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

JEE Main · 2017 · 02 Apr 2017 · Offline · Area Under Curves

The area (in sq. units) of the region

{(x, y) : x ⩾ 0, x + y \leq 3, x^{2} \leq 4 y

and

y \leq 1 + x}

is :

Answer: (A)

The region lies above the parabola

y = \frac{x ^{2}}{4}

and below whichever of

y = 1 + x

and

y = 3 - x

is the lower. Those two upper curves cross at

x = 1

, where both equal

2

, with

1 + x

lower on

[0, 1]

and

3 - x

lower on

[1, 2]

; the parabola meets the line

3 - x

x = 2

. So the area is

\int_{0}^{1} (1 + x - \frac{x ^{2}}{4}) d x + \int_{1}^{2} (3 - x - \frac{x ^{2}}{4}) d x = \frac{19}{12} + \frac{11}{12} = \frac{5}{2}

JEE Main · 2017 · 02 Apr 2017 · Offline · Single correct option