Find the area of the region enclosed by the curves
`y=x^(2)` and y = 2x

Answer

To find the area of the region enclosed by the curves \( y = x^2 \) and \( y = 2x \), we can follow these steps:

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

\[ x^2 = 2x \]

Rearranging gives:

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

Factoring out \( x \):

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

Thus, the solutions are:

\[ x = 0 \quad \text{and} \quad x = 2 \]

### Step 2: Determine the Area Between the Curves
The area \( A \) between the curves from \( x = 0 \) to \( x = 2 \) can be calculated using the integral of the upper curve minus the lower curve. Here, \( y = 2x \) is the upper curve and \( y = x^2 \) is the lower curve.

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

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

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

This can be split into two separate integrals:

\[ A = \int_{0}^{2} 2x \, dx - \int_{0}^{2} x^2 \, dx \]

Calculating each integral:

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

\[ \int 2x \, dx = x^2 \quad \text{(evaluated from 0 to 2)} \]

\[ = [2^2 - 0^2] = 4 \]

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

\[ \int x^2 \, dx = \frac{x^3}{3} \quad \text{(evaluated from 0 to 2)} \]

\[ = \left[\frac{2^3}{3} - \frac{0^3}{3}\right] = \frac{8}{3} \]

### Step 4: Combine the Results
Now, substituting back into the area equation:

\[ A = 4 - \frac{8}{3} \]

To combine these, we convert 4 into a fraction with a denominator of 3:

\[ 4 = \frac{12}{3} \]

Thus:

\[ A = \frac{12}{3} - \frac{8}{3} = \frac{4}{3} \]

### Final Answer
The area of the region enclosed by the curves \( y = x^2 \) and \( y = 2x \) is:

\[ \boxed{\frac{4}{3}} \text{ square units} \]