h(x) = the integral from 0 to x^2 of the function squareroot(1+r^3) dr


Is the answer = [squareroot(1+x^3)]* (2x)??

I tried using the fundamental theorem of calculus in conjunction with the chain rule but did I use it correctly?

This integral is not easily evaluated by elementary means. That's what leads me to

believe you're leaning toward the SECOND fundamental theorem of calculus.

\frac{d}{dx}\int_{h(x)}^{g(x)}f(r)dr=f(g(x))g'(x)-f(h(x))h'(x)

This gives \sqrt{1+x^{6}}(2x)-\sqrt{1+0^{3}}(0)


You're close. It's 2x\sqrt{x^{6}+1}

Oh wow.. I multiplied wrong haha. Thank you very much! It was very helpful!