Q. If $f(x) = \int_{x}^{x^{2}}{(t - 1)dt,\ 1 \leq x \leq 2,}$ then global maximum value of $f(x)$ is
A) 1
B) 2
C) 3
D) 4
Explanation
(d)
$f^{'}(x) = \left( x^{2} - 1 \right)(2x) - (x - 1) = (x - 1)(2x^{2} + 2x - 1)$
Which is positive for $x > 1$. Hence, $f$ increases in [1, 2]
Hence, global maximum of $f$ is $f(2) = 4$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.