Answer
$\dfrac{1}{6}$
Work Step by Step
Consider $f(x)=\sqrt x+\sqrt y=1$
This can be re-written as: $y=x-2\sqrt x+1$
Use formula such as:$\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+C$
This implies that $[\dfrac{x^{1+1}}{1+1}-\dfrac{x^{1/2+1}}{\dfrac{1}{2}+1}+x]_0^1=(\dfrac{1}{2}-0)-(\dfrac{4}{3}-0)+(1-0)$
Thus, $f(x)=\int_0^1 x-2\sqrt x+1 dx=\dfrac{1}{6}$
