$\int^1_0 \sqrt{\sqrt{x} – x } dx$

Solution:

$\begin{aligned} \text { Let } & I=\int_{0}^{1} \sqrt{\sqrt{x}-x} d x \\ & I=\int_{0}^{1} \sqrt{(\sqrt{x}-\sqrt{x} \cdot \sqrt{x})} d x \end{aligned}$

Take $\sqrt{x}$ common,

$I=\int_{0}^{1} x^{1 / 4} \sqrt{1-x^{1 / 2}} d x$

Put $\quad x^{1 / 2}=t$

Squaring both sides,

$\therefore x=t^{2}$

Differentiate w.r.t $x$ , $\therefore d x=2 t . d t$ Limits after substitution: $\lim \rightarrow[0,1]$ $\begin{aligned} \therefore & I=\int_{0}^{1} t^{1 / 2} \sqrt{1-t} .2 . t \text { dt } \ &=2 \int_{0}^{1} t^{3 / 2} \sqrt{1-t} d t \end{aligned}$ $=2 \beta\left(\frac{5}{2}, \frac{3}{2}\right)$ $\therefore \mathrm{I}=\frac{\pi}{8}$