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

Answer: (C)

$\frac{1}{4}$

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.