`x+4y = 8`

`y = (8-x)/4` ----(1)

`y^2 = x+4` ----(2)

(1)^2 = (2)

`(8-x)^2/16 = x+4`

`64-16x+x^2 = 16x+64`

`x^2-32x = 0`

`x(x-32) = 0 `

x = 0 and x = 32 are the points which the two graphs intercept.

So the difference in integral from x = 0 to x = 32 will give the enclosed area by the graphs.

We can rearrange the two graphs as;

`y = (8-x)/4`

`y = +-sqrt(x+4)`

`y = (8-x)/4`

`int^32_0 (8-x)/4dx`

`= 1/4[int^32_0 8dx-int^32_0xdx]`

`= 1/4[8x-x^2/2]^32_0`

`= -64`

This (-) value comes since the graph has a negative gradient. So the area is below the x-axis.

`y = +sqrt(x+4)`

`int^32_0sqrt(x+4)dx`

`= [(sqrt(x+4))^(1/2+1)/(1/2+1)]^32_0`

`= sqrt36-sqrt4`

`= 6-2`

`= 4`

`y = -sqrt(x+4)`

`int^32_0-sqrt(x+4)dx`

`= -int^32_0sqrt(x+4)dx`

`= -4`

So if we consider two graphs are` y = (8-x)/4` and `y = +sqrt(x+4)` ;

*Area enclosed by graphs = 6-(-64) = 70*

So if we consider two graphs are `y = (8-x)/4` and `y = -sqrt(x+4)` ;

*Area enclosed by graphs = -4-(-64) = 60*