The area enclosed between the parabola \( y^2 = 4x \) and the line \( y = 2x - 4 \) is:
Show Hint
To find the area between curves, first identify the points of intersection, then set up and evaluate the integral of the difference between the functions.
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:
\[
A = \frac{19}{3} \, \text{sq. units}
\]