Find the area enclosed with in the curve
`y=x^(2), y=x^(3)`

Answer

To find the area enclosed between the curves \( y = x^2 \) and \( y = x^3 \), we can 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:

\[ x^2 = x^3 \]

Rearranging gives:

\[ x^3 - x^2 = 0 \]

Factoring out \( x^2 \):

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

Setting each factor to zero gives us the solutions:

\[ x^2 = 0 \quad \Rightarrow \quad x = 0 \]
\[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]

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

### Step 2: Set up the integral for the area
The area \( A \) between the curves from \( x = 0 \) to \( x = 1 \) can be found using the integral:

\[ A = \int_{0}^{1} (x^2 - x^3) \, dx \]

### Step 3: Evaluate the integral
Now we compute the integral:

\[ A = \int_{0}^{1} (x^2 - x^3) \, dx = \int_{0}^{1} x^2 \, dx - \int_{0}^{1} x^3 \, dx \]

Calculating each integral separately:

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

\[ \int x^2 \, dx = \frac{x^3}{3} \Big|_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} \]

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

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

Now substituting back into the area formula:

\[ A = \frac{1}{3} - \frac{1}{4} \]

### Step 4: Simplify the expression
To subtract these fractions, we need a common denominator. The least common multiple of 3 and 4 is 12:

\[ A = \frac{4}{12} - \frac{3}{12} = \frac{1}{12} \]

### Conclusion
Thus, the area enclosed by the curves \( y = x^2 \) and \( y = x^3 \) is:

\[ \boxed{\frac{1}{12}} \]