# Math Help - Another area of shaded region

1. ## Another area of shaded region

find the area of the shaded region

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

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

$\left[\tfrac{1}{3}(1^3) - 3(1)- e^1\right] - \left[\tfrac{1}{3}(\text{-}1)^3 -3(\text{-}1)-e^{\text{-}1}\right]$

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

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

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

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

this is my final answer but I am wrong. Can someone suggest where I went wrong? Thanks!

2. ## Re: Another area of shaded region

just add the portion of the function e^y
i.e integrate the function y^2-3+e^y from -1 t0 1 and you will find e -1/e +16/3

3. ## Re: Another area of shaded region

That's exactly what he says he got!

Steelers72, did you notice that your answer is negative? "Area" cannot be negative. Your error is that you have subtracted $e^y$ from $y^2- 3$ when, because, for every y, $e^y$ is larger than $y^2- 3$, you should be integrating $\int_{-1}^1 e^y- (y^2- 3) dy$.