How to find the minimum length of $f(x)$ Let $f(0)=0, f(1)=t $ 
and
$$ \int_0^1 f(x) dx = 1 $$
How find the minimum length of $f(x)$ in $[0, 1]$ ?
For example when t=2, the minimum length of f(x) is the segment of OA
(O is the origin and  A=(1, 2))
But how find the minimum length when t=3 ?
 A: Following Gil Strang's calculus of variations notes online, you want to use a Lagrangian of the form
$$
    \mathscr{L}(y,y',\lambda) = \int_0^1 \sqrt{1+(y')^2}\,dx + \lambda \left(\int_0^1 y\,dx - 1\right)
    = \int_0^1\left(\sqrt{1+y'^2}+\lambda y\right)\,dx -\lambda
$$
Let the term in parentheses be $F(y,y',\lambda)$.
If $\dfrac{\delta \mathscr{L}}{\delta y} =0$, the Euler-Lagrange equations are
$$
    0 = \frac{\partial F}{\partial y} -\frac{d}{dx}\frac{\partial F}{\partial y'}
      = \lambda - \frac{d}{dx}\frac{y'}{\sqrt{1+(y')^2}}
$$
So we can integrate once to get:
$$
    \frac{y'}{\sqrt{1+y'^2}} = \lambda x + c
$$
for some constant $c$.  Solving,
$$
    y' = \frac{\lambda x + c}{\sqrt{1-(\lambda x + c)^2}}
$$
Let $u = \lambda x + c$.  Then
$$
    y = \int \frac{\lambda x + c}{\sqrt{1-(\lambda x + c)^2}}\,dx
      = \frac{1}{\lambda}\int \frac{u}{\sqrt{1-u^2}}\,du
      = -\frac{1}{\lambda} \sqrt{1-u^2} + d
$$
Or,
\begin{align*}
    (\lambda y - \lambda d)^2 &= 1-(\lambda x + c)^2\\
\implies \left(x + \frac{c}{\lambda}\right)^2 + (y-d)^2 &= \frac{1}{\lambda^2}
\end{align*}
The curve is a circular arc.  We can find $c$, $d$, and $\lambda$ by placing $(0,0)$ and $(1,t)$ on the curve, and solving $\int_0^1 y\,dx = 1$.
A: When you did $t=2$ the shortest distance was a straight line. If we increase $t$ then the shortest distance is likely to be an arc of a circle.
This is better justified with we rotate the graph so that $(1,t)$ is now on the x-axis at $(\sqrt{1+t^2},0)$.

The area can be found by considering the triangle take away the sector. The radius can be adjusted until the area enclosed is equal to $1$.
