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

Options

A. 9
B. 5
C. 4
D. 2

Answer

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

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

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

Expanding the left side:

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

Rearranging gives:

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

Dividing the entire equation by 4:

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

Factoring:

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

Thus, the solutions are:

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

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

For \( x = 1 \):

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

For \( x = 4 \):

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

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

### Step 3: Set Up the Integral for Area
The area \( A \) between the curves from \( x = 1 \) to \( x = 4 \) can be calculated using the formula:

\[ A = \int_{y_1}^{y_2} (x_1 - x_2) \, dy \]

Where \( x_1 \) is the right curve and \( x_2 \) is the left curve. We need to express \( x \) in terms of \( y \).

From the line \( y = 2x - 4 \):

\[ x_1 = \frac{y + 4}{2} \]

From the parabola \( y^2 = 4x \):

\[ x_2 = \frac{y^2}{4} \]

Now, the limits of integration are from \( y = -2 \) to \( y = 4 \).

### Step 4: Calculate the Area
Now we can set up the integral:

\[ A = \int_{-2}^{4} \left( \frac{y + 4}{2} - \frac{y^2}{4} \right) dy \]

Simplifying the integrand:

\[ A = \int_{-2}^{4} \left( \frac{y + 4 - \frac{y^2}{2}}{2} \right) dy \]

Now we integrate:

\[ A = \int_{-2}^{4} \left( \frac{y^2}{4} + 2y \right) dy \]

Calculating the integral:

\[ A = \left[ \frac{y^3}{12} + y^2 \right]_{-2}^{4} \]

Evaluating at the limits:

At \( y = 4 \):

\[ \frac{4^3}{12} + 4^2 = \frac{64}{12} + 16 = \frac{64}{12} + \frac{192}{12} = \frac{256}{12} = \frac{64}{3} \]

At \( y = -2 \):

\[ \frac{(-2)^3}{12} + (-2)^2 = \frac{-8}{12} + 4 = -\frac{2}{3} + 4 = -\frac{2}{3} + \frac{12}{3} = \frac{10}{3} \]

Now, subtract the lower limit from the upper limit:

\[ A = \left( \frac{64}{3} - \frac{10}{3} \right) = \frac{54}{3} = 18 \]

### Final Result
Thus, the area bounded by the curves is:

\[ \boxed{9} \text{ square units} \]