Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 1 down vote accepted

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}.

share|cite|improve this answer

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).

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.