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anonymous
 5 years ago
Find the area of the region between y=X^22x and the x axis for the interval from 0 to 2.
anonymous
 5 years ago
Find the area of the region between y=X^22x and the x axis for the interval from 0 to 2.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Integrate [f(x)  g(x)] on the given interval (in this case g(x) = 0 because it's the x axis): \[\int\limits_{0}^{2} (x ^{2}2x) dx\] \[[(x ^{3}/3)(2x ^{2}/2)]\] (from 0 to 2, I don't know how to make the editor do stacked #'s) \[[((2)^3/3))(2)^2]  [0]\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok same question but the second part is to find the area for the interval from 0 to 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Same method, just plug "3" in for x instead of 2 in the last step: \[[((3)3/3))−(3)2]−[0]\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oops, that came out all wrong! Here it is again: \[[((3)^3/3))−(3)^2]−[0]\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea for some reason i'm getting the wrong answer :S this method should give me 0 right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, it should equal some number greater than or equal to 0, since it's an area (you can get negative if the area is below the xaxis). So for the first one: \[[((2)^3/3))−(2)^2]−[0] = [(8/3)  4] = 4/3\] It's probably should have made it more clear that f(x) = (x^22x) and g(x) = 0. Sorry if that was confusing!
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