The least value of the function $f\left(x\right)={\int }_0^x\left(3\sin x+4\cos x\right)dx$ in the interval $\left[\frac{5\pi }{4},\frac{4\pi }{3}\right]$ is

We have, $f\left(x\right)={\int }_0^x\left(3\sin x+4\cos x\right)dx$

$f^{\text{'}}\left(x\right)=3\sin x+4\cos x=5\sin \left(x+{\tan }^{-1}\frac{4}{3}\right)$

$\frac{5\pi }{4}<x<\frac{4\pi }{3}$

$\pi <x+{\tan }^{-1}4/3<2\pi$

$\because f^{\text{'}}\left(x\right)<0$, i.e. $f\left(x\right)$ is decreasing.

$\because$ least value of $f\left(x\right)$ is $\left(\frac{5\pi }{4},\frac{4\pi }{3}\right)=f\left(\frac{4\pi }{3}\right)$

least $f\left(x\right)={\int }_0^{4\pi /3}\left(3\sin x+4\cos x\right)dx$

$={\left[-3\cos x+4\sin x\right]}_0^{4\pi /3}$

$=-3\left(-\frac{1}{2}-1\right)+4\left(-\frac{\sqrt{3}}{2}\right)$

$=\frac{9-4\sqrt{3}}{2}$