The area bounded by the parabola `y=x^(2)` and the line `y=2x is 4//3` sq. units.

Answer

To determine if the area bounded by the parabola \( y = x^2 \) and the line \( y = 2x \) is indeed \( \frac{4}{3} \) square units, we will follow these steps:

### Step 1: Find the points of intersection
We need to find the points where the parabola and the line intersect. This requires solving the equations \( y = x^2 \) and \( y = 2x \).

Set \( x^2 = 2x \):
\[ x^2 - 2x = 0 \]
Factoring gives:
\[ x(x - 2) = 0 \]
Thus, \( x = 0 \) or \( x = 2 \).

Now, substituting these \( x \)-values back into either equation to find the corresponding \( y \)-values:
- For \( x = 0 \): \( y = 0^2 = 0 \) → point \( (0, 0) \)
- For \( x = 2 \): \( y = 2^2 = 4 \) → point \( (2, 4) \)

### Step 2: Set up the integral for the area
The area \( A \) between the curves from \( x = 0 \) to \( x = 2 \) can be calculated using the integral:
\[ A = \int_{0}^{2} (2x - x^2) \, dx \]
Here, \( 2x \) is the line and \( x^2 \) is the parabola.

### Step 3: Calculate the integral
Now, we compute the integral:
\[ A = \int_{0}^{2} (2x - x^2) \, dx \]
Calculating the integral:
\[ A = \int_{0}^{2} 2x \, dx - \int_{0}^{2} x^2 \, dx \]

Calculating each integral separately:
1. \( \int 2x \, dx = x^2 \)
2. \( \int x^2 \, dx = \frac{x^3}{3} \)

Now evaluating from 0 to 2:
\[ \int_{0}^{2} 2x \, dx = [x^2]_{0}^{2} = 2^2 - 0^2 = 4 \]
\[ \int_{0}^{2} x^2 \, dx = \left[\frac{x^3}{3}\right]_{0}^{2} = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3} \]

Now substituting back into the area calculation:
\[ A = 4 - \frac{8}{3} \]
To combine these, convert 4 into a fraction:
\[ 4 = \frac{12}{3} \]
Thus,
\[ A = \frac{12}{3} - \frac{8}{3} = \frac{4}{3} \]

### Conclusion
The area bounded by the parabola \( y = x^2 \) and the line \( y = 2x \) is indeed \( \frac{4}{3} \) square units.