Answer
$| Error| \lt 0.000013$
Work Step by Step
We integrate the integral with respect to $ x $ taken a lower limit as $0$ and upper limit as $1$.
$ f(x)=\int_0^{x} \sin t^2 dt=\int_0^{x} [t^2-\dfrac{t^6}{3 !}+\dfrac{t^{10}}{5!}-...] dt $
or, $=\dfrac{x^3}{3}-\dfrac{x^{7}}{7 \cdot 3 !}+\dfrac{x^{11}}{11 \cdot 5!}- ....$
So, the error can be calculated as:
$| Error| \lt \dfrac{1}{15\cdot 7 !} \approx 0.000013$
