Single Variable Calculus, Chapter 5, 5.3, Section 5.3, Problem 28

Find the integral $\displaystyle \int^1_0 (3 + x \sqrt{x}) dx$

Using 2nd Fundamental Theorem of Calculus

$\displaystyle \int^b_a f(x) dx = F(b) - F(a)$, where $F$ is any anti-derivative of $f$.

Let $\displaystyle f(x) = 3 + x \sqrt{x}$ or $f(x) = 3 + (x)^{\frac{2}{3}}$, then


$\begin{equation} \begin{aligned} F(x) =& 3 \left( \frac{x^{0 + 1}}{0 + 1} \right) + \left( \frac{x^{\frac{3}{2} + 1}}{\displaystyle \frac{3}{2} + 1} \right) + C \\ \\ F(x) =& 3x + \frac{x^{\frac{5}{2}}}{\displaystyle \frac{5}{2}} + C \\ \\ F(x) =& 3x + \frac{2x^{\frac{5}{2}}}{5} + C \end{aligned} \end{equation}$



$\begin{equation} \begin{aligned} & \int^1_0 (3 + x \sqrt{x}) dx = F(1) - F(0) \\ \\ & \int^1_0 (3 + x \sqrt{x}) dx = 3(1) + \frac{2(1)^{\frac{5}{2}}}{5} + C - \left[ 3(0) + \frac{2(0)^{\frac{5}{2}}}{5} + C \right] \\ \\ & \int^1_0 (3 + x \sqrt{x}) dx = 3 + \frac{2}{5} + C - 0 - 0 - C \\ \\ & \int^1_0 (3 + x \sqrt{x}) dx = \frac{15 + 2}{5} \\ \\ & \int^1_0 (3 + x \sqrt{x}) dx = \frac{17}{5} \\ \\ & \text{ or } \\ \\ & \int^1_0 (3 + x \sqrt{x}) dx = 3.4 \end{aligned} \end{equation}$