Answer
$\dfrac{9}{14}$
Work Step by Step
Since, we have $ x^3+\sqrt y=1$
$\implies y=(1-x^3)^2=x^6-2x^3+1$
We know that $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+C$
where $C$ is a constant of proportionality.
$\int_0^1 x^6-2x^3+1 dx=[\dfrac{x^7}{7}-2x^4/4+x]_0^1$
or, $=(\dfrac{1}{7}-0)-2(\dfrac{1}{4}-0)+(1-0)$
or $=\dfrac{9}{14}$
