Question
Let $$I = \int\limits_0^1 {\frac{{\sin \,x}}{{\sqrt x }}dx} $$ and $$J = \int\limits_0^1 {\frac{{\cos \,x}}{{\sqrt x }}dx.} $$ Then which one of the following is true?
A.
$$I > \frac{2}{3}{\text{ and }}J > 2$$
B.
$$I < \frac{2}{3}{\text{ and }}J < 2$$
C.
$$I < \frac{2}{3}{\text{ and }}J > 2$$
D.
$$I > \frac{2}{3}{\text{ and }}J < 2$$
Answer :
$$I < \frac{2}{3}{\text{ and }}J < 2$$
Solution :
$$\eqalign{
& {\text{We know that }}\frac{{\sin \,x}}{x} < 1,\,{\text{for }}x\, \in \left( {0,\,1} \right) \cr
& \Rightarrow \frac{{\sin \,x}}{{\sqrt x }} < \sqrt x {\text{ on }}x\, \in \left( {0,\,1} \right) \cr
& \Rightarrow \int\limits_0^1 {\frac{{\sin \,x}}{{\sqrt x }}dx < } \int\limits_0^1 {\sqrt x \,dx = \left[ {\frac{{2{x^{\frac{3}{2}}}}}{3}} \right]_0^1} \cr
& \Rightarrow \int\limits_0^1 {\frac{{\sin \,x}}{{\sqrt x }}dx < } \frac{2}{3} \cr
& \Rightarrow I < \frac{2}{3}{\text{ Also }}\frac{{\cos \,x}}{{\sqrt x }} < \frac{1}{{\sqrt x }}\,\,{\text{for }}x\, \in \left( {0,\,1} \right) \cr
& \Rightarrow \int\limits_0^1 {\frac{{\cos \,x}}{{\sqrt x }}dx < \int\limits_0^1 {{x^{ - \frac{1}{2}}}} dx = \left[ {2\sqrt x } \right]_0^1 = 2} \cr
& \Rightarrow \int\limits_0^1 {\frac{{\cos \,x}}{{\sqrt x }}dx < 2} \,\,\, \Rightarrow J < 2 \cr} $$