The area (in square units) of the region $\left\{(x, y) : x ≥ 0, x + y ≤ 3, x^2 ≤ 4y\, and\, y ≤1+ \sqrt{x}\right\}$ is

$\frac{5}{2}$

Let A be the area of the shaded region. Then,

$A=\int\limits_0^1(y_2-y_1)dx+\int\limits_1^2(y_4-y_3)dx$

$⇒A=\int\limits_0^1\left(1+\sqrt{x}-\frac{x^2}{4}\right)dx+\int\limits_1^2\left(3-x-\frac{x^2}{4}\right)dx$

$⇒A=\left[x+\frac{2}{3}x^{3/2}-\frac{x^3}{12}\right]_0^1+\left[3x-\frac{x^2}{2}-\frac{x^3}{12}\right]_1^2$

$⇒A=\frac{19}{12}+\frac{10}{3}-\frac{29}{12}=\frac{5}{2}$