009B Sample Final 1, Problem 1 - Revision history

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MathAdmin at 22:09, 18 April 2016

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MathAdmin at 05:47, 27 February 2016

2016-02-27T05:47:29Z

MathAdmin: Created page with "Consider the region bounded by the following two functions: :::::::: $y=2(-x^2+9)$ and
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Created page with "Consider the region bounded by the following two functions: :::::::: <math style="vertical-align: -5px">y=2(-x^2+9)</math> and <math styl..."

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Consider the region bounded by the following two functions:
:::::::: <math style="vertical-align: -5px">y=2(-x^2+9)</math> and <math style="vertical-align: -4px">y=0</math>.

a) Using the lower sum with three rectangles having equal width, approximate the area.

b) Using the upper sum with three rectangles having equal width, approximate the area.

c) Find the actual area of the region.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|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.
|-
|'''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.
|-
|'''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>.
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|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>.
|-
|This means that we need to calculate the Riemann sums over the interval <math style="vertical-align: -5px">[-3,3]</math>.
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|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,
|-
|each rectangle will have width <math style="vertical-align: 0px">2</math> .
|-
|Thus, the lower Riemann sum is
|-
|
::<math>2(f(-3)+f(-1)+f(3))\,=\,2(0+16+0)\,=\,32.</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|As in Part (a), the length of our inteval is <math style="vertical-align: 0px">6</math> and
|-
|each rectangle will have width <math style="vertical-align: 0px">2</math>. (See Step 1 and 2 for '''(a)''')
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Thus, the upper Riemann sum is
|-
|
::<math>2(f(-1)+f(-1)+f(1))\,=\,2(16+16+16)\,=\,96.</math>
|}

'''(c)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|To find the actual area of the region, we need to calculate
|-
|
::<math>\int_{-3}^3 2(-x^2+9)~dx.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|We integrate to get
|-
|
::<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{-3^3}{3}+9\times 3\bigg)-2\bigg(\frac{-(-3)^3}{3}+9(-3)\bigg)}\\
&&\\
& = & \displaystyle{2(-9+27)-2(9-27)}\\
&&\\
& = & \displaystyle{2(18)-2(-18)}\\
&&\\
& = & \displaystyle{72}.\\
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)'''  <math style="vertical-align: 0px">32</math>
|-
|'''(b)'''  <math style="vertical-align: 0px">96</math>
|-
|'''(c)'''  <math style="vertical-align: 0px">72</math>
|}
[[009B_Sample_Final_1|'''Return to Sample Exam''']]

@@ Line 113: / Line 113: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)''' &nbsp;<math style="vertical-align: 0px">32</math>
+|&nbsp;&nbsp; '''(a)''' &nbsp;<math style="vertical-align: 0px">32</math>
 |-
-|'''(b)''' &nbsp;<math style="vertical-align: 0px">96</math>
+|&nbsp;&nbsp; '''(b)''' &nbsp;<math style="vertical-align: 0px">96</math>
 |-
-|'''(c)''' &nbsp;<math style="vertical-align: 0px">72</math>
+|&nbsp;&nbsp; '''(c)''' &nbsp;<math style="vertical-align: 0px">72</math>
 |}
 [[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]
 [[File:9BF1 1 GP.png|center|1000px]]

@@ 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> and <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.
+::<span class="exam">a) Using the lower sum with three rectangles having equal width, approximate the area.
-<span class="exam">b) Using the upper sum with three rectangles having equal width, approximate the area.
+::<span class="exam">b) Using the upper sum with three rectangles having equal width, approximate the area.
-<span class="exam">c) Find the actual area of the region.
+::<span class="exam">c) Find the actual area of the region.
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

← Older revision		Revision as of 05:47, 27 February 2016
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	[[009B_Sample_Final_1\|'''<u>Return to Sample Exam</u>''']]		[[009B_Sample_Final_1\|'''<u>Return to Sample Exam</u>''']]
		+
		+	[[File:9BF1 1 GP.png\|center\|1000px]]

@@ 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.
 |-
-|'''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.
 |-
-|'''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>.
 |}
@@ Line 29: / 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
 |-
-|This means that we need to calculate the Riemann sums over the interval <math style="vertical-align: -5px">[-3,3]</math>.
 |-
+|Solving for <math style="vertical-align: 0px">x,</math> we get <math style="vertical-align: 0px">x=\pm 3.</math>
 |}
@@ Line 41: / Line 53: @@
 |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,
 |-
-|each rectangle will have width <math style="vertical-align: 0px">2</math>&thinsp;.
+|each rectangle will have width <math style="vertical-align: 0px">2.</math>&thinsp;
 |-
 |Thus, the lower Riemann sum is
@@ Line 56: / Line 68: @@
 |As in Part (a), the length of our inteval is <math style="vertical-align: 0px">6</math> 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 <math style="vertical-align: 0px">2.</math> (See Step 1 and 2 for '''(a)''')
 |}