Q. Least value of the function $f(x) = \int_{0}^{x}{(1 - \cos{t)\ dt}}$ on the interval $\left\lbrack \frac{\pi}{2},\frac{3\pi}{2} \right\rbrack$ is
A) $\frac{3\pi}{2} + 1$
B) $\frac{\pi}{2} - 1$
C) $\frac{\pi}{2} + 1$
D) None of these
Explanation
(b)
$f^{'}(x) = 1 - \cos x$ always positive, so $f(x)$ is an increasing function so least value of $f(x)$ on $\left\lbrack \frac{\pi}{2},\frac{3\pi}{2} \right\rbrack\ is\ f\left( \frac{\pi}{2} \right)$
$\therefore\ f\left( \frac{\pi}{2} \right) = \int_{0}^{\pi/2}{(1 - \cos{t)dt}}$
$= \lbrack t - {\sin{t\rbrack}}_{0}^{\pi/2} = \left( \frac{\pi}{2} - 1 \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.