009B Sample Midterm 1, Problem 2 - Revision history

Kayla Murray at 23:01, 12 November 2017

2017-11-12T23:01:57Z

Kayla Murray at 01:51, 5 November 2017

2017-11-05T01:51:08Z

Kayla Murray at 17:54, 18 March 2017

2017-03-18T17:54:58Z

Kayla Murray at 20:31, 14 March 2017

2017-03-14T20:31:44Z

Kayla Murray at 00:54, 27 February 2017

2017-02-27T00:54:35Z

Kayla Murray at 00:08, 19 February 2017

2017-02-19T00:08:53Z

Kayla Murray at 23:36, 7 February 2017

2017-02-07T23:36:43Z

Kayla Murray at 16:05, 7 February 2017

2017-02-07T16:05:51Z

Kayla Murray at 16:04, 7 February 2017

2017-02-07T16:04:04Z

Kayla Murray at 17:50, 6 February 2017

2017-02-06T17:50:50Z

@@ Line 5: / Line 5: @@
 <span class="exam">(b) &nbsp; <math>\int _{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\cos(x)}{\sin^2(x)}~dx</math>
 <hr>
-(insert picture of handwritten solution)
+[[009B Sample Midterm 1, Problem 2 Solution|'''<u>Solution</u>''']]
 [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
-<span class="exam"> Otis Taylor plots the price per share of a stock that he owns as a function of time
+<span class="exam">Evaluate the indefinite and definite integrals.
-<span class="exam">and finds that it can be approximated by the function
+<span class="exam">(a) &nbsp; <math>\int x^2\sqrt{1+x^3}~dx</math>
-::<math>s(t)=t(25-5t)+18</math>
+<span class="exam">(b) &nbsp; <math>\int _{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\cos(x)}{\sin^2(x)}~dx</math>
-<span class="exam">where &nbsp;<math style="vertical-align: 0px">t</math>&nbsp; is the time (in years) since the stock was purchased.
+[[009B Sample Midterm 1, Problem 2 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']]
 [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 50: / Line 50: @@
 \displaystyle{s_{\text{avg}}} & = & \displaystyle{\frac{1}{5}\bigg[\frac{25(5)^2}{2}-\frac{5(5)^3}{3}+18(5)\bigg]-0}\\
 &&\\
-& = & \displaystyle{\frac{233}{6}.}
+& = & \displaystyle{\frac{233}{6}}\\
 \end{array}</math>
@@ Line 59: / Line 61: @@
 !Final Answer: &nbsp;
 |-
-| &nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{233}{6}</math>
+| &nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{233}{6}\approx $38.83</math>
 |-
+|
 |}
 [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 35: / Line 35: @@
 |First, we distribute to get
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>s_{\text{avg}}=\frac{1}{5} \int_0^5 25t-t^2+18~dt.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>s_{\text{avg}}=\frac{1}{5} \int_0^5 25t-5t^2+18~dt.</math>
 |-
 |Then, we integrate to get

@@ Line 5: / Line 5: @@
 ::<math>s(t)=t(25-5t)+18</math>
-<span class="exam">where <math style="vertical-align: 0px">t</math> is the time (in years) since the stock was purchased.
+<span class="exam">where &nbsp;<math style="vertical-align: 0px">t</math>&nbsp; is the time (in years) since the stock was purchased.
 <span class="exam">Find the average price of the stock over the first five years.
@@ Line 13: / Line 13: @@
 !Foundations: &nbsp;
 |-
-|The average value of a function <math style="vertical-align: -5px">f(x)</math> on an interval <math style="vertical-align: -5px">[a,b]</math> is given by
+|The average value of a function &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; on an interval &nbsp;<math style="vertical-align: -5px">[a,b]</math>&nbsp; is given by
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -18px">f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)~dx.</math>
@@ Line 23: / Line 23: @@
 !Step 1: &nbsp;
 |-
-|This problem wants us to find the average value of <math style="vertical-align: -5px">s(t)</math> over the interval <math style="vertical-align: -5px">[0,5].</math>
+|This problem wants us to find the average value of &nbsp;<math style="vertical-align: -5px">s(t)</math>&nbsp; over the interval &nbsp;<math style="vertical-align: -5px">[0,5].</math>
 |-
 |Using the average value formula, we have

@@ Line 15: / Line 15: @@
 |The average value of a function <math style="vertical-align: -5px">f(x)</math> on an interval <math style="vertical-align: -5px">[a,b]</math> is given by
 |-
-|&nbsp; &nbsp; <math style="vertical-align: -18px">f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)~dx.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -18px">f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)~dx.</math>
 |}
@@ Line 25: / Line 25: @@
 |This problem wants us to find the average value of <math style="vertical-align: -5px">s(t)</math> over the interval <math style="vertical-align: -5px">[0,5].</math>
 |-
-|Using the formula given in Foundations, we have:
+|Using the average value formula, we have
 |-
-| &nbsp; &nbsp;<math style="vertical-align: 0px">s_{\text{avg}}=\frac{1}{5-0} \int_0^5 t(25-5t)+18~dt.</math>
+| &nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: 0px">s_{\text{avg}}=\frac{1}{5-0} \int_0^5 t(25-5t)+18~dt.</math>
 |}
@@ Line 35: / Line 35: @@
 |First, we distribute to get
 |-
-|&nbsp; &nbsp; <math>s_{\text{avg}}=\frac{1}{5} \int_0^5 25t-t^2+18~dt.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>s_{\text{avg}}=\frac{1}{5} \int_0^5 25t-t^2+18~dt.</math>
 |-
 |Then, we integrate to get
 |-
-|&nbsp; &nbsp; <math>s_{\text{avg}}=\left. \frac{1}{5}\bigg[\frac{25t^2}{2}-\frac{5t^3}{3}+18t\bigg]\right|_0^5.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>s_{\text{avg}}=\left. \frac{1}{5}\bigg[\frac{25t^2}{2}-\frac{5t^3}{3}+18t\bigg]\right|_0^5.</math>
 |}
@@ Line 47: / Line 47: @@
 |We now evaluate to get
 |-
-|&nbsp; &nbsp; <math>\begin{array}{rcl}
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 \displaystyle{s_{\text{avg}}} & = & \displaystyle{\frac{1}{5}\bigg[\frac{25(5)^2}{2}-\frac{5(5)^3}{3}+18(5)\bigg]-0}\\
 &&\\
@@ Line 59: / Line 59: @@
 !Final Answer: &nbsp;
 |-
-| &nbsp; &nbsp; <math>\frac{233}{6}</math>
+| &nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{233}{6}</math>
 |-
+|
 |}
 [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]