Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises - Page 320: 7

$1$

$\int_{0}^{1} (x^2+\sqrt x)dx=[\frac{x^3}{3}+\frac{2}{3}\cdot x^{\frac{3}{2}}]_{0}^{1}=[\frac{1^3}{3}+\frac{2}{3}\cdot 1^{\frac{3}{2}}]-[\frac{0^3}{3}+\frac{2}{3}\cdot 0^{\frac{3}{2}}]=[\frac{1}{3}+\frac{2}{3}\cdot 1]-0=[\frac{1}{3}+\frac{2}{3}]=1$