# Evaluate this integral

• Feb 22nd 2013, 12:34 PM
Steelers72
Evaluate this integral

Ok my work:

b/c of abs value law x { x when x>/= to 0
{ -x when x<0

Intergral from -2 to 0 (x-2(-x))dx + Integral from 0 to 3 (x-2(x)) dx

= integral from -2 to 0 (3x) dx + integral from 0 to 3 -x dx

Antiderivative= (3x^2)/2 | -2 to 0 intervals

Antiderivative= -x^2/2 | 0 to 3 intervals

you plug in each and i got

= (0 - (3(4))/2) + ( 9/2 - 0) = -1.5 which they said was wrong?! and they wanted as an improper fraction so that is -3/2 but I only have one more try to enter in an answer and I want to be 100% sure I am correct.

Any help greatly appreciated!!

• Feb 22nd 2013, 02:16 PM
BobP
Re: Evaluate this integral
Look again at the second of the two integrals, I think that you've lost a negative sign. And yes, if the question asks for the answer to be given as an improper fraction then that is what you should do.
• Feb 22nd 2013, 08:48 PM
Steelers72
Re: Evaluate this integral
ok so it should be (0 - (3(4))/2) + (-9/2 - 0) = 6+ (-9/2) which is 1.5 positive sign ==> 3/2. Do you agree?
• Feb 23rd 2013, 06:45 AM
BobP
Re: Evaluate this integral
No, now you've lost a negative sign from the first integral.
• Feb 23rd 2013, 04:00 PM
Steelers72
Re: Evaluate this integral
Quote:

Originally Posted by BobP
No, now you've lost a negative sign from the first integral.

(0 - (3(4))/2) + (-9/2 - 0) = -6+ (-9/2) which is -10.5 or -21/2 ? right?
• Feb 24th 2013, 12:27 AM
BobP
Re: Evaluate this integral
Yes.