Answer
$\int_{4}^{9}\sqrt xdx=\frac{38}{3}$
Work Step by Step
$$\lim\limits_{N\to \infty}\frac{5}{N}\sum_{j=1}^{N}\sqrt {4+5\frac{j}{N}}$$
$$\lim\limits_{N\to \infty}\sum_{j=1}^{N}\sqrt {4+5\frac{j}{N}}\cdot \frac{5}{N}$$
$$\lim\limits_{N\to \infty}\sum_{j=1}^{N}\sqrt {4+j\frac{5}{N}}\cdot \frac{5}{N}$$
$$\lim\limits_{N\to \infty}\sum_{j=1}^{N}\sqrt {4+j\frac{9-4}{N}}\cdot \frac{9-4}{N}$$
Where $x_j=4+j\frac{9-4}{N}$,$f(x)=\sqrt x$,$a=4$,$b=9$
$$\lim\limits_{N\to \infty}\sum_{j=1}^{N}\sqrt {4+j\frac{9-4}{N}}\cdot \frac{9-4}{N}=\int_{4}^{9}\sqrt xdx$$
Using the $FTC I$ it follows:
$$\int_{4}^{9}\sqrt xdx=[\frac{2}{3}x^{\frac{3}{2}}]_{4}^{9}=\frac{2}{3}\cdot 9^{\frac{3}{2}}-\frac{2}{3}\cdot 4^{\frac{3}{2}}=\frac{38}{3}$$
