If the area (in sq. units) of the region $$\{(x, y): y^2 \leq 4x, x + y \leq 1, x \geq 0, y \geq 0\}$$ is $$a\sqrt{2} + b$$, then a - b is

The region is described by four simultaneous conditions:

$$y^2 \le 4x,\qquad x+y \le 1,\qquad x \ge 0,\qquad y \ge 0.$$

Because of $$x \ge 0,\; y \ge 0,$$ we remain in the first quadrant. The inequality $$y^2 \le 4x$$ can be rewritten as

$$x \ge \dfrac{y^2}{4},$$

which is the right-opening parabola $$x=\dfrac{y^2}{4}.$$ The linear inequality $$x+y \le 1$$ gives

$$x \le 1-y,$$

which is the straight line $$x = 1-y$$ with negative slope, restricted to the first quadrant. Hence, for each admissible value of $$y$$ the $$x$$-coordinate is trapped between

$$x_{\text{left}} = \dfrac{y^2}{4}\quad\text{and}\quad x_{\text{right}} = 1-y.$$

For such an $$x$$-interval to exist we must have

$$\dfrac{y^2}{4} \le 1-y.$$

Solving this inequality will give the vertical range of the region.

Multiplying by $$4$$ we obtain

$$y^2 + 4y - 4 \le 0.$$

The quadratic equation $$y^2 + 4y - 4 = 0$$ has solutions

$$y = \dfrac{-4 \pm \sqrt{4^2-4(1)(-4)}}{2} = \dfrac{-4 \pm \sqrt{16+16}}{2} = \dfrac{-4 \pm 4\sqrt2}{2} = -2 \pm 2\sqrt2.$$

Only the positive root is relevant in the first quadrant, so

$$y_{\max} = -2 + 2\sqrt2 = 2(\sqrt2-1).$$

Therefore $$y$$ varies from $$0$$ to $$2\!\left(\sqrt2-1\right).$$

Using the formula “Area between two curves expressed as $$x=f(y)$$ is $$A=\displaystyle\int_{y_1}^{y_2} \bigl(x_{\text{right}}-x_{\text{left}}\bigr)\,dy,$$” we write

$$A = \int_{0}^{\,2(\sqrt2-1)}\Bigl[(1-y)-\dfrac{y^2}{4}\Bigr]\,dy.$$

Inside the integral the simplified integrand is

$$1-y-\dfrac{y^2}{4}.$$

Now we integrate term by term:

$$\int 1\,dy = y,\qquad \int (-y)\,dy = -\dfrac{y^2}{2},\qquad \int \!\left(-\dfrac{y^2}{4}\right)dy = -\dfrac{y^3}{12}.$$

Hence

$$A = \Bigl[y - \dfrac{y^2}{2} - \dfrac{y^3}{12}\Bigr]_{0}^{\,2(\sqrt2-1)}.$$

Let us denote $$\alpha = 2(\sqrt2-1)=2\sqrt2-2.$$ We evaluate every power of $$\alpha$$ explicitly.

First power: $$\alpha = 2\sqrt2-2.$$

Second power: $$\alpha^2 = \bigl(2(\sqrt2-1)\bigr)^2 = 4(\sqrt2-1)^2 = 4(2-2\sqrt2+1) = 4(3-2\sqrt2) = 12-8\sqrt2.$$

Third power: $$\alpha^3 = \alpha^2\alpha = (12-8\sqrt2)\bigl(2\sqrt2-2\bigr) = (12-8\sqrt2)\,2(\sqrt2-1) = 2\!\left[12\sqrt2-12-16+8\sqrt2\right] = 2\!\left(20\sqrt2-28\right) = 40\sqrt2-56.$$

Substituting these expressions back into the bracket:

$$\begin{aligned} A &= \Bigl[\alpha - \dfrac{\alpha^2}{2} - \dfrac{\alpha^3}{12}\Bigr] - 0 \\[4pt] &= \Bigl[\,\bigl(2\sqrt2-2\bigr) - \dfrac{12-8\sqrt2}{2} - \dfrac{40\sqrt2-56}{12}\Bigr]. \end{aligned}$$

Compute each fraction step by step:

$$\dfrac{\alpha^2}{2} = \dfrac{12-8\sqrt2}{2} = 6-4\sqrt2,$$

$$\dfrac{\alpha^3}{12} = \dfrac{40\sqrt2-56}{12} = \dfrac{10\sqrt2-14}{3}.$$

So

$$\begin{aligned} A &= \Bigl[(2\sqrt2-2) - (6-4\sqrt2) - \dfrac{10\sqrt2-14}{3}\Bigr] \\[4pt] &= \Bigl[(2\sqrt2-2) -6 +4\sqrt2\Bigr] - \dfrac{10\sqrt2-14}{3} \\[4pt] &= \bigl(6\sqrt2-8\bigr) - \dfrac{10\sqrt2-14}{3}. \end{aligned}$$

Bring everything to the common denominator $$3$$:

$$6\sqrt2-8 = \dfrac{18\sqrt2-24}{3},$$

so

$$A = \dfrac{18\sqrt2-24 - (10\sqrt2-14)}{3} = \dfrac{18\sqrt2-24-10\sqrt2+14}{3} = \dfrac{8\sqrt2-10}{3}.$$

Thus the area can be written in the required form

$$A = \dfrac{8}{3}\sqrt2 - \dfrac{10}{3}.$$

Comparing with $$a\sqrt2 + b$$ we identify $$a=\dfrac{8}{3}$$ and $$b=-\dfrac{10}{3}.$$

Finally,

$$a-b = \dfrac{8}{3} - \bigl(-\dfrac{10}{3}\bigr) = \dfrac{18}{3} = 6.$$

Hence, the correct answer is Option A.