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

1. Find intersection points:

Set \( y = 3x \) equal to \( y = x^2 \): \[ x^2 = 3x \implies x(x - 3) = 0 \implies x = 0 \text{ or } x = 3 \]

2. Set up the integral:

The area between the curves is: \[ \int_{0}^{3} (3x - x^2) dx \]

3. Evaluate the integral:

\[ \int_{0}^{3} (3x - x^2) dx = \left[ \frac{3x^2}{2} - \frac{x^3}{3} \right]_{0}^{3} = \left( \frac{27}{2} - 9 \right) - 0 = \frac{9}{2} \]

Correct Answer: (B) \( \frac{9}{2} \)

Answer 2

We Know that,

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

This gives us three solutions:

$$ x = 0, \quad x = \sqrt{3}, \quad x = -\sqrt{3} $$

Since the curves intersect at three points, we'll need to set up two integrals to find the total area. We need to determine which function is "on top" in each interval:

• Interval $[- \sqrt{3}, 0]$: In this interval, $ x^3 $ is greater than $ 3x $.
• Interval $[0, \sqrt{3}]$: In this interval, $ 3x $ is greater than $ x^3 $.

Therefore, the area is:

$$ \text{Area} = \int_{-\sqrt{3}}^{0} (x^3 - 3x) \, dx + \int_{0}^{\sqrt{3}} (3x - x^3) \, dx $$

Now, we evaluate the integrals:

$ \int (x^3 - 3x) \, dx = \frac{x^4}{4} - \frac{3x^2}{2} + C $

$$ \int_{-\sqrt{3}}^{0} (x^3 - 3x) \, dx = \left[ \frac{0^4}{4} - \frac{3(0)^2}{2} \right] - \left[ \frac{(\sqrt{3})^4}{4} - \frac{3(-\sqrt{3})^2}{2} \right] $$ $$ = 0 - \left[ \frac{9}{4} - \frac{9}{2} \right] $$ $$ = -\left( \frac{9}{4} - \frac{18}{4} \right) $$ $$ = -\left( -\frac{9}{4} \right) $$ $$ = \frac{9}{4} $$

$ \int (3x - x^3) \, dx = \frac{3x^2}{2} - \frac{x^4}{4} + C $

$$ \int_{0}^{\sqrt{3}} (3x - x^3) \, dx = \left[ \frac{3(\sqrt{3})^2}{2} - \frac{(\sqrt{3})^4}{4} \right] - \left[ \frac{3(0)^2}{2} - \frac{(0)^4}{4} \right] $$ $$ = \left[ \frac{9}{2} - \frac{9}{4} \right] - 0 $$ $$ = \frac{18}{4} - \frac{9}{4} $$ $$ = \frac{9}{4} $$

Add the areas from the two intervals:

$$ \text{Total Area} = \frac{9}{4} + \frac{9}{4} = \frac{18}{4} = \frac{9}{2} $$