Answer
$ f(x) =\dfrac{x^3}{3}-\dfrac{x^{5}}{5}+\dfrac{x^{7}}{7 \cdot 2!}-\dfrac{x^{9}}{9 \cdot 3!}+\dfrac{x^{11}}{11 \cdot 4!} ....$
$| Error| \lt 0.00064$
Work Step by Step
We integrate the integral with respect to $ x $ taking a lower limit as $0$ and upper limit as $1$.
$ f(x)=\int_0^{x} t^2 e^{-t^2} dt=\int_0^{x} [t^2-t^4+\dfrac{t^6}{2 !} -\dfrac{t^{8}}{3!}-...] dt $
or, $ f(x) =\dfrac{x^3}{3}-\dfrac{x^{5}}{5}+\dfrac{x^{7}}{7 \cdot 2!}-\dfrac{x^{9}}{9 \cdot 3!}+\dfrac{x^{11}}{11 \cdot 4!} ....$
So, the first five terms of the series would give the accuracy.
And the error can be calculated as:
$| Error| \lt \dfrac{1}{13\cdot 5 !} \approx 0.00064$
