Find the area of the region between y=X^2-2x and the x axis for the interval from 0 to 2.

Integrate [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) \[[((2)^3/3))-(2)^2] - [0]\]

thank you!

ok same question but the second part is to find the area for the interval from 0 to 3

Same method, just plug "3" in for x instead of 2 in the last step: \[[((3)3/3))−(3)2]−[0]\]

Oops, that came out all wrong! Here it is again: \[[((3)^3/3))−(3)^2]−[0]\]

yea for some reason i'm getting the wrong answer :S this method should give me 0 right?

No, 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 x-axis). 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^2-2x) and g(x) = 0. Sorry if that was confusing!

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