The area of the region given by $\mathrm{A}=\left\{(\mathrm{x}, \mathrm{y}): \mathrm{x}^{2} \leq \mathrm{y} \leq \min \{\mathrm{x}+2,4-3 \mathrm{x}\}\right\}$ is :

JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Area Under Curves

The area of the region given by

A = {(x, y) : x^{2} \leq y \leq min {x + 2, 4 - 3 x}}

is :

Answer: (B)

The region is bounded below by

y = x^{2}

and above by

y = min {x + 2, 4 - 3 x}

.

The two lines intersect where

x + 2 = 4 - 3 x ⟹ x = \frac{1}{2}

.

The parabola

y = x^{2}

intersects

y = x + 2

where

x^{2} = x + 2 ⟹ x = - 1

x = 2

. Since we need

x^{2} \leq x + 2

, we have

x \in [- 1, 2]

.

The parabola intersects

y = 4 - 3 x

where

x^{2} = 4 - 3 x ⟹ x = 1

x = - 4

.

The region splits at

x = \frac{1}{2}

A = \int_{- 1}^{1/2} (x + 2 - x^{2}) d x + \int_{1/2}^{1} (4 - 3 x - x^{2}) d x

= [\frac{x ^{2}}{2} + 2 x - \frac{x ^{3}}{3}]_{- 1}^{1/2} + [4 x - \frac{3 x ^{2}}{2} - \frac{x ^{3}}{3}]_{1/2}^{1}

Evaluating gives

\frac{17}{6}

JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Single correct option