Crocodile Forest

$\huge\sqrt{x\sqrt{x\sqrt{x\cdots}}}=x^{\small{\color{#D61F06}{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\cdots}}}$ $\\ (\mathcal{\color{#D61F06}{Infinite~ GP}})$ $\\ \huge=x^{\small{\color{#D61F06}{\frac{\frac{1}{2}}{1-\frac{1}{2}}}}}=x$ Hence integration simplifies to:- $\huge \mathfrak{I}=\int_{3}^{5} x \, dx$ $\huge =\dfrac{x^2}{2}|_{\small 3}^{\small 5}$ $\huge =\color{goldenrod}{\boxed{\color{#20A900}{\boxed{\color{#007fff}{8}}}}}$

Let $\large\displaystyle\ y =\sqrt{x\sqrt{x\sqrt{x\cdots}}}$

As the series is infinite, we can say that $\large\displaystyle\ y = \sqrt{x*y}$

which gives us $\large\displaystyle\ y =x$ [as $y=0$ only for $x=0$ ]

Therefore, the required integration now is: $\large\displaystyle\int_{3}^{5} \ x \, dx =\, 8$

Let $u = \sqrt{x \sqrt{x\sqrt{x \cdots}}}$ . Then

$\begin{aligned} u & = \sqrt {xu} & \small \color{#3D99F6} \text{Squaring both sides} \\ u^2 & = xu \\ u & = x \end{aligned}$

Therefore, $\displaystyle \int_3^5 \sqrt{x \sqrt{x\sqrt{x \cdots}}}\ dx = \int_3^5 x \ dx = \dfrac {x^2}2 \ \bigg|_3^5 = \dfrac {25-9}2 = \boxed 8$ .