Solution:

$I=\int_0^1 (xy +\frac {1}{2}y'^2)dx$

where $F=xy+ \frac {1}{2}y'^2$-----(2)

Assume the trial solution ,

$\bar y(x)=c_0+c_1x+c_2 x^2$ ----(3)

put $x=0$ in eqn(3)

$\bar y(0)=c_0+c_1(0)+c_2 (0)$

But $\bar y(0)=0$

$0=c_0+(0)+(0)$

$0=c_0$

Put $x=1$ in eqn(3)

$y_(1) = c_0+c_1(1)+c_2(1)$

but $\bar y (1) =0 $ and $c_0 =0$

$0 =0+c_1+c_2$

$c_2=-c_1$

Put $c_0=0$ and $c_2 =-c_1$ in eqn (3)

$\bar y (x) = 0+c_1x-c_1x^2$

$\bar y (x) = c_1(x-x^2)----(4)$

diff.(4) w.r.t x

$\bar y'(x)=c_1(1-2x)----(5)$

Put these values in eqn(1)

$I=\int_0^1 \{x[c_1(x-x^2)]+\frac{1}{2}c_1^2(1-2x)^2\} dx$

$I=c_1 \int_0^1 \{(x^2-x^3)]+\frac{1}{2}c_1(1-4x+4x^2)\} dx$

$c_1\{[\frac{x^3}{3} -\frac{x^4}{4}]+\frac{1}{2} c_1[1-\frac{4x^2}{2}+\frac{4x^3}{3}]\}_0^1 $

upper limit -lower limit

$c_1\{[\frac{1}{3} -\frac{1}{4}]+\frac{1}{2} c_1[1-\frac{4}{2}+\frac{4}{3}]\} -c_1\{[0-0]-\frac{1}{2} c_1[0-0+0]\}$

= $c_1\{\frac{1}{12} +\frac{1}{2} c_1 (\frac{1}{3})\}-c_1[0]$

= $c_1\{\frac{1}{12} +\frac{1}{2} c_1 \frac{1}{3)}\}$

$I=c_1(\frac{1}{12}) +c_1^2(\frac{1}{6})------(6) $

diff eqn (6) w.r.t $c_1$

$\frac {dI}{dc_1} =\frac{1}{12} +2c_1(\frac{1}{6})$

but $\frac {dI}{dc_1}$ is stationary value ,

hence $\frac{dI}{dc_1}=0$

$0= \frac{1}{12} +c_1(\frac{1}{3}) $

$\frac{-1}{12} = c_1(\frac{1}{3}) $

$c_1 = \frac{-1}{4}$

put $c_1 = \frac{-1}{4}$ in eqn (4)

$\bar y(x) = -\frac{1}{4} (x-x^2)$

$\bar y(x) = -\frac{1}{4} x(1-x)$