## University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{9}{14}$
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 $=\dfrac{9}{14}$