The task is to determine the area bounded by the parabola \( y^2 = 4x \) and the line \( y = 2x - 4 \).
Step 1: Determine Intersection Points
To find where the curves intersect, substitute the line's equation into the parabola's equation.
\[
y = 2x - 4
\]
Substitute \( y \) into \( y^2 = 4x \):
\[
(2x - 4)^2 = 4x
\]
Expand the equation:
\[
4x^2 - 16x + 16 = 4x
\]
Simplify the equation:
\[
4x^2 - 20x + 16 = 0
\]
Divide the equation by 4:
\[
x^2 - 5x + 4 = 0
\]
Factor the quadratic:
\[
(x - 4)(x - 1) = 0
\]
The intersection points occur at \( x = 4 \) and \( x = 1 \).
Step 2: Formulate the Area Integral
The area between the curves is found by integrating the difference of the functions with respect to \( x \) from 1 to 4. The formula for the area is:
\[
A = \int_{1}^{4} \left( (2x - 4) - \sqrt{4x} \right) dx
\]
Step 3: Compute the Integral
First, integrate the linear term \( 2x - 4 \):
\[
\int (2x - 4) dx = x^2 - 4x
\]
Next, integrate the term \( \sqrt{4x} \), which simplifies to \( 2\sqrt{x} \):
\[
\int 2\sqrt{x} dx = \frac{4}{3} x^{3/2}
\]
Now, apply the limits of integration:
\[
A = \left[ (x^2 - 4x) - \frac{4}{3} x^{3/2} \right]_{1}^{4}
\]
Evaluate at \( x = 4 \):
\[
A = \left( (16 - 16) - \frac{4}{3} (8) \right) = 0 - \frac{32}{3}
\]
Evaluate at \( x = 1 \):
\[
A = \left( (1 - 4) - \frac{4}{3} (1) \right) = -3 - \frac{4}{3} = -\frac{13}{3}
\]
The total area is the difference between the evaluations:
\[
A = \left( 0 - \frac{32}{3} \right) - \left( -3 - \frac{4}{3} \right) = -\frac{32}{3} + 3 + \frac{4}{3} = -\frac{28}{3} + 3 = \frac{-28 + 9}{3} = -\frac{19}{3}
\]
Correction: The order of subtraction should be top curve minus bottom curve.
We need to determine which function is "above" the other in the interval [1, 4].
Let's test x=2.5:
Line: y = 2(2.5) - 4 = 5 - 4 = 1
Parabola: y^2 = 4(2.5) = 10 > y = sqrt(10) approx 3.16
So the parabola is above the line.
The integral should be:
\[
A = \int_{1}^{4} \left( \sqrt{4x} - (2x - 4) \right) dx
\]
\[
A = \int_{1}^{4} \left( 2\sqrt{x} - 2x + 4 \right) dx
\]
\[
A = \left[ \frac{4}{3} x^{3/2} - x^2 + 4x \right]_{1}^{4}
\]
Evaluate at \( x = 4 \):
\[
\left( \frac{4}{3} (4^{3/2}) - 4^2 + 4(4) \right) = \left( \frac{4}{3} (8) - 16 + 16 \right) = \frac{32}{3}
\]
Evaluate at \( x = 1 \):
\[
\left( \frac{4}{3} (1^{3/2}) - 1^2 + 4(1) \right) = \left( \frac{4}{3} - 1 + 4 \right) = \frac{4}{3} + 3 = \frac{4+9}{3} = \frac{13}{3}
\]
Subtract the lower limit from the upper limit:
\[
A = \frac{32}{3} - \frac{13}{3} = \frac{19}{3}
\]
Thus, the total area is:
\[
A = \frac{19}{3} \, \text{sq. units}
\]