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

1. anonymous

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

2. anonymous

thank you!

3. anonymous

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

4. anonymous

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

5. anonymous

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

6. anonymous

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

7. anonymous

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!