# What is the area under the curve y=sqrtt from 0 to 1?

The area is given by the integral ${\int}_{0}^{1} \sqrt{t} \mathrm{dt} = {\int}_{0}^{1} \left({t}^{\frac{3}{2}} / \left(\frac{3}{2}\right)\right) ' \mathrm{dt} = \frac{2}{3}$