The given problem asks us to find the area enclosed between the parabola y^2 = 2x and the line x + y = 4. To solve this problem, we can follow the steps below:

1. Find the points of intersection:
- To find the points of intersection between the parabola and the line, we can substitute y = 4 - x into the equation of the parabola:
(4 - x)^2 = 2x
- Simplifying the equation:
16 - 8x + x^2 = 2x
x^2 + 10x - 16 = 0
- Solving this quadratic equation, we find two values of x: -8 and 2.
- Substituting these values back into the equation of the line, we can find the corresponding y-values:
For x = -8, y = 4 - (-8) = 12
For x = 2, y = 4 - 2 = 2
- Therefore, the points of intersection are (-8, 12) and (2, 2).

2. Determine the limits of integration:
- Since the parabola is below the line between these intersection points, we need to find the x-values where the parabola and the line intersect.
- We can set up the integral by integrating the difference between the two functions with respect to x within these limits:
∫[x1, x2] (y2 - y1) dx, where x1 = -8 and x2 = 2.

3. Calculate the area:
- We need to find the difference between the y-values of the parabola and the line at each x-value and integrate it with respect to x.
- The equation of the line is y = 4 - x, and the equation of the parabola is y^2 = 2x.
- Substituting y = 4 - x into the equation of the parabola, we get (4 - x)^2 = 2x.
- Expanding and simplifying the equation:
16 - 8x + x^2 = 2x
x^2 + 10x - 16 = 0
- Solving this quadratic equation, we find two values of x: -8 and 2.
- Therefore, the area enclosed between the parabola and the line is given by:
Area = ∫[-8, 2] (4 - x - (4 - x)^2) dx
= ∫[-8, 2] (x^2 - 6x + 8) dx
= [x^3/3 - 3x^2 + 8x]∣[-8, 2]
= [(2^3/3 - 3(2^2) + 8(2))] - [(-8^3/3 - 3(-8^2) + 8(-8))]
= [8/3 - 12 + 16] - [(-512/3 - 3(64) - 64)]
= [8/3 - 12 + 16] - [(-512/3 - 192 - 64)]
= (8