Difference between revisions of "Math 22 The Area of a Region Bounded by Two Graphs"

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   <math>A=\int_a^b [f(x)-g(x)]dx</math>
 
   <math>A=\int_a^b [f(x)-g(x)]dx</math>
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'''Exercises'''
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'''1)''' Find the area of the region bounded by the graph of <math>f(x)=x^2</math> and the graph of <math>g(x)=x^3</math>.
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{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
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!Solution: &nbsp;
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|-
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|Find the bound of the region by setting <math>f(x)=g(x)</math>, so <math>x^2=x^3</math>, hence <math>x^3-x^2=0</math>, then <math>x^2(x-1)=0</math>, therefore <math>x=0</math> and <math>x=1</math>
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|-
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|Check which function is the top function by choosing one number in between the bound and plug in:
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|-
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|Pick <math>x=\frac{1}{2}</math>, so <math>f(\frac{1}{2})=\frac{1}{4}</math>, and <math>g(\frac{1}{2})=\frac{1}{8}</math>. Therefore, <math>f(x)</math> will be the top function.
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|<math>\text{Area}=\int_0^1 [f(x)-g(x)]dx=\int_0^1 [x^2-x^3]dx=[\frac{1}{3}x^3-\frac{1}{4}x^4]\Biggr_0^1=\frac{1}{3}(1)^3-\frac{1}{4}(1)^4=\frac{1}{12}</math>
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|}
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[[Math_22| '''Return to Topics Page''']]
 
[[Math_22| '''Return to Topics Page''']]
  
 
'''This page were made by [[Contributors|Tri Phan]]'''
 
'''This page were made by [[Contributors|Tri Phan]]'''

Revision as of 07:52, 17 August 2020

Area of a Region Bounded by Two Graphs

Region 2 graph.png

 If  and  are continuous on  and  for all  in , 
 then the area of the region bounded by the graphs of  given by
 
 

Exercises

1) Find the area of the region bounded by the graph of and the graph of .

Solution:  
Find the bound of the region by setting , so , hence , then , therefore and
Check which function is the top function by choosing one number in between the bound and plug in:
Pick , so , and . Therefore, will be the top function.
Failed to parse (syntax error): {\displaystyle \text{Area}=\int_0^1 [f(x)-g(x)]dx=\int_0^1 [x^2-x^3]dx=[\frac{1}{3}x^3-\frac{1}{4}x^4]\Biggr_0^1=\frac{1}{3}(1)^3-\frac{1}{4}(1)^4=\frac{1}{12}}


Return to Topics Page

This page were made by Tri Phan