Answer
$\approx0.517$
Work Step by Step
We are given $f(x)=\frac{1}{1-x^3}$
with $a_0=1$
The Taylor series for $\frac{1}{1-x}$
is $1+x+x^2+x^3+...+x^n+...$
which converges for $x$ in $(-1,1)$; replace each $x$ with $x^3$ to get
$\frac{1}{1-x}=1+x^3+(x^3)^2+(x^3)^3+(x^3)^4...+(x^3)^n+...$
$=1+x^3+x^6+x^9+x^{12}$
$\int^{\frac{1}{2}}_0 \frac{1}{1-x^3}dx \approx \int^{\frac{1}{2}}_0(1+x^3+x^6+x^9+x^{12})dx$
$=(x+\frac{x^4}{4}+\frac{x^7}{7}+\frac{x^{10}}{10}+\frac{x^{13}}{13})|^{\frac{1}{2}}_0$
$\approx0.517$
