• Apr 4th 2013, 03:33 PM
Steelers72
I solved that the integral for this is Integral -1 to 1 (y^2 -3 -e^y) dy

antiderivative of that is y^3/3 -3y -e^y from -1 to 1

plug in those intervals, you get

(1^3/3-3(1)-e^1) - (-1^3 -3(-1)-e^-1)

What would be the final answer of that? I'm terrible with calculating the end result, but I understand the problem as a whole though. Thanks!
• Apr 4th 2013, 04:39 PM
Soroban
Hello, Steelers72!

Quote:

I solved: . $\displaystyle \int^1_{\text{-}1}\left(y^2 - 3 - e^y)\,dy$

and got: . $\tfrac{1}{3}y^3 -3y - e^y\,\bigg]^1_{\text{-}1}$

Plug in those limits and get: . $\left[\tfrac{1}{3}(1^3) - 3(1)- e^1\right] - \left[\tfrac{1}{3}(\text{-}1)^3 -3(\text{-}1)-e^{\text{-}1}\right]$

What would be the final answer of that? . . . . . . . . . . LOL!
I'm terrible with calculating the end result, . . . . . . . . Your calculus is excellent,
but I understand the problem as a whole though. . . . but you suck at arithmetic ??

We have: . $\left(\tfrac{1}{3} - 3 - e\right) - \left(\text{-}\tfrac{1}{3} + 3 - \tfrac{1}{e}\right)$

. . . . . . $=\;\tfrac{1}{3} - 3 - e + \tfrac{1}{3} - 3 + \tfrac{1}{e}\right)$

. . . . . . $=\;\tfrac{2}{3} - 6 + \tfrac{1}{e} - e$

. . . . . . $=\;-\frac{16}{3} + \frac{1-e^2}{e}$
• Apr 8th 2013, 08:06 PM
Steelers72