# Length of a curve y = 1 - √x

I want to know the length of the curve described by

$$f(x) = 1 - \sqrt{x},\quad x \in [0,1].$$

When I build the derivative and plug it in the length formula:

$$\int_0^1 \sqrt{1 +\frac{1}{4x}} dx = x+\frac{\log(1)}{4} - x+\frac{\log(0)}{4}$$

I get a problem because of $\log(0)$. I have no idea what to do now. Thanks for your help in advance!

Edit: In the original formulation of the question the square root in the integral was missing.

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I'm so unconcentrated today... yes, you are right, and now it's easy to solve. Thanks! –  BernardF Jul 6 '11 at 14:03
Ok, maybe it's good to get the correct length formula when searching the web for "Lenght of a curve". ;-) –  BernardF Jul 6 '11 at 14:09

Generally Arc Length for your curve is given by the formula \begin{align*} \ell = \int\limits_{0}^{1} \sqrt{1+(f'(x))^{2}} \ dx \end{align*}

$$f'(x) = -\frac{1}{2\sqrt{x}}$$ Squaring you have $(f'(x))^{2} = \frac{1}{4x}$. Hence the integral is $$\int\limits_{0}^{1} \sqrt{1 + \frac{1}{4x}} \ dx$$

So this is an improper integral. You will have to evaluate it as it is done in the Wikipedia link for $\sqrt{x}$. Try giving the trigonometric substitution $x= \frac{1}{4}\tan^{2}{t}$. {After taking l.c.m inside the square root}.

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According to Wolfram Online Integrator, $$\int {\sqrt {1 + \frac{1}{{4x}}} } dx = F(x) + C,$$ where $$F(x) = \frac{1}{8}\bigg(4\sqrt {\frac{1}{x} + 4} x + \log \bigg(4\bigg(\sqrt {\frac{1}{x} + 4} + 2\bigg)x + 1 \bigg)\bigg).$$ Noting that $\mathop {\lim }\nolimits_{x \to 0^ + } F(x) = 0$, it thus follows that $$\int_0^1 {\sqrt {1 + \frac{1}{{4x}}} dx} = F(1) = \frac{{4\sqrt 5 + \log (4\sqrt 5 + 9)}}{8} \approx 1.47894$$ (confirmed using Wolfram Definite Integral Calculator).

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It often pays to consider an equivalent problem for which the algebra looks a bit easier. At least it provides an alternative check on doing things the hard way.

Here the length of the curve $y = 1 - \sqrt{x}$ on $[0,1]$ is by vertical translation and reflection in the $x$-axis the same as for $y = \sqrt{x}$ on the unit interval.

Now exchange the roles of $x$ and $y$, which amounts to reflection in $y = x$, and we would have the same length for $y = x^2$ on $[0,1]$.

Then applying the arclength formula that Chandru has nicely formatted:

\begin{align*} \ell = \int\limits_{0}^{1} \sqrt{1+4x^2} \ dx \end{align*}

one gets a proper integral that yields to trigonometric substitution:

$$x = \frac{1}{2} \tan{\theta}$$

$$dx = \frac{1}{2} \sec^{2}{\theta} d\theta$$

with respective limits of integration for $\theta \in [0,\tan^{-1}(2)]$.

Thus: \begin{align*} \ell = \int\limits_{0}^{1} \sqrt{1+4x^2} \ dx = \frac{1}{2} \int\limits_{0}^{\tan^{-1}(2)} \sec^3{\theta} \ d\theta \end{align*}

Consulting a table of trigonometric identities, I find $\sec^{3}{\theta}$ has antiderivative:

$$\frac{1}{2} ( \sec{\theta} \tan{\theta} + \ln{| \sec{\theta} + \tan{\theta}|} ) + C$$

Fortunately the nonconstant terms are zero when $\theta = 0$, so we only have the value at $\theta = \tan^{-1}(2)$ to simplify:

$$\sec(\tan^{-1}(2)) = \sqrt{5}$$

$$\ell = \frac{1}{2} (\sqrt{5} + \frac{1}{2} \ln( 2 + \sqrt{5}) )$$

Numerically this gives $\ell = 1.47894...$ or slightly more than the straightline distance $\sqrt{2}$, which seems plausible.

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