Evaluate the double integral for $f(x,y)=ye^x$ given $R = [2,4] \times [1,9]$

Question

So I did first did it by integrating with respect to $y$ first then $x$ and eventually got the answer of $40(e^4-e^2)$, which is correct. But when I attempted to apply Fubini's theorem and switch the order of integration, I should get the same answer but I'm not. Possible made an error somewhere?

What I did:

\begin{align}\text{Double integral} &= \int_1^9 \int_2^4 ye^x\, dx\, dy \\ &= \int_1^9 \left.ye^x \right\vert_{x=2}^{x=4}\, dy \\ &= \int_1^9 (ye^4 - ye^2) \,dy\\ &= \left[\left[\frac{y^2}2\right]e^4 - \left[\frac{y^2}2\right]e^2 \right]_ {y=1}^{y=9}\\ &= \end{align} and I got similar numbers as to when I integrated with $dx$ then $dy$, but couldn't get $40(e^4-e^2)$.

Can someone show their algebra for after integrating?

Answer 1

$$\left. \frac{y^2}{2}\right\vert_{y=1}^{y=9}=\frac{9^2-1^2}{2}=\frac{80}{2}=40$$

Hence the value is indeed $40(e^4-e^2)$