Area under the shaded region answer check please?

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!

Re: Area under the shaded region answer check please?

Hello, Steelers72!

Quote:

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

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

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

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

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

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

. . . . . . $\displaystyle =\;-\frac{16}{3} + \frac{1-e^2}{e}$

Re: Area under the shaded region answer check please?

Haha yes I absolutely suck at doing that type of stuff without a calculator; no lies. I appreciate your help so much!!