To find the area 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 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:
\[
x(x - 1)(x + 1) = 0
\]
Thus, the points of intersection are:
\[
x = 0, \quad x = 1, \quad x = -1
\]
### Step 2: Determine the Area
The area between the curves from \( x = -1 \) to \( x = 1 \) can be split into two parts: from \( x = -1 \) to \( x = 0 \) and from \( x = 0 \) to \( x = 1 \).
The area \( A \) can be expressed as:
\[
A = \int_{-1}^{0} (x^3 - x) \, dx + \int_{0}^{1} (x - x^3) \, dx
\]
### Step 3: Calculate the First Integral
Calculating the first integral:
\[
\int_{-1}^{0} (x^3 - x) \, dx
\]
This simplifies to:
\[
\int_{-1}^{0} (x^3) \, dx - \int_{-1}^{0} (x) \, dx
\]
Calculating each part:
1. For \( \int x^3 \, dx \):
\[
\frac{x^4}{4} \bigg|_{-1}^{0} = \frac{0^4}{4} - \frac{(-1)^4}{4} = 0 - \frac{1}{4} = -\frac{1}{4}
\]
2. For \( \int x \, dx \):
\[
\frac{x^2}{2} \bigg|_{-1}^{0} = \frac{0^2}{2} - \frac{(-1)^2}{2} = 0 - \frac{1}{2} = -\frac{1}{2}
\]
Thus, the first integral becomes:
\[
-\frac{1}{4} + \frac{1}{2} = -\frac{1}{4} + \frac{2}{4} = \frac{1}{4}
\]
### Step 4: Calculate the Second Integral
Now for the second integral:
\[
\int_{0}^{1} (x - x^3) \, dx
\]
This simplifies to:
\[
\int_{0}^{1} x \, dx - \int_{0}^{1} x^3 \, dx
\]
Calculating each part:
1. For \( \int x \, dx \):
\[
\frac{x^2}{2} \bigg|_{0}^{1} = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2}
\]
2. For \( \int x^3 \, dx \):
\[
\frac{x^4}{4} \bigg|_{0}^{1} = \frac{1^4}{4} - \frac{0^4}{4} = \frac{1}{4}
\]
Thus, the second integral becomes:
\[
\frac{1}{2} - \frac{1}{4} = \frac{2}{4} - \frac{1}{4} = \frac{1}{4}
\]
### Step 5: Combine the Areas
Now, we combine the areas from both integrals:
\[
A = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}
\]
### Final Answer
The area bounded by the curves \( y = x \) and \( y = x^3 \) is:
\[
\boxed{\frac{1}{2}} \text{ square units}
\]
