Chapter 5 - Section 5.3 - The Fundamental Theorem of Calculus - 5.3 Exercises - Page 400: 45

$= \frac{16}{3}$

$\int^{4}_0 (\sqrt x - 0) dx$ $= \int^{4}_0 \sqrt x dx$ $= \int^{4}_0 x^{\frac{1}{2}} dx$ $= \frac{x^{\frac{3}{2}}}{\frac{3}{2}} |^{4}_0$ $= [\frac{4^{\frac{3}{2}}}{\frac{3}{2}}]-[\frac{0^{\frac{3}{2}}}{\frac{3}{2}}]$ $= [\frac{4^{\frac{3}{2}}}{\frac{3}{2}}]-0$ $= [\frac{8}{\frac{3}{2}}]$ $= \frac{2\times8}{3}$ $= \frac{16}{3}$