Answer
$A = \sqrt{3}-\frac{1}{3}$
Work Step by Step
We can calculate the area $A$ of the region that lies under the curve:
$A = \int_{0}^{1}\sqrt{2x+1}~dx$
Let $u = 2x+1$
$\frac{du}{dx} = 2$
$dx = \frac{du}{2}$
When $x = 0$, then $u = 1$
When $x = 1$, then $u = 3$
$\int_{1}^{3}\frac{1}{2}\sqrt{u}~du$
$= \frac{1}{2}~[\frac{2}{3}(u)^{3/2}\vert_{1}^{3}~]$
$= \frac{1}{3}~[(u)^{3/2}\vert_{1}^{3}~]$
$= \frac{1}{3}~[(3)^{3/2}-(1)^{3/2}]$
$= \frac{1}{3}~(3\sqrt{3}-1)$
$= \sqrt{3}-\frac{1}{3}$
