## anonymous 5 years ago Integrate b=0 a=2 x(x-3) dx=??

1. anonymous

$\int\limits_{0}^{2} x(x-3) dx=?$

2. anonymous

Multiply it out and integrate term by term. $= \int\limits_{0}^{2} x^2 - 3x\ dx$

3. anonymous

(x^3) -(3x^2)/2 and then use the integration limits

4. anonymous

(x^3)/3 -(3x^2)/2 *

5. anonymous

i know the answer is -10/3 but i cant seem to get it. Idk what im missing

6. anonymous

ok so we have (2^3)/3 = 8/3 and 3(2^2)/2 = 3*4/3 = 4 so for the limit 2 we get 8/3 - 4 = and limit 0 will just be 0 for x so it will turn out to be 0. so final answer is 8/3 -4 = 8/3 - 12/3 = -4/3

7. anonymous

woops hold on lol error

8. anonymous

the answer is -10/3 according to the book. I think where you have 4, it should be 6 right? that works out i think

9. anonymous

it shuld be 8/3 - 6 = 8/3 - 18/6 = -10/3

10. anonymous

8/3 - 18/3 = -10/3 *

11. anonymous

thank you for the help... im new to the site... anybody want to give me a quick overview?

12. anonymous

in general, we post doubts and other people try to help...

13. anonymous

i see that... i dont know how i have never heard of this site... it would have been a lifesaver for calculus this semester

14. anonymous

hahaha!