# How do you find the area of the region under the given curve y = (2x + 2)^1/2 on the interval [0,1]?

$\frac{8}{3} - 2 \frac{\sqrt{2}}{3}$
The area of the region bounded by the curve and the x-axis would be ${\int}_{0}^{1} y \mathrm{dx}$ = ${\int}_{0}^{1} \sqrt{2 x + 2} \mathrm{dx}$ = 2/3 sqrt2 (x+1)^(3/2)]_o^1
=$\frac{2}{3} \sqrt{2} \left({2}^{\frac{3}{2}} - 1\right)$= $\frac{8}{3} - 2 \frac{\sqrt{2}}{3}$