Difference between revisions of "Math 22 The Area of a Region Bounded by Two Graphs"
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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. | |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> | + | |<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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Revision as of 07:53, 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}} |
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