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anonymous
 4 years ago
find the area enclosed between the curve y=x^3 and the line y=x
anonymous
 4 years ago
find the area enclosed between the curve y=x^3 and the line y=x

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0did you sketch the region?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0im not sure on how to

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you're not sure how to sketch x^3 ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no i mean i sketched it, i dont know what to do after

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0say, for example, you're asking about the area enclosed for positive values of x..: dw:1355190249043:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you can see that it's the area under the line x (from 0 to 1) minus the area under x^3 (from 0 to 1)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0aka:\[\int\limits_{0}^{1} xx ^{3} dx\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can you evaluate that integral?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and then how do we find the anti derivitve of that hat

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the usual way:\[\int\limits_{ }^{ } x^n dx = \frac{ x^{n+1} }{ n+1 }\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so for \[\int\limits_{ }^{} x^1 dx\] \[\int\limits_{ }^{} x^1 dx = \frac{ x^{1+1} }{ 1+1}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so \[\frac{ x^2 }{ 2 } \frac{ x^4 }{4 } \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for the region on the interval x =0...+inf ( 1/2 1/4 )  (0/2 0/4)
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