What is the Length of a Curve with Integrals?

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Homework Statement



How can we found the length of the curve:
f(x) = \frac {1}{12}(x - 48)\sqrt x

where x\ge0 and the vertical line x=48.



Homework Equations





The Attempt at a Solution


I tried to use the formula L=\int^{48}_{0}\sqrt {1 + \ [f'(x)]^2} dx
But I think that there is a problem where x=0 because the first derivative is:
f'(x) = \frac {(x - 16) \sqrt {x}}{8x} and because x is in the denominator cannot take the value 0.

How can solve this issue??
Any ideas??
Thanks anyone in advance.
 
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What is the final integral? Do you know what the function looks like after the integral has been performed?
 
djeitnstine said:
What is the final integral? Do you know what the function looks like after the integral has been performed?

No I don't
 
You should try performing the integral before you ask whether or not there is division by zero. Perhaps there may be division by zero, in that case you will have to use the improper integral method (limit) to find out whether it converges or diverges at that point.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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