Double Integrals From 15.8
asked 2014-07-17 11:07:00 -0600
So I am having trouble with #9 from 15.8:
Compute $$\int_0^1\int_\sqrt{y}^1 \sqrt{x^3+1} \hspace{2 mm} dxdy$$ here is my work $$\int_0^1\int_0^{x^2} \sqrt{x^3+1} \hspace{2 mm} dydx$$ $$\int_0^1 y\sqrt{x^3+1} \biggr|_0^{x^2} dx$$ $$\int_0^1 x^2\sqrt{x^3+1} \hspace{2 mm} dx$$ I used $u$ substitution here
$\hspace{14 pc}$ Let $ u=x^3+1$ $$du=3x^2 dx$$ $$1/3 du=x^2 dx$$ $$1/3 \int_0^1 \sqrt{u} \hspace{2 mm} du$$ $$1/3(2/3 \sqrt{u^3}) \biggr|_0^1$$ $$=2/9$$ what am I missing here?
