For the function $f\left(x\right)=\underset{0}{\overset{x}{\int }}\frac{\sin t}{t}dt$, where $x>0$

We have, $f\text{'}\left(x\right)=\frac{\sin x}{x}$ and $f\text{'}\text{'}\left(x\right)=\frac{x\cos x-\sin x}{x^2}$

For maximum or minimum, $f\text{'}\left(x\right)=0$

$\Rightarrow \frac{\sin x}{x}=0\Rightarrow \sin x=0; x\ne 0$

$\therefore x=n\pi ; n=1,2,\mathrm{3........}(\because x>0)$

At $x=n\pi ; f\text{'}\left(x\right)=\frac{n\pi \cos n\pi -\sin n\pi }{{\left(n\pi \right)}^2}=\frac{\cos n\pi }{n\pi }$

$=\frac{{(-1)}^n}{n\pi }$

$\therefore$ Extreme points are $x=n\pi , n=1,2,\mathrm{3...........}$ where the maximum occurs at $x=\pi ,3\pi ,5\pi \mathrm{.......}$ and the minimum occurs at $x=2\pi ,4\pi ,6\pi ,\mathrm{.......}$