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

From Math Wiki
Jump to navigation Jump to search
Line 20: Line 20:
 
|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.
 
|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.
 
|-
 
|-
|<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>\textit{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>
 
|}
 
|}
  

Revision as of 07:53, 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 \textit{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