integrate 2t + cos t from -pi/2 to pi/2 - Wolfram|Alpha
Your answer is 2? looks fine.
so at the very end when i integrated (2t + cost) i got (t^2 + sint)..... then i plugged in positive and negative Pi/2...
in which i got [(Pi/2)^2 + sin(Pi/2) - (-Pi/2)^2 + sin(-Pi/2)]......
then i got [((Pi^2/4) + 1) - (-Pi^2/4) + -1]..... IS THAT RIGHT?
integrate 2t + cos t from -pi/2 to pi/2 - Wolfram|Alpha
Your answer is 2? looks fine.
Well you're supposed to get 2 as your evaluated integral from -pi/2 to pi/2 and what you have looks like 2pi^2/4+2. When you squared the negative pi/2 the negative goes away and you're left with pi^2/4 so you can cancel out with pi^2/4 .