Question

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

• A. $0.2$ sq. units
• B. $0.3$ sq. units
• C. $0.4$ sq. units
• D. $0.5$ sq. units

Answer 1

The Correct Option is D

1. Find intersection points:

Set \( y = x \) equal to \( y = x^3 \): \[ x^3 = x \implies x(x^2 - 1) = 0 \implies x = -1, 0, 1 \]

2. Set up the integral:

The area between the curves is symmetric, so compute from 0 to 1 and double: \[ 2 \int_{0}^{1} (x - x^3) dx \]

3. Evaluate the integral:

\[ 2 \left[ \frac{x^2}{2} - \frac{x^4}{4} \right]_{0}^{1} = 2 \left( \frac{1}{2} - \frac{1}{4} \right) = 2 \cdot \frac{1}{4} = 0.5 \]

Correct Answer: (D) 0.5 sq. units

Answer 2

We got, 

$$ x = x^3 $$ $$ x^3 - x = 0 $$ $$ x(x^2 - 1) = 0 $$ $$ x(x - 1)(x + 1) = 0 $$

This gives us three intersection points:

$$ x = 0, \quad x = 1, \quad x = -1 $$

• Interval $[-1, 0]$: In this interval, $ x^3 $ is greater than $ x $.
• Interval $[0, 1]$: In this interval, $ x $ is greater than $ x^3 $.

Therefore, the area is:

$$ \text{Area} = \int_{-1}^{0} (x^3 - x) \, dx + \int_{0}^{1} (x - x^3) \, dx $$

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

$$ \int_{-1}^{0} (x^3 - x) \, dx = \left[ \frac{0^4}{4} - \frac{0^2}{2} \right] - \left[ \frac{(-1)^4}{4} - \frac{(-1)^2}{2} \right] $$ $$ = 0 - \left[ \frac{1}{4} - \frac{1}{2} \right] $$ $$ = -\left( \frac{1}{4} - \frac{2}{4} \right) $$ $$ = -\left( -\frac{1}{4} \right) $$ $$ = \frac{1}{4} $$

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

$$ \int_{0}^{1} (x - x^3) \, dx = \left[ \frac{1^2}{2} - \frac{1^4}{4} \right] - \left[ \frac{0^2}{2} - \frac{0^4}{4} \right] $$ $$ = \left[ \frac{1}{2} - \frac{1}{4} \right] - 0 $$ $$ = \frac{1}{2} - \frac{1}{4} $$ $$ = \frac{1}{4} $$

$$ \text{Total Area} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} $$