Definite integrals of absolute values

[Hurricane3]

Homework Statement

\int|x^{2}+x-2|dx from -2 to 2

Homework Equations

The integral of f(x) from a to b = F(b) - F(a)
|x| = { x if x >= 0; -x if x < 0

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Ok, I don't know how to do the definite integrals of absolute values.. was never shown an example of it in class, but I kind of have an idea...

we know that
|x| = { x if x >= 0; -x if x < 0,

and then I am lost from here...

btw, x= 1, -2

 

[Dick]

If you want to apply your definition of abs, you'll first have to figure out where x^2+x-2 is positive and negative.

 

[rock.freak667]

quick sketches of the function will help you...

 

[Hurricane3]

ok so from -2 to 1, it is negative and from 1 to 2 it is positive...

do i just split up the integral from here?
like from -2 to 1, its negative, so the integral would be negative, and from 1 to 2, it is postive so the integral would be postive...

and just from inspection, would the integral be equal to 0 because its symmetric and one is a negative the other is postive...

 

[Dick]

The integral of x^2+x-2 from -2 to 1 IS negative. The integral of |x^2+x-2| is not. Do you see why?

 

[Hurricane3]

nvm. i got 19/3...

 

[Hurricane3]

The integral of x^2+x-2 from -2 to 1 IS negative. The integral of |x^2+x-2| is not. Do you see why?

I don't know if I do..

is it because the absolute value of any function is always positive?

 

[rock.freak667]

Well from 2 to one it is +ve so you want to get
\int_1 ^{2} (x^2+x-2)dx

and from 1 to -2 it is -ve so you want to find

\int_{-2} ^{1} -(x^2+x-2)dx

and sum them up