If the area (in sq. units) of the region {(x, | vedclass

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(B) The region is bounded by the parabola $y^2 = 4x$ and the line $x + y = 1$ in the first quadrant.To find the intersection point,substitute $x = 1 -

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$6$

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(B) The region is bounded by the parabola $y^2 = 4x$ and the line $x + y = 1$ in the first quadrant.To find the intersection point,substitute $x = 1 - y$ into $y^2 = 4x$:$y^2 = 4(1 - y) \implies y^2 + 4y - 4 = 0$.Using the quadratic formula,$y = \frac{-4 \pm \sqrt{16 - 4(1)(-4)}}{2} = \frac{-4 \pm \sqrt{32}}{2} = -2 \pm 2\sqrt{2}$.Since $y \ge 0$,we have $y = 2\sqrt{2} - 2$. Then $x = 1 - y = 1 - (2\sqrt{2} - 2) = 3 - 2\sqrt{2}$.The area is given by $\int_{0}^{3-2\sqrt{2}} 2\sqrt{x} dx + \int_{3-2\sqrt{2}}^{1} (1 - x) dx$.$= \left[ \frac{4}{3} x^{3/2} \right]_{0}^{3-2\sqrt{2}} + \left[ x - \frac{x^2}{2} \right]_{3-2\sqrt{2}}^{1}$.$= \frac{4}{3} (3-2\sqrt{2})^{3/2} + \left( (1 - 1/2) - ((3-2\sqrt{2}) - \frac{(3-2\sqrt{2})^2}{2}) \right)$.Note that $(3-2\sqrt{2}) = (\sqrt{2}-1)^2$,so $(3-2\sqrt{2})^{3/2} = (\sqrt{2}-1)^3 = 2\sqrt{2} - 3(2) + 3\sqrt{2} - 1 = 5\sqrt{2} - 7$.Area $= \frac{4}{3}(5\sqrt{2} - 7) + \frac{1}{2} - (3 - 2\sqrt{2} - \frac{9 + 8 - 12\sqrt{2}}{2}) = \frac{20\sqrt{2} - 28}{3} + \frac{1}{2} - (3 - 2\sqrt{2} - 8.5 + 6\sqrt{2}) = \frac{20\sqrt{2} - 28}{3} + \frac{1}{2} - (-5.5 + 4\sqrt{2}) = \frac{20\sqrt{2} - 28}{3} + 6 - 4\sqrt{2} = \frac{20\sqrt{2} - 28 + 18 - 12\sqrt{2}}{3} = \frac{8\sqrt{2} - 10}{3} = \frac{8}{3}\sqrt{2} - \frac{10}{3}$.Thus,$a = \frac{8}{3}$ and $b = -\frac{10}{3}$.$a - b = \frac{8}{3} - (-\frac{10}{3}) = \frac{18}{3} = 6$.

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

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