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.
 A: 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}. 
A: 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).
A: 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.
