Find the area of the region bounded by curve $4x^2 = y$ and the line $y = 8x + 12$, using integration.

$\frac{128}{3}$ sq units

The correct answer is Option (2) → $\frac{128}{3}$ sq units

Given curve is $4x^2 = y$ and line is $y = 8x + 12$.

On solving both equation, we get

$4x^2 = 8x + 12$

$⇒x^2 = 2x + 3$

$⇒x^2 - 2x - 3 = 0$

$⇒x = 3, -1$

Required area $= \int\limits_{-1}^{3} \{(8x + 12) - 4x^2\} dx$

$= 4 \int\limits_{-1}^{3} (2x + 3 - x^2) dx$

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

$= 4 \left[ (9 + 9 - 9) - \left( 1 - 3 + \frac{1}{3} \right) \right]$

$= 4 \left( 9 + 2 - \frac{1}{3} \right)$

$= 4 \left( 11 - \frac{1}{3} \right)$

$= 4 \times \frac{32}{3} = \mathbf{\frac{128}{3} \text{ sq. units}}$