Calcuate the integral $\int_1^{-1}dx\int_x^{2x}e^{x+y}dy$
Here is my working out:
Split the integral into their respective derivate parts: -
$$\int_1^{-1}e^xdx\int_x^{2x}e^ydy$$
The RHS is given as :
$$e^x\left(e^{2x}-e^x\right)$$
Finally:
$$\int^{-1}_1e^{3x}dx-\int^{-1}_1e^2xdy$$
Which is equal to:
$$\frac{1}{3}\left(e^{-3}-e^3 \right)+\frac{1}{2}\left(e^{2}-e^{-2} \right)$$
I know that $\frac{1}{2}\left(e^{2}-e^{-2} \right) = \sinh(2)$
However, in my solutons it says that: $\frac{1}{3}\left(e^{-3}-e^3 \right) = -\frac{2}{3}\sinh(3)$, although I cannot seem to figure out how this was achieved, any clarification on this will be very helpful!
$$\because\sinh(x)=\frac{e^x-e^{-x}}{2}$$ $$e^3-e^{-3}=2\sinh(3)$$
So $$\Longrightarrow-\frac{e^3-e^{-3}}{3} =(-2\sinh(3))\frac{1}{3} $$