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

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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>
 
|<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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==Consumer Surplus and Producer Surplus==
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Revision as of 07:58, 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.

Consumer Surplus and Producer Surplus

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This page were made by Tri Phan