Answer
$\dfrac{1}{6}$
Work Step by Step
Our aim is to find the area between the two given functions $y_1=1$ and $y_2=\sqrt x$.
This can be computed as;
$Area=\int_0^1 (1-\sqrt x)^2 \ dx \\=\int_0^1 (1-2\sqrt x+x) \ dx\\=[ x-\dfrac{2x^{3/2}}{3/2}+\dfrac{x^2}{3}]_0^1 \\=[1-\dfrac{2(1)^{3/2}}{3/2}+\dfrac{(1)^2}{3}]\\=\dfrac{1}{6}$
