Probability double integration 
My work:
I solved first one using $$\int\limits_{0}^{1}\int\limits_{0}^{x}ce^{x^2}dydx = 1$$ and obtained $$c=\frac{2}{e-1}$$
SO for next question I did the same,
$$P = \int\limits_{0}^{1}\int\limits_{0}^{2x}\frac{2}{e-1}e^{x^2}dy dx=2  \text{     not possible  ?}$$
I am not sure where I went wrong, kindly tell me
 A: With Fubini's theorem, because everything is measurable and positive:
\begin{align}
\int_{[0,1]\times[0,1]} f(x,y)\mathrm{d}\lambda &= \int_{[0,1]\times[0,1]} ce^{x^2}\mathbb{1}_{\{y\leqslant x\}}(x,y)\mathrm{d}\lambda \\
&= \int_{[0,1]}\left(\int_{[0,1]}c e^{x^2}\mathbb{1}_{\{y\leqslant x\}}(x,y) \mathrm{d}y \right)\mathrm{d}x \\
&=\int_0^1\left(\int_0^xc e^{x^2}\mathrm{d}y \right)\mathrm{d}x\\
&=\int_0^1 cxe^{x^2}\mathrm{d}x
\end{align}
Thus it equals $1$ if and only if $c = \left(\int_0^1xe^{x^2}\mathrm{d}x \right)^{-1}$ which you can easily compute thanks to an antiderivative.
Now you can calculate:
\begin{align}
\mathbb{P}(y\leqslant 2x) &=\mathbb{E}[\mathbb{1}_{\{y\leqslant 2x\}}] \\
&= \int_{[0,1]\times[0,1]} \mathbb{1}_{\{y\leqslant 2x\}} f(x,y) \mathrm{d}\lambda
\end{align}
You can remark that $f = f \times \mathbb{1}_{\{y\leqslant 2x\}}$! So the integral is $1$, and $y \leqslant 2x$ almost surely.
A: There may be an issue in question 2,  because $x \ge 0$ implies $x \le 2x$, in which case $y \le x$ implies $y \le 2x$, and $P(Y \le 2X)=1$ as you say.  
If question 2 was instead to repeat question 1 but with $y \le 2x$ instead of $y \le x$ then you would be looking for the $c$ satisfying  $$\int\limits_{0}^{1}\int\limits_{0}^{\min(1,2x)}ce^{x^2}dy\,dx = 1$$  and you might find this easier to consider $$\int\limits_{0}^{1/2}\int\limits_{0}^{2x}ce^{x^2}dy\,dx + \int\limits_{1/2}^{1}\int\limits_{0}^{1}ce^{x^2}dy\,dx = 1$$  though the second of those integrals is not simple
