Q. Let $I = \int_{0}^{1}{\frac{\sin x}{\sqrt{x}}\ dx}\ and\ J = \int_{0}^{1}{\frac{\cos x}{\sqrt{x}}\ dx}$. Then, which one of the following is true?
A) $I > \frac{2}{3}\ and\ J < 2$
B) $I > \frac{2}{3}\ and\ J > 2$
C) $I < \frac{2}{3}\ and\ J < 2$
D) $I < \frac{2}{3}\ and\ \ J > 2$
Explanation
(c)
Since, $I = \int_{0}^{1}\frac{\sin x}{\sqrt{x}}dx < \int_{0}^{1}\frac{x}{\sqrt{x}}dx$
Because in $x\epsilon(0,1),x > \sin x$
$I < \int_{0}^{1}\sqrt{x}dx = \frac{2}{3}\left\lbrack x^{\frac{3}{2}} \right\rbrack_{0}^{1} \Rightarrow I < \frac{2}{3}$
For $x\epsilon(0,1),\frac{\cos x}{\sqrt{x}} < \frac{1}{\sqrt{x}}$
And $J = \int_{0}^{1}\frac{\cos x}{\sqrt{x}}dx < \int_{0}^{1}x^{- \frac{1}{2}}dx = 2 \Rightarrow J < 2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.