Approximating $\int_0^1 x\sin xdx$ Check whether $$\int_0^1 x\sin xdx>0.3$$

The anti-derivative of $x\sin x$ is $\sin x-x\cos x$
So the integral is $$I\approx\sin(57^\circ)-\cos(57^\circ)$$
Since $\sin x $ is increasing in the first quadrant and $\cos x$ is decreasing,
$$I<\sin(60^\circ)-\cos(60^\circ)$$
$$I<\frac{\sqrt3-1}{2}$$
$$I<0.37$$
Which also means $$I>0.3$$
Is there any other, preferably graphical, method?

$I < 0.37$ does not necesarrily imply that $I > 0.3$. By Taylor series again, you would have:
$$I = \sin 1 - \cos 1 \approx 1 - \frac{1^3}{3!} + \frac{1^5}{5!} - \frac{1^7}{7!} - \left(1 - \frac{1^2}{2!} + \frac{1^4}{4!} - \frac{1^6}{6!}\right) = \frac{253}{840} \approx 0.3012.$$
and the maximum error of a Taylor series is given by the next term:
$$\text{Maximum error} = \left|\frac{1}{9!}\right| + \left|\frac{1}{8!}\right| = \frac{1}{36288}$$
but $\frac{253}{840} - \frac{1}{36288} > 0.3$. Hence $\sin 1 - \cos 1 = \int_0^1 x \sin x \ dx > 0.3$.
The series approximation gives $0.3011629 \cdots$, where the actual value of $\sin 1 - \cos 1$ is $0.3011687 \cdots$, off by only $6 \cdot 10^{-6}$.
