
 |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>.
 |}
@@ Line 39: / Line 57: @@
 !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>. Also, we need to change the bounds of integration.
+|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> 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>, 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>.
+|Plugging in our values into the equation <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>
 |-
-|Therefore, the integral becomes <math style="vertical-align: -19px">\int_{\frac{\sqrt{2}}{2}}^1 \frac{1}{u^2}~du</math>.
+|Therefore, the integral becomes
 |-
+|
 |}
@@ Line 52: / Line 71: @@
 |-
 |We now have:
 |-
+|
 |}