Draw a rough sketch of the region enclosed between the curve `y ^(2) = 4x` and the line `y = 2x.` Also, determine the area of the region.

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the curves
The curves we need to analyze are:
1. The parabola given by the equation \( y^2 = 4x \).
2. The line given by the equation \( y = 2x \).

### Step 2: Find the points of intersection
To find the area enclosed between the two curves, we first need to determine the points where they intersect. We can do this by substituting the equation of the line into the equation of the parabola.

Substituting \( y = 2x \) into \( y^2 = 4x \):
\[ (2x)^2 = 4x \]
\[ 4x^2 = 4x \]
Dividing both sides by 4 (assuming \( x \neq 0 \)):
\[ x^2 = x \]
This gives us:
\[ x(x - 1) = 0 \]
Thus, \( x = 0 \) or \( x = 1 \).

Now, substituting these \( x \) values back into the line equation to find the corresponding \( y \) values:
- For \( x = 0 \): \( y = 2(0) = 0 \) → Point (0, 0)
- For \( x = 1 \): \( y = 2(1) = 2 \) → Point (1, 2)

### Step 3: Sketch the curves
We can now sketch the curves:
- The parabola \( y^2 = 4x \) opens to the right and passes through the origin (0, 0).
- The line \( y = 2x \) also passes through the origin and has a slope of 2.

The intersection points (0, 0) and (1, 2) are marked, and the area we are interested in is between these two curves from \( x = 0 \) to \( x = 1 \).

### Step 4: Set up the integral for the area
The area \( A \) between the curves can be found using the integral:
\[ A = \int_{a}^{b} (f(x) - g(x)) \, dx \]
where \( f(x) \) is the upper curve and \( g(x) \) is the lower curve.

Here, from \( x = 0 \) to \( x = 1 \):
- The upper curve is the line \( y = 2x \).
- The lower curve is the parabola \( y = \sqrt{4x} = 2\sqrt{x} \).

Thus, the area can be expressed as:
\[ A = \int_{0}^{1} (2x - 2\sqrt{x}) \, dx \]

### Step 5: Calculate the integral
Now we calculate the integral:
\[ A = \int_{0}^{1} (2x - 2\sqrt{x}) \, dx = 2 \int_{0}^{1} (x - \sqrt{x}) \, dx \]

Calculating each part:
1. For \( \int x \, dx \):
\[ \int x \, dx = \frac{x^2}{2} \Big|_{0}^{1} = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2} \]

2. For \( \int \sqrt{x} \, dx \):
\[ \int \sqrt{x} \, dx = \frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2} \Big|_{0}^{1} = \frac{2}{3}(1^{3/2}) - \frac{2}{3}(0^{3/2}) = \frac{2}{3} \]

Putting it all together:
\[ A = 2 \left( \frac{1}{2} - \frac{2}{3} \right) = 2 \left( \frac{3}{6} - \frac{4}{6} \right) = 2 \left( -\frac{1}{6} \right) = -\frac{1}{3} \]

Since area cannot be negative, we take the absolute value:
\[ A = \frac{1}{3} \text{ square units} \]

### Final Answer
The area of the region enclosed between the curve \( y^2 = 4x \) and the line \( y = 2x \) is \( \frac{1}{3} \) square units.

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