009B Sample Final 2, Problem 2 - Revision history

Kayla Murray at 23:05, 4 March 2017

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Kayla Murray at 22:59, 4 March 2017

2017-03-04T22:59:51Z

Kayla Murray at 22:59, 4 March 2017

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Kayla Murray at 22:57, 4 March 2017

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Kayla Murray at 22:42, 4 March 2017

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Kayla Murray at 20:44, 3 March 2017

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Kayla Murray at 02:22, 19 February 2017

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Kayla Murray at 03:09, 18 February 2017

2017-02-18T03:09:01Z

Kayla Murray: Created page with "Consider the region bounded by the following two functions: :::::::: $y=2(-x^2+9)$ and
2017-02-18T02:43:20Z

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..."

New page
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.

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'''Solution:'''

'''(a)'''

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'''(b)'''

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'''(c)'''

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[[009B_Sample_Final_2|'''Return to Sample Exam''']]

@@ Line 37: / Line 37: @@
 \end{array}</math>
 |-
-|Therefore, we get that these two curves intersect at &nbsp;<math>x=-2,~x=0,~x=2.</math>
+|Therefore, we get that these two curves intersect at &nbsp;<math style="vertical-align: -4px">x=-2,~x=0,~x=2.</math>
 |-
-|Hence, the region we are interested in occurs between &nbsp;<math>x=-2</math>&nbsp; and &nbsp;<math>x=2.</math>
+|Hence, the region we are interested in occurs between &nbsp;<math style="vertical-align: 0px">x=-2</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">x=2.</math>
 |}
@@ Line 45: / Line 45: @@
 !Step 2: &nbsp;
 |-
-|Since the curves intersect also intersect at &nbsp;<math>x=0,</math>&nbsp; this breaks our region up into two parts,
+|Since the curves intersect also intersect at &nbsp;<math style="vertical-align: -4px">x=0,</math>&nbsp; this breaks our region up into two parts,
 |-
-|which correspond to the interval &nbsp;<math>[-2,0]</math>&nbsp; and &nbsp;<math>[0,2].</math>
+|which correspond to the intervals &nbsp;<math style="vertical-align: -5px">[-2,0]</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">[0,2].</math>
 |-
-|Now, in each of the regions we need to determine which curve has the higher &nbsp;<math>y</math>&nbsp; value.
+|Now, in each of the regions we need to determine which curve has the higher &nbsp;<math style="vertical-align: -4px">y</math>&nbsp; value.
 |-
 |To figure this out, we use test points in each interval.
 |-
-|For &nbsp;<math>x=-1,</math>&nbsp; we have
+|For &nbsp;<math style="vertical-align: -5px">x=-1,</math>&nbsp; we have
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math> y=3(-1)-(-1)^2=-4</math>&nbsp; and &nbsp;<math>y=2(-1)^3-(-1)^2-5(-1)=2.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px"> y=3(-1)-(-1)^2=-4</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">y=2(-1)^3-(-1)^2-5(-1)=2.</math>
 |-
-|For &nbsp;<math>x=1,</math>&nbsp; we have
+|For &nbsp;<math style="vertical-align: -5px">x=1,</math>&nbsp; we have
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math> y=3(1)-(1)^2=2</math>&nbsp; and &nbsp;<math>y=2(1)^3-(1)^2-5(1)=-4.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px"> y=3(1)-(1)^2=2</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">y=2(1)^3-(1)^2-5(1)=-4.</math>
 |-
-|Hence, the area &nbsp;<math>A</math>&nbsp; of the region bounded by these two curves is given by
+|Hence, the area &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; of the region bounded by these two curves is given by
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp;<math>A=\int_{-2}^0 (2x^3-x^2-5x)-(3x-x^2)~dx+\int_0^2 (3x-x^2)-(2x^3-x^2-5x)~dx.</math>

@@ Line 55: / Line 55: @@
 |For &nbsp;<math>x=-1,</math>&nbsp; we have
 |-
-|&nbsp;<math> y=3(-1)-(-1)^2=-4</math>&nbsp; and &nbsp;<math>y=2(-1)^3-(-1)^2-5(-1)=2.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math> y=3(-1)-(-1)^2=-4</math>&nbsp; and &nbsp;<math>y=2(-1)^3-(-1)^2-5(-1)=2.</math>
 |-
 |For &nbsp;<math>x=1,</math>&nbsp; we have
 |-
-|&nbsp;<math> y=3(1)-(1)^2=2</math>&nbsp; and &nbsp;<math>y=2(1)^3-(1)^2-5(1)=-4.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math> y=3(1)-(1)^2=2</math>&nbsp; and &nbsp;<math>y=2(1)^3-(1)^2-5(1)=-4.</math>
 |-
 |Hence, the area &nbsp;<math>A</math>&nbsp; of the region bounded by these two curves is given by

