Answer
$\dfrac{1}{6}$
Work Step by Step
Since, we have $\sqrt x+\sqrt y=1$
$\implies y=x-2\sqrt x+1$
We know that $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+C$
where $C$ is a constant of proportionality.
$\int_0^1 x-2\sqrt x+1 dx=[\dfrac{x^2}{2}-\dfrac{x^{1/2+1}}{1/2+1}+x]_0^1$
or, $=(\dfrac{1}{2}-0)-(\dfrac{4}{3}-0)+(1-0)$
or $=\dfrac{1}{6}$
