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

To find the area bounded by the line $y = x$ and the curve $y = x^3$, first determine their points of intersection by solving the equation $x = x^3$. This simplifies to $x(x^2 - 1) = 0$, yielding $x = 0$ and $x = \pm1$. We will calculate the area between $x = 0$ and $x = 1$: \[ A = \int_{0}^{1} (x - x^3) \, dx \] Evaluating the integral: \[ \int (x - x^3) \, dx = \frac{x^2}{2} - \frac{x^4}{4} \] Substituting the limits: \[ A = \left[\frac{x^2}{2} - \frac{x^4}{4}\right]_{0}^{1} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4} \text{ square units} \] **Note:** The calculated area is $\frac{1}{4}$ square units. However, the closest provided option is (D) $0.5$ square units.