To find the area of the region bounded by the curves \(y = x\) and \(y = x^3\), we will follow these steps:
### Step 1: Find the points of intersection
To find the points where the two curves intersect, we set the equations equal to each other:
\[
x = x^3
\]
Rearranging gives:
\[
x^3 - x = 0
\]
Factoring out \(x\):
\[
x(x^2 - 1) = 0
\]
This gives us:
\[
x = 0 \quad \text{or} \quad x^2 - 1 = 0
\]
Solving \(x^2 - 1 = 0\) gives:
\[
x = 1 \quad \text{and} \quad x = -1
\]
Thus, the points of intersection are \(x = -1\), \(x = 0\), and \(x = 1\).
### Step 2: Determine the area between the curves
The area \(A\) between the curves from \(x = -1\) to \(x = 1\) can be calculated using the integral:
\[
A = \int_{-1}^{1} (y_{\text{upper}} - y_{\text{lower}}) \, dx
\]
For the interval \([-1, 0]\), \(y = x\) is above \(y = x^3\). For the interval \([0, 1]\), \(y = x\) is still above \(y = x^3\). Thus, we can express the area as:
\[
A = \int_{-1}^{1} (x - x^3) \, dx
\]
### Step 3: Evaluate the integral
We can split the integral into two parts:
\[
A = \int_{-1}^{1} x \, dx - \int_{-1}^{1} x^3 \, dx
\]
Calculating the first integral:
\[
\int x \, dx = \frac{x^2}{2} \quad \text{evaluated from } -1 \text{ to } 1
\]
\[
\left[ \frac{1^2}{2} - \frac{(-1)^2}{2} \right] = \left[ \frac{1}{2} - \frac{1}{2} \right] = 0
\]
Now for the second integral:
\[
\int x^3 \, dx = \frac{x^4}{4} \quad \text{evaluated from } -1 \text{ to } 1
\]
\[
\left[ \frac{1^4}{4} - \frac{(-1)^4}{4} \right] = \left[ \frac{1}{4} - \frac{1}{4} \right] = 0
\]
### Step 4: Combine the results
Thus, the area is:
\[
A = 0 - 0 = 0
\]
However, we need to consider the absolute values of the areas between the curves since we are looking for the total area. The area from \(x = 0\) to \(x = 1\) is positive, and we can calculate it as:
\[
A = 2 \int_{0}^{1} (x - x^3) \, dx
\]
Calculating:
\[
\int_{0}^{1} (x - x^3) \, dx = \int_{0}^{1} x \, dx - \int_{0}^{1} x^3 \, dx
\]
Calculating each part:
\[
\int_{0}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{1}{2}
\]
\[
\int_{0}^{1} x^3 \, dx = \left[ \frac{x^4}{4} \right]_{0}^{1} = \frac{1}{4}
\]
Thus:
\[
A = 2 \left( \frac{1}{2} - \frac{1}{4} \right) = 2 \left( \frac{1}{4} \right) = \frac{1}{2}
\]
### Final Answer
The area of the region bounded by the curves \(y = x\) and \(y = x^3\) is:
\[
\boxed{\frac{1}{2}} \text{ square units}
\]
