We have the following definite integral $$I = \int_{\pi}^{3\pi}\frac{\left | sinx+cosx \right |}{x}$$
Now obviously this is a non elementary function. Thus we need to approximate the area using inequalities. Now I had learned that: $$m(b-a)\leq \int_{a}^{b}f(x)\leq M(b-a)$$ Where m and M correspond to the maximum and minimum value of $f(x)$ in the domain of $(a,b)$. But I have no idea how to calculate these aforesaid maximum and minimum values. Thanks in advance.
On
That depends on how sharp a bound you want. Simple bounds you'd get from just $0 < \lvert \sin x + \cos x \rvert < 2$. Somewhat better is to note that in your range:
$\begin{align*} \frac{\lvert \sin x + \cos x\rvert}{x} &= \sqrt{\frac{(\sin x + \cos x)^2}{x^2}} \end{align*}$
getting minima/maxima of the quantity under the root is easier than futzing around with the many different cases from the absolute value. Or cut up the range and get bounds on each part, sum them up.
Hint: On the interval $[\pi,3\pi]$
$$0\le\frac{\left | \sin x+\cos x \right |}{x}\le \frac{\sqrt{2}}{\pi}$$
therefore
$$0\le I\le \frac{\sqrt{2}}{\pi}\int_{\pi}^{3\pi}\,dx=\frac{\sqrt{2}}{\pi}\big(3\pi-\pi\big)=2\sqrt{2}$$