Question

The area (in sq. units) of the region $\left\{(x,y):x\ge 0,x+y\le 3,x^2\le 4y\text{and}y\le 1+x\right\}$ is:

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

Answer 1

The correct option is C

$\frac{5}{2}$

Explanation for correct answer:

Determining the area:

We have a region $\left\{(x,y):x\ge 0,x+y\le 3,x^2\le 4y\text{and}y\le 1+x\right\}$

The graph of the given curves is shown in the figure.

The area of a shaded region is given below.

$$\begin{gathered} A={\int }_0^1\left(1+\sqrt{x}\right)dx+{\int }_1^2\left(3-x\right)dx-{\int }_0^2\frac{x^2}{4}dx \\ \Rightarrow A={\left[x+\frac{x^{{\displaystyle \frac{3}{2}}}}{{}^{\frac{3}{2}}}\right]}_0^1+{\left[3x-\frac{x^2}{2}\right]}_1^2-{\left[\frac{x^3}{12}\right]}_0^2 \\ =\left[1+\frac{2}{3}-0\right]+\left[6-2-3+\frac{1}{2}\right]-\left[\frac{8}{12}-0\right] \\ =\frac{5}{3}+\frac{3}{2}-\frac{8}{12} \\ =\frac{20+18-8}{12} \\ =\frac{30}{12} \\ =\frac{5}{2} \end{gathered}$$

Therefore the area of the shaded portion is equal to $\frac{5}{2}\text{square units}$

Hence, the correct answer is option (C).