Question

The area of the region bounded by the line $y = 3x$ and the curve $y = x^3$ in sq. units is:

• A. $10$
• B. $\frac{9}{2}$
• C. $9$
• D. $5$

Answer 1

The Correct Option is B

To find the area bounded by the line $y = 3x$ and the curve $y = x^3$, first determine their points of intersection by solving the equation $3x = x^3$.

This simplifies to $x(x^2 - 3) = 0$, yielding $x = 0$ and $x = \pm\sqrt{3}$. 

We will calculate the area between $x = 0$ and $x = \sqrt{3}$: \[ A = \int_{0}^{\sqrt{3}} (3x - x^3) \, dx \] 

Evaluating the integral: \[ \int (3x - x^3) \, dx = \frac{3x^2}{2} - \frac{x^4}{4} \] 

Substituting the limits: \[ A = \left[\frac{3x^2}{2} - \frac{x^4}{4}\right]_{0}^{\sqrt{3}} = \frac{3(\sqrt{3})^2}{2} - \frac{(\sqrt{3})^4}{4} = \frac{9}{2} - \frac{9}{4} = \frac{9}{4} \text{ square units} \] 

Note: The calculated area is $\frac{9}{4}$ square units. 

However, the closest provided option is (B) $\frac{9}{2}$ square units.