Statement 1: On the interval $\left[\frac{5\pi }{4},\frac{4\pi }{3}\right]$, the least value of the function $f\left(x\right)={\int }_{\frac{5\pi }{4}}^x (3\sin t+4\cos t)dt\text{ is }0$
Statement 2: If f (x) is a decreasing function on the interval [a, b], then the least value of $f\left(x\right)\text{ is }f\left(b\right)$
 

$f\left(x\right)={\int }_{\frac{5\pi }{4}}^x (3\sin t+4\cos t)dt\Rightarrow f^{\mathrm{\prime }}\left(x\right)=3\sin x+4\cos x,x\in \left[\frac{5\pi }{4},\frac{4\pi }{3}\right]$
These values of x is are in third quadrant where both sin x and cos x are negative
$\text{ Then }{f}^{\mathrm{\prime }}\left(x\right)<0\text{ for }x\in \left[\frac{5\pi }{4},\frac{4\pi }{3}\right]\text{, }$
Hence, f(x) is decreasing for these values of x. Then the least value of function occurs at $x=\frac{4\pi }{3}.$
$\Rightarrow f_{min\text{imum }}={\int }_{\frac{5\pi }{4}}^{\frac{4\pi }{3}} (3\sin t+4\cos t)dt=\frac{3}{2}+\frac{1}{\sqrt{2}}-2\sqrt{3}$