Area of the region lying between the curve `y=x^(3)` and the line `y=x` is ….............

Answer

To find the area of the region lying between the curve \( y = x^3 \) and the line \( y = x \), we will follow these steps:

### Step 1: Find the Points of Intersection
To determine the area between the two curves, we first need to find the points where they intersect. We set the equations equal to each other:

\[ x^3 = x \]

Rearranging gives us:

\[ x^3 - x = 0 \]

Factoring out \( x \):

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

This gives us:

\[ x = 0, \quad x^2 - 1 = 0 \implies x = 1 \quad \text{and} \quad x = -1 \]

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

### Step 2: Set Up the Integral for Area Calculation
The area \( A \) between the curves from \( x = 0 \) to \( x = 1 \) can be calculated using the integral:

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

Here, \( x \) is the upper curve and \( x^3 \) is the lower curve in this interval.

### Step 3: Calculate the Integral
Now we compute the integral:

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

Calculating each integral separately:

1. For \( \int_0^1 x \, dx \):

\[ \int x \, dx = \frac{x^2}{2} \bigg|_0^1 = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2} \]

2. For \( \int_0^1 x^3 \, dx \):

\[ \int x^3 \, dx = \frac{x^4}{4} \bigg|_0^1 = \frac{1^4}{4} - \frac{0^4}{4} = \frac{1}{4} \]

Now substituting back into our area calculation:

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

### Step 4: Double the Area for Symmetry
Since the area is symmetric about the y-axis, we need to double the area calculated for \( x = 0 \) to \( x = 1 \):

\[ \text{Total Area} = 2A = 2 \times \frac{1}{4} = \frac{1}{2} \]

### Final Answer
Thus, the area of the region lying between the curve \( y = x^3 \) and the line \( y = x \) is:

\[ \boxed{\frac{1}{2}} \text{ square units} \]