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

Question

anonymous · Answer

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)
$$[((2)^3/3))-(2)^2] - [0]$$

anonymous · Answer

thank you!

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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!