The area bounded by `y^(2)=4xandy=2x-4` is

Answer

To find the area bounded by the parabola \( y^2 = 4x \) and the line \( y = 2x - 4 \), we will follow these steps:

### Step 1: Find the Points of Intersection
To find the points of intersection between the parabola and the line, we substitute \( y = 2x - 4 \) into the parabola's equation \( y^2 = 4x \).

\[ (2x - 4)^2 = 4x \]

Expanding this gives:

\[ 4x^2 - 16x + 16 = 4x \]

Rearranging the equation:

\[ 4x^2 - 20x + 16 = 0 \]

Dividing the entire equation by 4:

\[ x^2 - 5x + 4 = 0 \]

### Step 2: Factor the Quadratic Equation
Now we can factor the quadratic equation:

\[ (x - 1)(x - 4) = 0 \]

Thus, the solutions are:

\[ x = 1 \quad \text{and} \quad x = 4 \]

### Step 3: Find the Corresponding y-coordinates
Next, we find the y-coordinates for these x-values using the line equation \( y = 2x - 4 \).

For \( x = 1 \):

\[ y = 2(1) - 4 = -2 \]

For \( x = 4 \):

\[ y = 2(4) - 4 = 4 \]

Thus, the points of intersection are \( (1, -2) \) and \( (4, 4) \).

### Step 4: Set Up the Integral for Area
The area \( A \) between the curves from \( x = 1 \) to \( x = 4 \) can be expressed as:

\[ A = \int_{1}^{4} (y_{\text{line}} - y_{\text{parabola}}) \, dx \]

We need to express both curves in terms of \( x \):

1. For the line: \( y = 2x - 4 \) (already in terms of \( x \))
2. For the parabola: \( y^2 = 4x \Rightarrow y = \sqrt{4x} = 2\sqrt{x} \)

Thus, the area becomes:

\[ A = \int_{1}^{4} ((2x - 4) - (2\sqrt{x})) \, dx \]

### Step 5: Evaluate the Integral
Now we compute the integral:

\[ A = \int_{1}^{4} (2x - 4 - 2\sqrt{x}) \, dx \]

Breaking it down:

\[ A = \int_{1}^{4} 2x \, dx - \int_{1}^{4} 4 \, dx - \int_{1}^{4} 2\sqrt{x} \, dx \]

Calculating each integral:

1. \( \int 2x \, dx = x^2 \) evaluated from 1 to 4 gives:

\[ [4^2 - 1^2] = [16 - 1] = 15 \]

2. \( \int 4 \, dx = 4x \) evaluated from 1 to 4 gives:

\[ [4(4) - 4(1)] = [16 - 4] = 12 \]

3. \( \int 2\sqrt{x} \, dx = \frac{4}{3}x^{3/2} \) evaluated from 1 to 4 gives:

\[ \left[\frac{4}{3}(4^{3/2}) - \frac{4}{3}(1^{3/2})\right] = \left[\frac{4}{3}(8) - \frac{4}{3}(1)\right] = \left[\frac{32}{3} - \frac{4}{3}\right] = \frac{28}{3} \]

### Step 6: Combine the Results
Now substituting back into the area formula:

\[ A = 15 - 12 - \frac{28}{3} \]

Converting 15 and 12 into fractions with a common denominator of 3:

\[ A = \frac{45}{3} - \frac{36}{3} - \frac{28}{3} = \frac{45 - 36 - 28}{3} = \frac{-19}{3} \]

### Step 7: Final Area Calculation
The area is:

\[ A = \frac{19}{3} \text{ square units} \]