Question

The area (in s units) of the region $\{ (x , y) : x \geq 0 , x + y \leq 3, x^2 \leq 4 y$ and $y \leq 1 + \sqrt{x} \}$ is

• A. $\frac{3}{2}$
• B. $\frac{7}{3}$
• C. $\frac{5}{2}$
• D. $\frac{59}{12}$

Answer 1

The Correct Option is C

To find the area of the region defined by the inequalities:

• x \geq 0
• x + y \leq 3
• x^2 \leq 4y
• y \leq 1 + \sqrt{x}

We first need to graph the inequalities and identify the enclosed region:

1. x \geq 0: This indicates the region is to the right of the y-axis.
2. x + y \leq 3: This is a line y = 3 - x and the region below it.
3. x^2 \leq 4y: This rearranges to y \geq \frac{x^2}{4}, a parabola opening upwards.
4. y \leq 1 + \sqrt{x}: This rearranges to y = 1 + \sqrt{x}, the region below this curve.

We need to find the intersections of these curves:

• The line x + y = 3 intersects the parabola y = \frac{x^2}{4} at points found by solving:

\begin{align*} y &= 3 - x,\\ y &= \frac{x^2}{4}.\\ \Rightarrow 3 - x &= \frac{x^2}{4},\\ \Rightarrow 12 - 4x &= x^2,\\ \Rightarrow x^2 + 4x - 12 &= 0. \end{align*}

Solving this quadratic equation:

\begin{align*} x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},\\ x &= \frac{-4 \pm \sqrt{16 + 48}}{2},\\ x &= \frac{-4 \pm 8}{2}. \end{align*}

This gives x = 2 and x = -6 (discard -6 as it doesn't satisfy x \geq 0).

At x = 2, y = 1.

The boundary of the region is thus bounded by:

• Between x = 0 to x = 2, using the parabola and line intersection.

Let's compute the area:

\begin{align*} \text{Area} &= \int_{0}^{2} \left((1 + \sqrt{x}) - \frac{x^2}{4}\right) \, dx. \end{align*}

Calculating the integral:

\begin{align*} \int (1 + \sqrt{x} - \frac{x^2}{4}) \, dx &= \left[x + \frac{2}{3}x^{3/2} - \frac{x^3}{12}\right]_{0}^{2},\\ &= \left[2 + \frac{2}{3}(2\sqrt{2}) - \frac{2^3}{12}\right] - \left[0\right],\\ &= \left[2 + \frac{4\sqrt{2}}{3} - \frac{8}{12}\right],\\ &= \left[2 + \frac{4\sqrt{2}}{3} - \frac{2}{3}\right],\\ &= \left[2 - \frac{2}{3} + \frac{4\sqrt{2}}{3}\right],\\ &= \frac{5}{2}. \end{align*}

Thus, the area of the region is \frac{5}{2} square units.