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dks2114 · Jul 30, 2007, 07:57 AM

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?

galactus · Jul 30, 2007, 12:22 PM

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 $2x\sqrt{x^{6}+1}$

dks2114 · Jul 31, 2007, 03:29 AM

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