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?
On
$I < 0.37$ 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 $6 \cdot 10^{-6}$.
On
What you could do is a Taylor series around $\frac \pi 3$. This gives $$\sin(x)-\cos(x)=\sum_{n=0}^\infty \frac{\left(\sqrt{3}+1\right) \sin \left(n\frac{\pi }{2}\right)+\left(\sqrt{3}-1\right) \cos \left(n\frac{\pi }{2}\right)}{2 n!}\left(x-\frac{\pi }{3}\right)^n$$ $$\sin(1)-\cos(1)=\sum_{n=0}^\infty \frac{\left(\sqrt{3}+1\right) \sin \left(n\frac{\pi }{2}\right)+\left(\sqrt{3}-1\right) \cos \left(n\frac{\pi }{2}\right)}{2 n!}\left(1-\frac{\pi }{3}\right)^n$$
Computing the partial sum from $n=0$ to $n=p$, you have $$\left( \begin{array}{cc} p & S_p \\ 0 & 0.366025 \\ 1 & 0.301552 \\ 2 & 0.301145 \\ 3 & 0.301169 \end{array} \right)$$
With $\sin x>x-\frac {x^3}{3!}$, we have: \begin{align} \int _0^1x\sin x \, dx&>\int _0^1x^2-\frac {x^4}{6}\, dx\\ &=\big[\frac {x^3}3-\frac {x^5}{30}\big]_0^1\\ &=0.3 \end{align}