Find the area of the region bounded by the curve $y = x^3, y = x + 6$ and $x = 0$.

10

The correct answer is Option (2) → 10

We have,

$y = x^3 \quad \dots(i)$

$y = x + 6 \quad \dots(ii)$

$\text{and} \quad x = 0 \quad \dots(iii)$

On solving Eqs. (i) and (ii), we get

$x^3 = x + 6$

$⇒x^3 - x - 6 = 0$

$⇒x^2(x - 2) + 2x(x - 2) + 3(x - 2) = 0$

$⇒(x - 2)(x^2 + 2x + 3) = 0$

$⇒x = 2, \text{ and two imaginary points}$

Let $A_1 = \text{Area under the line, } y = x + 6 = \text{Area of OABDO}$

and $A_2 = \text{Area under the curve, } y = x^3 = \text{Area of OCBDO}$

$∴\text{Area of shaded region} = A_1 - A_2$

$= \int\limits_{0}^{2} (x + 6) \, dx - \int\limits_{0}^{2} x^3 \, dx$

$= \int\limits_{0}^{2} (x + 6 - x^3) \, dx = \left[ \frac{x^2}{2} + 6x - \frac{x^4}{4} \right]_{0}^{2} = \left[ \frac{4}{2} + 12 - \frac{16}{4} - 0 \right]$

$= [2 + 12 - 4] = 10 \text{ sq. units}$