Area between $x=y^2-7$ and $x=e^y$ for $-1\leq y\leq 1$

Question

I need to find the area of the shaded region here

I thought the area would be $\int_{-1}^{1}{y^2-7-e^y}dy$,which got me $-\frac{40}{3}-3+\frac1e$, but I was marked wrong.

By inspection it looks like the area under the blue curve has a symmetry that makes its total area $0$ from $x=-7$ to $x=\frac12$. So I figured the area I'm looking for is the area under the red curve from $x=1/2$ to $x=3$. I tried this and got an answer of $\frac13 - 2$ which was also marked wrong.

What do I do?

Answer 1

Note the red curve is on the right and the blue is on the left. so the resulting area would be $$ \int_{-1}^{1} \left(e^y - \left(y^2-7\right) \right)dy $$ and if you did the calculations correctly, you would negate your first answer...

As Namburu Karthik points out below, you did make an arithmetic mistake evaluating the integral, which you can check at WolframAlpha

Answer 2

Of course it is wrong. You got a negative number!

Not that $y$ can take any value from $-1$ to $1$ and that, for each $y\in[-1,1]$, $x$ tan take any values from $y^2-7$ up to $e^y$. So, the area is equal to$$\int_{-1}^1\int_{y^2-t}^{e^y}1\,\mathrm dx\,\mathrm dy=\int_{-1}^1e^y-(y^2-t)\,\mathrm dy.$$

Answer 3

Consider interchanging axes that is easier to evaluate

$$\int_{-1}^1 \left[ e^x -\left(x^2-7\right)\right] dx $$