009B Sample Final 1, Problem 1 - Revision history

Kayla Murray at 16:45, 28 February 2017

2017-02-28T16:45:59Z

Kayla Murray at 16:42, 28 February 2017

2017-02-28T16:42:20Z

Kayla Murray at 20:19, 27 February 2017

2017-02-27T20:19:51Z

Kayla Murray at 17:24, 27 February 2017

2017-02-27T17:24:54Z

Kayla Murray at 17:23, 27 February 2017

2017-02-27T17:23:11Z

Kayla Murray at 02:10, 19 February 2017

2017-02-19T02:10:46Z

Kayla Murray at 02:08, 19 February 2017

2017-02-19T02:08:46Z

Grad at 19:34, 25 February 2016

2016-02-25T19:34:50Z

Grad: /* Temp4 */

2016-02-25T19:34:35Z

Temp4

Grad: /* Temp3 */

2016-02-25T19:33:43Z

Temp3

@@ Line 46: / Line 46: @@
 |'''3.''' The height of each rectangle in the midpoint Riemann sum is given by
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -5px">\frac{f(a)+f(b)}{2}</math>&nbsp; where <math>a</math> is the left endpoint of the interval and <math>b</math> is the right endpoint of the interval.
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -14px">\frac{f(a)+f(b)}{2}</math>&nbsp; where &nbsp;<math>a</math>&nbsp; is the left endpoint of the interval and &nbsp;<math style="vertical-align: -1px">b</math>&nbsp; is the right endpoint of the interval.
 |}
@@ Line 69: / Line 69: @@
 !Step 2: &nbsp;
 |-
-|Let <math>s(t)</math> be the speed of the bee during the experiement.
+|Let &nbsp;<math style="vertical-align: -5px">s(t)</math>&nbsp; be the speed of the bee during the experiment.
 |-
 |Then, the left-hand Riemann sum is
@@ Line 98: / Line 98: @@
 !Step 2: &nbsp;
 |-
-|Let <math>s(t)</math> be the speed of the bee during the experiement.
+|Let &nbsp;<math style="vertical-align: -5px">s(t)</math>&nbsp; be the speed of the bee during the experiment.
 |-
 |Then, the Riemann sum using the midpoint rule is

