To find the area enclosed between the curves \( y = x \) and \( y = x^3 \), we will follow these steps:
### Step 1: Find the points of intersection
To determine the area between the curves, we first need 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:
\[
x = 0, \quad x^2 - 1 = 0 \quad \Rightarrow \quad x = 1 \quad \text{and} \quad x = -1
\]
Thus, the points of intersection are \( (-1, -1) \), \( (0, 0) \), and \( (1, 1) \).
### Step 2: Determine the area between the curves
The area between the curves from \( x = 0 \) to \( x = 1 \) can be calculated using the formula for the area between two curves:
\[
\text{Area} = \int_{a}^{b} (f(x) - g(x)) \, dx
\]
Here, \( f(x) = x \) (the upper curve) and \( g(x) = x^3 \) (the lower curve). Therefore, we need to compute:
\[
\text{Area} = \int_{0}^{1} (x - x^3) \, dx
\]
### Step 3: Compute the integral
Now we evaluate the integral:
\[
\text{Area} = \int_{0}^{1} (x - x^3) \, dx = \int_{0}^{1} x \, dx - \int_{0}^{1} x^3 \, dx
\]
Calculating each integral separately:
1. For \( \int_{0}^{1} x \, dx \):
\[
\int x \, dx = \frac{x^2}{2} \Big|_{0}^{1} = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2}
\]
2. For \( \int_{0}^{1} x^3 \, dx \):
\[
\int x^3 \, dx = \frac{x^4}{4} \Big|_{0}^{1} = \frac{1^4}{4} - \frac{0^4}{4} = \frac{1}{4}
\]
### Step 4: Combine the results
Now, substituting back into the area formula:
\[
\text{Area} = \left( \frac{1}{2} - \frac{1}{4} \right) = \frac{1}{2} - \frac{1}{4} = \frac{2}{4} - \frac{1}{4} = \frac{1}{4}
\]
### Step 5: Total area between the curves
Since the area from \( x = -1 \) to \( x = 0 \) is identical to the area from \( x = 0 \) to \( x = 1 \), we multiply by 2:
\[
\text{Total Area} = 2 \times \frac{1}{4} = \frac{1}{2}
\]
Thus, the area enclosed between the curves \( y = x \) and \( y = x^3 \) is \( \frac{1}{2} \) square units.
