Thread: Evaluate this integral

Evaluate the integral. (Give your answer as an improper fraction.)


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!!

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.

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?

No, now you've lost a negative sign from the first integral.

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?

Yes.