@@ Line 40: / Line 40: @@
 !Foundations: &nbsp;
 |-
-|Recall:
+|'''1.''' The height of each rectangle in the left-hand Riemann sum is given by choosing
 |-
-|'''1.''' The height of each rectangle in the lower Riemann sum is given by choosing
+|&nbsp; &nbsp; &nbsp; &nbsp; the left endpoints of each interval.
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; the minimum &nbsp;<math style="vertical-align: -5px">y</math>&nbsp; value of the left and right endpoints of the rectangle.
+|'''3.''' The height of each rectangle in the midpoint Riemann sum is given by
 |-
-|'''2.''' The height of each rectangle in the upper Riemann sum is given by choosing
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -5px">\frac{f(a)+f(b)}{2}</math>&nbsp; where <math>a</math> is the left endpoint of the interval and <math>b</math> is the right endpoint of the interval.
 |}
@@ Line 63: / Line 57: @@
 !Step 1: &nbsp;
 |-
-|We need to set these two equations equal in order to find the intersection points of these functions.
+|To estimate the distance the bee moved during this experiment,
 |-
-|So, we let &nbsp;<math style="vertical-align: -5px">2(-x^2+9)=0</math>.&nbsp; Solving for &nbsp;<math style="vertical-align: -4px">x,</math>&nbsp; we get &nbsp;<math style="vertical-align: 0px">x=\pm 3</math>.
+|we need to calculate the left-hand Riemann sum over the interval &nbsp;<math style="vertical-align: -5px">[0,10].</math>
 |-
-|This means that we need to calculate the Riemann sums over the interval &nbsp;<math style="vertical-align: -5px">[-3,3]</math>.
+|Based on the information given in the table, we will have &nbsp;<math style="vertical-align: 0px">5</math>&nbsp; rectangles and
 |-
+|each rectangle will have width &nbsp;<math style="vertical-align: 0px">2.</math>
 |}
@@ Line 75: / Line 69: @@
 !Step 2: &nbsp;
 |-
-|Since the length of our interval is &nbsp;<math style="vertical-align: 0px">6</math>&nbsp; and we are using &nbsp;<math style="vertical-align: 0px">3</math>&nbsp; rectangles,
+|Let <math>s(t)</math> be the speed of the bee during the experiement.
 |-
-|each rectangle will have width &nbsp;<math style="vertical-align: 0px">2.</math>
+|Then, the left-hand Riemann sum is
 |-
+|
-::<math>2(f(-3)+f(-1)+f(3))\,=\,2(0+16+0)\,=\,32.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |}
@@ Line 90: / Line 86: @@
 !Step 1: &nbsp;
 |-
-|As in Part (a), the length of our interval is &nbsp;<math style="vertical-align: 0px">6</math>&nbsp; and
+|To estimate the distance the bee moved during this experiment,
 |-
+|we need to calculate the Riemann sum using the midpoint rule over the interval &nbsp;<math style="vertical-align: -5px">[0,10].</math>
 |-
-|To find the actual area of the region, we need to calculate
+|Based on the information given in the table, we will have &nbsp;<math style="vertical-align: 0px">5</math>&nbsp; rectangles and
 |-
+|each rectangle will have width &nbsp;<math style="vertical-align: 0px">2.</math>
 |}
@@ Line 118: / Line 98: @@
 !Step 2: &nbsp;
 |-
-|We integrate to get
+|Let <math>s(t)</math> be the speed of the bee during the experiement.
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\int_{-3}^3 2(-x^2+9)~dx} & = & \displaystyle{2\bigg(\frac{-x^3}{3}+9x\bigg)\bigg|_{-3}^3}\\
+\displaystyle{2\bigg(\frac{s(0)+s(2)}{2}+\frac{s(2)+s(4)}{2}+\frac{s(4)+s(6)}{2}+\frac{s(6)+s(8)}{2}+\frac{s(8)+s(10)}{2}\bigg)} & = & \displaystyle{1170\text{ cm}.}
 \end{array}</math>
 |}
@@ Line 138: / Line 112: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp;<math style="vertical-align: 0px">32</math>
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp;<math style="vertical-align: 0px">1182\text{ cm}</math>
 |-
-|&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp;<math style="vertical-align: 0px">72</math>
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp;<math style="vertical-align: 0px">1170\text{ cm}</math>
 |}
 [[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
-<span class="exam">Consider the region bounded by the following two functions:
+<span class="exam">Suppose the speed of a bee is given in the table.
-<span class="exam">(a) Using the lower sum with three rectangles having equal width, approximate the area.
+<table border="1" cellspacing="0" cellpadding="6" align = "center">
-<span class="exam">(b) Using the upper sum with three rectangles having equal width, approximate the area.
+</table>
-<span class="exam">(c) Find the actual area of the region.
+<span class="exam">(a) Using the given measurements, find the left-hand estimate for the distance the bee moved during this experiment.
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 21: / Line 21: @@
 |&nbsp; &nbsp; &nbsp; &nbsp; the maximum &nbsp;<math style="vertical-align: -5px">y</math>&nbsp; value of the left and right endpoints of the rectangle.
 |-
-|'''3.''' The area of the region is given by
+|'''3.''' The area of the region is given by &nbsp;<math style="vertical-align: -14px">\int_a^b y~dx</math>
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; for appropriate values &nbsp;<math style="vertical-align: -4px">a,b</math>.
@@ Line 38: / Line 36: @@
 |We need to set these two equations equal in order to find the intersection points of these functions.
 |-
-|So, we let &nbsp;<math style="vertical-align: -5px">2(-x^2+9)=0</math>.&nbsp; Solving for &nbsp;<math style="vertical-align: 0px">x,</math>&nbsp; we get &nbsp;<math style="vertical-align: 0px">x=\pm 3</math>.
+|So, we let &nbsp;<math style="vertical-align: -5px">2(-x^2+9)=0</math>.&nbsp; Solving for &nbsp;<math style="vertical-align: -4px">x,</math>&nbsp; we get &nbsp;<math style="vertical-align: 0px">x=\pm 3</math>.
 |-
 |This means that we need to calculate the Riemann sums over the interval &nbsp;<math style="vertical-align: -5px">[-3,3]</math>.

