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evaluate the definite integral

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?

Re: evaluate the definite integral

Re: evaluate the definite integral

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 .

Re: evaluate the definite integral

You could write:

$\displaystyle 2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}t\,dt+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(t)\,dt$

Using the odd function rule on the first integral, and the even function rule on the second we have:

$\displaystyle 2\cdot0+2\int_{0}^{\frac{\pi}{2}}\cos(t)\,dt=2 \left[\sin(t) \right]_{0}^{\frac{\pi}{2}}=2\left(\sin\left(\frac{\pi}{2 } \right)-\sin(0) \right)=2(1-0)=2$

Re: evaluate the definite integral

so you plugged in -Pi/2 into cos(t) thats why you got 0? well why didn't you do the same thing for the positive Pi/2???

Re: evaluate the definite integral

so you plugged in -Pi/2 into cos(t) thats why you got 0? well why didn't you do the same thing for the positive Pi/2??????

Re: evaluate the definite integral

Quote:

Originally Posted by

**asilvester635** [((Pi^2/4) + 1) - (-Pi^2/4) + -1]..... IS THAT RIGHT?

Actually, line 2 is incorrect. You need to watch your signs more carefully: It should be

[(Pi/2)^2 + sin(Pi/2)**]** - **[**(-Pi/2)^2 + sin(-Pi/2)]

Leading to

Pi^2/4 + 1 - Pi^2/4 + 1

-Dan

Geez! A lot of posts were made while I wrote that!

Re: evaluate the definite integral

why did the exponent turn into 2/4?? what did you do with it?

Re: evaluate the definite integral

It **didn't**. If the exponent were 2/4, it would be written pi^(2/4) or pi^(1/2). pi^2/4 means (pi^2)/4.