#### Answer

$| Error| \lt 0.000004960$

#### Work Step by Step

We integrate the integral with respect to $ x $ taking a lower limit as $0$ and upper limit as $1$.
$\int_0^{1} \cos \sqrt {t} dt=\int_0^{1} [1-\dfrac{t}{2}+\dfrac{t^{2}}{4 !}-...] dx $
or, $=[t-\dfrac{t^2}{4}+\dfrac{t^{3}}{3 \cdot 4 !}-...]_0^{1} ....$
So, the error can be calculated as:
$| Error| \lt \dfrac{1}{5\cdot 8 !} \approx 0.000004960$