Find the arc length of $f(x)=4x^\frac{1}{2}+9$

Question

Question:

If the arc length formula of a function $f$ on an interval $[a,b]$ is given by $L_a^b=\int_a^b \sqrt{1+[f'(x)]^2} \ dx$. Find the arc length of $f(x)=4x^\frac{1}{2}+9$ on $[0,1]$.

We have $[f'(x)]^2=[2x^\frac{-1}{2}]^2=\frac{4}{x}$. Then,

$$L_0^1=\int_0^1 \sqrt{1+\frac{4}{x}} \ dx=\int_0^1 \sqrt{\frac{x+4}{x}} \ dx = \int_0^1 \frac{\sqrt{x+4}}{\sqrt{x}} \ dx=\int_0^1 (x+4)^\frac{1}{2} \ x^\frac{-1}{2} dx$$

I am stuck at this integral, I tried change of variable but it didn't work. Any suggestions? Thanks!

Answer 1

Let $u = \sqrt{1 + \frac{4}{x}}$, $x= \frac{4}{u^2-1}$

$\frac{du}{dx} = \frac{-2}{x^2 \cdot\sqrt{1 + \frac{4}{x}} }$

$dx = - \frac{x^2 \cdot\sqrt{1 + \frac{4}{x}}} {2}du =- \frac{8u}{(u^2-1)^2} du$

$\sqrt{1 + \frac{4}{x}} dx = - \frac{8u^2}{(u^2-1)^2} du$

$I=\int\limits_{\sqrt{5}}^{\infty} \frac{8u^2}{(u^2-1)^2} du$

Then just use partial fraction to compute this integral.

Answer 2

$I= \int_0^1 \frac{\sqrt {x+4}}{\sqrt x}\, dx$

Put $\frac{\sqrt {x+4}}{\sqrt x}=u$

$I = \int_2^{\sqrt5} \frac{8 }{u^4-2u^2+1}\,du$

you can use partial fractions and proceed from here

$I={2}\ln(\frac{ \sqrt {x+4} + \sqrt x}{\sqrt {x+4}}) - {2}\ln(\frac{ \sqrt {x+4} - \sqrt x}{\sqrt {x+4}}) + \sqrt x \sqrt{x+4}$

putting the bounds;

$I=2 {\ln(\sqrt5+1)-2\ln(\sqrt5-1)+\sqrt5} \approx 4.2$

Answer 3

Sometimes, if the arc-length integrand for a function looks troublesome (or hopeless) in one variable, the inverse function may be easier to deal with. For $ \ y \ = \ 4\sqrt{x} \ + \ 9 \ \ $ on $ \ [0 \ , \ 1] \ \ , $ that inverse would be $ \ x \ = \ \left(\frac{y \ - \ 9}{4} \right)^2 \ \ $ on $ \ [9 \ , \ 13] \ \ . \ $ The arc-length integral becomes $$ \ \int_9^{13} \ \sqrt{ \ \left(\frac{dx}{dy} \right)^2 \ + \ 1} \ \ dy \ \ $$

and we obtain by implicit differentiation $ \ \frac{d}{dy} [ \ y \ ] \ = \ \frac{d}{dy} [ \ 4·x^{1/2} + 9 \ ] \ \Rightarrow \ 1 \ = \ 2·x^{-1/2}·\frac{dx}{dy} $ $ \Rightarrow \ \frac{dx}{dy} \ = \ \frac{\sqrt{x}}{2} \ \ . \ $ The integral is now $$ \ \int_9^{13} \ \sqrt{ \ \left(\frac{\sqrt{x}}{2} \right)^2 \ + \ 1} \ \ dy \ \ = \ \ \int_9^{13} \ \sqrt{ \ \frac{ x \ + \ 4}{4} } \ \ dy $$ $$ = \ \ \ \int_9^{13} \ \sqrt{ \ \frac{[ \ (y \ - \ 9)^2 \ / \ 4^2 \ ] \ + \ 4}{4} } \ \ dy $$ $$ = \ \ \frac12·\frac14 \ \int_9^{13} \ \sqrt{ \ (y \ - \ 9)^2 \ + \ 64 } \ \ dy \ \ \rightarrow_{u \ = \ y - 9} \ \ \frac18 \ \int_0^4 \ \sqrt{ \ u^2 \ + \ 64 } \ \ du \ \ , $$ which now at least looks like a more familiar integral if you've worked with trigonometric substitutions.

[Note, incidentally, that this is an improper integral (of the type where the integrand is undefined somewhere on the interval of integration). We see, however, that it does "converge" to a finite value.]