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
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. |
Consumer Surplus and Producer Surplus
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