Answer
$| Error| \lt 0.000004960$
Work Step by Step
Integrate the integral with respect to $ x $ as follows:
$\int_0^{1} \cos \sqrt {t} \space dt=\int_0^{1} [1-\dfrac{t}{2}+\dfrac{t^{2}}{4 !}-...] \space dt \\=[t-\dfrac{t^2}{4}+\dfrac{t^{3}}{3 \cdot 4 !}-...]_0^{1} ....$
Now, $| Error| \lt \dfrac{1}{5\cdot 8 !} \approx 0.000004960$
or, $| Error| \lt 0.000004960$
