The Statement of an Example Problem

Consider the problem of finding the curve y(x) of shortest length that connects the two points (0, 0) and (1, 1) in the plane. Letting ds be an element of arc length, the arc length of a curve y(x) from x = 0 to x = 1 is $\int_{0}^{1}$  ds. We can use Pythagoras' theorem to relate ds to dx and dy: drawing a triangle with sides of length dx and dy at right angles to one another, the hypotenuse is $\approx$ ds and so ds² = dx² + dy² and s = $\sqrt{{\mathrm{d}x^2 + \mathrm{d}y^2}}$ = $\sqrt{{1+ \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}}$dx. This means the arc length equals $\int_{0}^{1}$$\sqrt{{1+y'^2}}$dx.

The curve y(x) we are looking for minimizes the functional

I(y) = $$\int_{0}^{1}$$ ds = $$\int_{0}^{1}$$ (1 + (y')²)^1/2  dx    subject to boundary conditions  y(0) = 0,  y(1) = 1

which means that I(y) has an extremum at y(x). It seems obvious that the solution is y(x) = x, the straight line joining (0, 0) and (1, 1), but how do we prove this?