Mathematics - Definite Integration Question with Solution | TestHub

$$\begin{array}{rl}& \mathrm{I}={\int }_0^1\frac{\sin x}{\sqrt{x}}\mathrm{dx}<{\int }_0^1\frac{\mathrm{x}}{\sqrt{\mathrm{x}}}\mathrm{dx}={\int }_0^1\sqrt{\mathrm{x}}\mathrm{dx}={\frac{2}{3}{\mathrm{x}}^{3/2}|}_0^1=\frac{2}{3}\\ & \Rightarrow \mathrm{I}<\frac{2}{3}\\ & \mathrm{J}={\int }_0^1\frac{\cos x}{\sqrt{\mathrm{x}}}\mathrm{dx}<{\int }_0^1\frac{1}{\sqrt{\mathrm{x}}}\mathrm{dx}={2\sqrt{\mathrm{x}}|}_0^1=2\\ & \therefore \mathrm{J}\le 2.\end{array}$$