On the interval $\left[\frac{5 \pi}{4}, \frac{4 \pi}{3}\right]$ the least value of the function $f(x

Question

On the interval $\left[\frac{5 \pi}{4}, \frac{4 \pi}{3}\right]$ the least value of the function $f(x)=\int_{5 M / 4}^x(3 \sin t+4 \cos t) d t$ is

Professor PC Thomas Classes · Answer

$f(x)=\int_{5\pi/4}^{x}(3~sin~t+4~cos~t)dt,x\in[\frac{5\pi}{4},\frac{4\pi}{3}]$
$f^{\prime}(x)=3~sin~x+4~cos~x$
$f^{\prime}(x)<0$ as $\sin x, \cos x$ are negative for $x\in[\frac{5\pi}{4},\frac{4\pi}{3}]\Rightarrow f(x)|_{min}=f(\frac{4\pi}{3})$
$=\int_{5\pi/4}^{4\pi/3}(3~sin~t+4~cos~t)dt=\frac{3}{2}+\frac{1}{\sqrt{2}}-2\sqrt{3}$
Hence, (A) is the correct answer.