Chapter 6.4, Problem 29E

To determine

To evaluate the given integral.

Answer to Problem 29E

  ${\displaystyle \underset{0}{\overset{1}{\int }}\left(x^2+\sqrt{x}\right)dx}=1$

Explanation of Solution

Given:

  ${\displaystyle \underset{0}{\overset{1}{\int }}\left(x^2+\sqrt{x}\right)dx}$

Calculation:

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{1}{\int }}\left(x^2+\sqrt{x}\right)dx}={\left[\frac{x^3}{3}+\frac{x^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]}_0^1\\ =\left[\frac{{\left(1\right)}^3}{3}+\frac{2}{3}\cdot {\left(1\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]-\left[\frac{{\left(0\right)}^3}{3}+\frac{2}{3}\cdot {\left(0\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]\\ =\frac{1}{3}+\frac{2}{3}\\ =1\end{array}$