The least value of the function f ( x ) = ∫ _ 0 ^ x (3 x + 4 x)dx...

We have, $f\left(x\right) = {∫}_{0}^{x}\left(3\sin x + 4\cos x\right)dx$

${f}^{'}\left(x\right) = 3\sin x + 4\cos x = 5sin$\left(x + \tan^{- 1}\dfrac{4}{3}\right)$</p><p>$\dfrac{5\pi}{4} < x < \dfrac{4\pi}{3}$</p><p>$\pi < x + \tan^{- 1}4 / 3 < 2\pi$</p><p>$∵ {f}^{'}\left(x\right) < 0$, i.e.$f\left(x\right)$is decreasing.</p><p>$∵$least value of$f\left(x\right) is \left(\dfrac{5\pi}{4}$,$\dfrac{4\pi}{3}\right) = f \left(\dfrac{4\pi}{3}\right)

least $f\left(x\right)$ = {∫}_{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}$</p><p>$= - 3 \left(- \dfrac{1}{2} -1\right)$+ 4 \left(- \dfrac{\sqrt{3}}{2}\right)$

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