@@ Line 1: / Line 1: @@
 <span class="exam">Consider the region bounded by the following two functions:
-::<span class="exam"> <math style="vertical-align: -5px">y=2(-x^2+9)</math> and <math style="vertical-align: -4px">y=0</math>.
+::<span class="exam"> <math style="vertical-align: -5px">y=2(-x^2+9)</math>&nbsp; and &nbsp;<math style="vertical-align: -4px">y=0</math>.
 <span class="exam">(a) Using the lower sum with three rectangles having equal width, approximate the area.
@@ Line 13: / Line 13: @@
 |Recall:
 |-
-|'''1.''' The height of each rectangle in the lower Riemann sum is given by choosing the minimum <math style="vertical-align: -5px">y</math> value of the left and right endpoints of the rectangle.
+|'''1.''' The height of each rectangle in the lower Riemann sum is given by choosing
 |-
-|'''2.''' The height of each rectangle in the upper Riemann sum is given by choosing the maximum <math style="vertical-align: -5px">y</math> value of the left and right endpoints of the rectangle.
+|&nbsp; &nbsp; &nbsp; &nbsp; the minimum &nbsp;<math style="vertical-align: -5px">y</math>&nbsp; value of the left and right endpoints of the rectangle.
 |-
-|'''3.''' The area of the region is given by <math style="vertical-align: -14px">\int_a^b y~dx</math> for appropriate values <math style="vertical-align: -4px">a,b</math>.
+|'''2.''' The height of each rectangle in the upper Riemann sum is given by choosing
 |}
@@ Line 30: / Line 38: @@
 |We need to set these two equations equal in order to find the intersection points of these functions.
 |-
-|So, we let <math style="vertical-align: -5px">2(-x^2+9)=0</math>. Solving for <math style="vertical-align: 0px">x</math>, we get <math style="vertical-align: 0px">x=\pm 3</math>.
+|So, we let &nbsp;<math style="vertical-align: -5px">2(-x^2+9)=0</math>.&nbsp; Solving for &nbsp;<math style="vertical-align: 0px">x,</math>&nbsp; we get &nbsp;<math style="vertical-align: 0px">x=\pm 3</math>.
 |-
-|This means that we need to calculate the Riemann sums over the interval <math style="vertical-align: -5px">[-3,3]</math>.
+|This means that we need to calculate the Riemann sums over the interval &nbsp;<math style="vertical-align: -5px">[-3,3]</math>.
 |-
+|
@@ Line 40: / Line 48: @@
 !Step 2: &nbsp;
 |-
-|Since the length of our interval is <math style="vertical-align: 0px">6</math> and we are using <math style="vertical-align: 0px">3</math> rectangles,
+|Since the length of our interval is &nbsp;<math style="vertical-align: 0px">6</math>&nbsp; and we are using &nbsp;<math style="vertical-align: 0px">3</math>&nbsp; rectangles,
 |-
-|each rectangle will have width <math style="vertical-align: 0px">2</math>&thinsp;.
+|each rectangle will have width &nbsp;<math style="vertical-align: 0px">2.</math>
 |-
 |Thus, the lower Riemann sum is
@@ Line 55: / Line 63: @@
 !Step 1: &nbsp;
 |-
-|As in Part (a), the length of our inteval is <math style="vertical-align: 0px">6</math> and
+|As in Part (a), the length of our interval is &nbsp;<math style="vertical-align: 0px">6</math>&nbsp; and
 |-
-|each rectangle will have width <math style="vertical-align: 0px">2</math>. (See Step 1 and 2 for '''(a)''')
+|each rectangle will have width &nbsp;<math style="vertical-align: 0px">2.</math> (See Step 1 and 2 for (a))
 |}
@@ Line 103: / Line 111: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)''' &nbsp;<math style="vertical-align: 0px">32</math>
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp;<math style="vertical-align: 0px">32</math>
 |-
-|'''(b)''' &nbsp;<math style="vertical-align: 0px">96</math>
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp;<math style="vertical-align: 0px">96</math>
 |-
-|'''(c)''' &nbsp;<math style="vertical-align: 0px">72</math>
+|&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp;<math style="vertical-align: 0px">72</math>
 |}
 [[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 77: / Line 77: @@
 |-
+|
-::<math>\int_{-3}^3 2(-x^2+9)~dx</math>
+::<math>\int_{-3}^3 2(-x^2+9)~dx.</math>
 |}
@@ Line 95: / Line 95: @@
 & = & \displaystyle{2(18)-2(-18)}\\
 &&\\
-& = & \displaystyle{72}\\
+& = & \displaystyle{72}.\\
 \end{array}</math>
 |}
 == Temp4 ==
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"