Question

The area (in square units) of the region bounded by curves \( y = x \) and \( y = x^3 \) is:

• A. 0
• B. \( \frac{1}{2} \)
• C. \( \frac{1}{4} \)
• D. 4

Answer 1

The Correct Option is C

To find the area bounded by the curves \(y = x\) and \(y = x^3\), we first need to determine their points of intersection. We set the equations equal: 

\(x = x^3\)

This simplifies to:

\(x^3 - x = 0\)

Factor out \(x\):

\(x(x^2 - 1) = 0\)

Further factorize:

\(x(x - 1)(x + 1) = 0\)

Thus, the points of intersection are \(x = 0\), \(x = 1\), and \(x = -1\).

We will consider the area between the curves from \(x = 0\) to \(x = 1\), because the region is symmetric about the origin.

The area \(A\) is given by the integral of the difference of the functions:

\[ A = \int_{0}^{1} (x - x^3) \, dx \]

Calculate the integral:

\[ A = \int_{0}^{1} x \, dx - \int_{0}^{1} x^3 \, dx \]

Compute \(\int x \, dx\):

\[ \int x \, dx = \frac{x^2}{2} \] evaluated from 0 to 1 is \(\frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2}\)

Compute \(\int x^3 \, dx\):

\[ \int x^3 \, dx = \frac{x^4}{4} \] evaluated from 0 to 1 is \(\frac{1^4}{4} - \frac{0^4}{4} = \frac{1}{4}\)

Thus, the area is:

\[ A = \frac{1}{2} - \frac{1}{4} = \frac{1}{4} \]

Therefore, the area of the region bounded by the curves is \(\frac{1}{4}\) square units.

Answer 2

The given curves are \( y = x \) and \( y = x^3 \). To find the area of the region between these curves, first find the points of intersection:

\[ x = x^3 \implies x(x^2 - 1) = 0 \implies x = 0 \text{ or } x = \pm 1. \]

The region lies between \( x = 0 \) and \( x = 1 \) (since negative values will mirror the same area). The area between the curves is:

\[ \text{Area} = \int_0^1 (x - x^3) \, dx. \]

Evaluate the integral:

\[ \int_0^1 (x - x^3) \, dx = \int_0^1 x \, dx - \int_0^1 x^3 \, dx. \]

Compute each term:

\[ \int_0^1 x \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2}, \] \[ \int_0^1 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4}. \]

Subtract the results:

\[ \text{Area} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}. \]

Thus, the area of the region is \( \frac{1}{4} \) square units.