Answer
$ 0.3461$
Work Step by Step
We are given $f(x)=e^{x^2}$
with $a_0=1$
The Taylor series for $e^x$
is $1+x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+...+\frac{1}{n!}x^n+...$
which converges for $x$ in $(-\infty,\infty)$; replace each $x$ with $x^2$ to get
$e^{x^2}=1+x^2+\frac{1}{2!}(x^2)^2+\frac{1}{3!}(x^2)^3+\frac{1}{4!}(x^2)^4$
$=1+x^2+\frac{1}{2}x^4+\frac{1}{6}x^6+\frac{1}{24}x^8$
$\int^{\frac{1}{3}}_0 e^{x^2}dx \approx \int^{\frac{1}{3}}_0(1+x^2+\frac{1}{2}x^4+\frac{1}{6}x^6+\frac{1}{24}x^8)dx$
$=(x+\frac{x^3}{3}+\frac{x^5}{10}+\frac{x^{7}}{42}+\frac{x^{9}}{216})|^{\frac{1}{3}}_0$
$\approx 0.3461$