@@ Line 49: / Line 49: @@
 |which correspond to the interval &nbsp;<math>[-2,0]</math>&nbsp; and &nbsp;<math>[0,2].</math>
 |-
-|Now, in each of the regions we need to determine which curve has the higher <math>y</math> value.
+|Now, in each of the regions we need to determine which curve has the higher &nbsp;<math>y</math>&nbsp; value.
 |-
 |To figure this out, we use test points in each interval.
 |-
-|For <math>x=-1,</math> we have
+|For &nbsp;<math>x=-1,</math>&nbsp; we have
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math> y=3(-1)-(-1)^2=-4</math>&nbsp; and &nbsp;<math>y=2(-1)^3-(-1)^2-5(-1)=2.</math>
+|&nbsp;<math> y=3(-1)-(-1)^2=-4</math>&nbsp; and &nbsp;<math>y=2(-1)^3-(-1)^2-5(-1)=2.</math>
 |-
-|For <math>x=1,</math> we have
+|For &nbsp;<math>x=1,</math>&nbsp; we have
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math> y=3(1)-(1)^2=2</math>&nbsp; and &nbsp;<math>y=2(1)^3-(1)^2-5(1)=-4.</math>
+|&nbsp;<math> y=3(1)-(1)^2=2</math>&nbsp; and &nbsp;<math>y=2(1)^3-(1)^2-5(1)=-4.</math>
 |-
-|Hence, the area <math>A</math> of the region bounded by these two curves is given by
+|Hence, the area &nbsp;<math>A</math>&nbsp; of the region bounded by these two curves is given by
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp;<math>A=\int_{-2}^0 (2x^3-x^2-5x)-(3x-x^2)~dx+\int_0^2 (3x-x^2)-(2x^3-x^2-5x)~dx.</math>

@@ Line 21: / Line 21: @@
 !Step 1: &nbsp;
 |-
+|First, we need to find the intersection points of these two curves.
 |-
+|To do this, we set
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>3x-x^2=2x^3-x^2-5x.</math>
 |-
+|Getting all the terms on one side of the equation, we get
 |}
@@ Line 33: / Line 45: @@
 !Step 2: &nbsp;
 |-
+|Since the curves intersect also intersect at &nbsp;<math>x=0,</math>&nbsp; this breaks our region up into two parts,
 |-
+|Hence, the area <math>A</math> of the region bounded by these two curves is given by
 |-
+|Now, we integrate to get
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |}
@@ Line 46: / Line 88: @@
 !Final Answer: &nbsp;
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>16</math>
 |}
 [[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
-<span class="exam"> Find the area of the region between the two curves <math>y=3x-x^2</math> and <math>y=2x^3-x^2-5x.</math>
+<span class="exam"> Find the area of the region between the two curves &nbsp;<math style="vertical-align: -4px">y=3x-x^2</math>&nbsp; and &nbsp;<math style="vertical-align: -4px">y=2x^3-x^2-5x.</math>
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 '''Solution:'''
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@@ Line 42: / Line 40: @@
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
+!Final Answer: &nbsp;
 |-
+|
 |}
 [[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 4: / Line 4: @@
 !Foundations: &nbsp;
 |-
+|'''1.''' You can find the intersection points of two functions, say &nbsp;<math style="vertical-align: -5px">f(x),g(x),</math>
 |-
+|
 |-
+|'''2.''' The area between two functions, &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">g(x),</math>&nbsp; is given by &nbsp;<math>\int_a^b f(x)-g(x)~dx</math>
 |-
+|
 |}

← Older revision		Revision as of 03:09, 18 February 2017
Line 1:		Line 1:
−	<span class="exam">~~Consider~~ the region ~~bounded by~~ the ~~following~~ two ~~functions:~~	+	<span class="exam"> Find the area of the region between the two curves <math>y=3x-x^2</math> and <math>y=2x^3-x^2-5x.</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">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.~~

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