First slide

Area of bounded Regions

Question

The area(in sq. units) of the region $\{(x, y) \in R^{2} | 4 x^{2} \leq y \leq 8 x + 12\}$ is _____

Moderate

A

$\frac{127}{3}$
B

$\frac{128}{3}$
C

$\frac{124}{3}$
D

$\frac{125}{3}$

Solution

We have $4 x^{2} = y$
$\Rightarrow y = 8 x + 12$
$\Rightarrow 4 x^{2} = 8 x + 12$
$\Rightarrow x^{2} - x - 3 = 0$
$\Rightarrow x^{2} - 2 x - 3 = 0$
$\Rightarrow x^{2} - 3 x + x - 3 = 0$
$\Rightarrow (x + 1) (x - 3) = 0$
$\Rightarrow x = - 1, 3$
$∴ A r e a = A = \int_{- 1}^{3} (8 x + 12 - 4 x^{2}) d x$
$= \frac{8 x^{2}}{2} + 12 x - {\frac{4 x^{3}}{3}|}_{- 1}^{3} = (4 (9) + 36 - 36) - (4 - 12 + \frac{4}{3}) = 36 + 8 - \frac{4}{3}$
$= 44 - \frac{4}{3} = \frac{132 - 4}{3} = \frac{128}{3}$

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