The least value of the function $F\left(x\right)={\int }_0^x$ $(3 \sin u+4 \cos u) du$

on the interval $(5\pi /4, 4\pi /3]$ is

We have $F\text{'}\left(x\right)=3 \sin x+4 \cos x.$ Since sin $x$ and cos $x$ assume negative values in the third quadrant, 

we have $F^{\mathrm{\prime }}\left(x\right)<0$ for all $x\in (5\pi /4, 4\pi /3)$ so $F\left(x\right)$ assumes the least value at the point $x = 4\mathrm{\pi }/3.$

Thus the least value is $F(4\pi /3)={\int }_0^{4\pi /3} (3\sin u+4\cos u)\mathrm{d}u$

$\begin{array}{l}={\left.(-3\cos u+4\sin u)\right|}_0^{4\pi /3}\\ =-3\cos \frac{4\pi }{3}+4\sin \frac{4\pi }{3}-(-3)\\ =\frac{9}{2}-\frac{4\sqrt{3}}{2}=\frac{9-4\sqrt{3}}{2}.\end{array}$