Chapter 4 - Integrals - 4.3 The Fundamental Theorem of Calculus - 4.3 Exercises - Page 328: 39

$$\frac{16}{3} $$

Since the region bounded by $$ y=\sqrt{x},\ \ \ y=0 ,\ \ \ x=4 $$ and $$ \sqrt{x}\geq 0 \ \ \ \text{for}\ \ \ 0\leq x\leq 4 $$ Then area is given by \begin{aligned} \text { Area }&=\int_{0}^{4} \sqrt{x} d x\\ &=\int_{0}^{4} x^{1 / 2} d x\\ &=\left[\frac{2}{3} x^{3 / 2}\right]_{0}^{4}\\ &=\frac{2}{3}(8)-0\\ &=\frac{16}{3} \end{aligned}