# Integrals with Functional Limits

• Nov 15th 2007, 12:00 PM
Del
Integrals with Functional Limits
• Nov 15th 2007, 12:44 PM
PaulRS
$f(x^{1/4})=\int_{-3}^x{\sqrt[]{t^2+3}dt}$ for $x\geq{0}$

Differentiating (with the chain rule): $\frac{x^{-3/4}\cdot{f'(x^{1/4})}}{4}=\sqrt[]{x^2+3}$

Then: $f'(x^{1/4})=4\cdot{\sqrt[]{x^2+3}}\cdot{x^{3/4}}$

Let: $z=x^{1/4}$

We get $f'(z)=4\cdot{\sqrt[]{z^8+3}}\cdot{z^{3}}$ with $z\geq{0}$

In the original function we can see that it has the following property $f(x)=f(-x)$

Then it follows that $f'(-x)=-f'(x)$ (imagine the graph of f(x)) which clearly holds in our solution $f'(z)=4\cdot{\sqrt[]{z^8+3}}\cdot{z^{3}}$

Thus $f'(x)=4\cdot{\sqrt[]{x^8+3}}\cdot{x^{3}}$

I hope it's ok :rolleyes:
• Nov 17th 2007, 03:45 PM
Krizalid
Quote:

Originally Posted by Del

Here's another attempt:

By the FTC we have

$f'(x)=\sqrt{(x^4)^2+3}\cdot(x^4)'=4x^3\sqrt{x^8+3} ,$

Of course, if we wish to complicate us the existence, we can integrate $\sqrt{t^2+3}$ then evaluate between $t\in[x^4,-3].$