## Thomas' Calculus 13th Edition

$3.71728$
Here, we have $\int dy=\int 2xe^{x^2} dx$ Now, $y=e^{x^2}+c$ Further, apply the initial conditions. $2=e^{(0^2)}+c$ or, $c=1$ Thus, $y=e^{x^2}+c \implies y=e^{x^2}+1$ Therefore, $y(1)=e^{(1)^2}+1 =e+1 \approx 3.71728$