Question

INTEGRATE ABSOLUTE VALUE- 2x-x^2 from -4 t0 3

Answer 1

this was my approach, x(2-x) =0
x =2
but then I ten to realise that as the numbers get to be >2, the fucntion is negative, same when they <2.....or am i missing something?

Answer 2

integrated:
x^2 - x^3/3

evaluate at 3
3^2 - 3^3/3 = 0

evaluate at -4
-4^2 + 64/3 = 16/3

Answer 3

i think you're trying to integrate this :

\(\large \int\limits_{-4}^{3}|2x-x ^{2}|dx\)

?

Answer 4

yes, I forgot the bars, thank you

Answer 5

x(2-x) =0

x = 0, x = 2


x >2 : negative
x < 2 : positive
x < 0 : negative

Answer 6

so,

-4->0 : |2x-x^2| = x^2-2x
0->2 : |2x-x^2| = 2x - x^2
2 -> 3 : |2x-x^2| = x^2-2x

Answer 7

wow, hey, sorry I was waayy off! i'll keep my mouth shut and learn from the masters!

Answer 8

you need to split it into 3 integrals

Answer 9

ok, let me try that. because when its less than 2, its not at all positive

Answer 10

you're right it is positive only between 2 and 0
x < 0 is negative .... thats the reason, we need 3 integrals

Answer 11

spoiler alert : you should get :

http://www.wolframalpha.com/input/?i=%5Cint+-4+to+0+%28x%5E2-2x%29+dx+%2B+%5Cint++0+to+2+%282x-x%5E2%29+dx+%2B+%5Cint+2+to+3+%28x%5E2-2x%29+dx

Answer 12

yes, i got 40. I went wolfram first, but then trying to figure out reverse wise was becoming difficult. Thank you, but then explain to me why we broke it into three.

Answer 13

good question :)

Answer 14

we broke it cuz we dont know how to take integral of absolute value