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> | ||
+ | |||
+ | '''Exercises''' | ||
+ | |||
+ | '''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>. | ||
+ | {| class = "mw-collapsible mw-collapsed" style = "text-align:left;" | ||
+ | !Solution: | ||
+ | |- | ||
+ | |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> | ||
+ | |- | ||
+ | |Check which function is the top function by choosing one number in between the bound and plug in: | ||
+ | |- | ||
+ | |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_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
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: |
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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}} |
This page were made by Tri Phan