Step 1: Understanding the Concept:
We need to find the area of the region bounded by a parabola \( y^2 = 4x \) and a modulus function \( y = |x| \).
First, we should find the points of intersection to determine the limits of integration.
Step 2: Key Formula or Approach:
The area between two curves \( f(x) \) and \( g(x) \) from \( x=a \) to \( x=b \) where \( f(x) \ge g(x) \) is given by \( \int_a^b (f(x) - g(x)) dx \).
Step 3: Detailed Explanation:
The given curves are:
1) \( y^2 = 4x \) (A parabola opening rightwards. Since \( y^2 \ge 0 \), we must have \( x \ge 0 \)).
2) \( y = |x| \) (A V-shaped graph. For \( x \ge 0 \), it is \( y = x \)).
Because the first curve only exists for \( x \ge 0 \), we only need to consider the intersection in the first quadrant where \( y = x \).
Let's find the intersection points by substituting \( y = x \) into the parabola's equation:
\[ x^2 = 4x \]
\[ x^2 - 4x = 0 \]
\[ x(x - 4) = 0 \]
This gives \( x = 0 \) and \( x = 4 \).
For \( x = 0 \), \( y = 0 \).
For \( x = 4 \), \( y = 4 \).
In the interval \( [0, 4] \), the parabola \( y = \sqrt{4x} = 2\sqrt{x} \) lies above the line \( y = x \).
Let's verify by picking a point, say \( x=1 \): \( y_{parabola} = 2\sqrt{1} = 2 \), \( y_{line} = 1 \). So parabola is upper curve.
The required area \( A \) is:
\[ A = \int_{0}^{4} (\text{Upper Curve} - \text{Lower Curve}) dx \]
\[ A = \int_{0}^{4} (2\sqrt{x} - x) dx \]
\[ A = \int_{0}^{4} (2x^{1/2} - x) dx \]
Integrate term by term:
\[ A = \left[ 2 \cdot \frac{x^{3/2}}{3/2} - \frac{x^2}{2} \right]_{0}^{4} \]
\[ A = \left[ \frac{4}{3} x^{3/2} - \frac{x^2}{2} \right]_{0}^{4} \]
Now, apply the limits:
\[ A = \left( \frac{4}{3} (4)^{3/2} - \frac{4^2}{2} \right) - \left( 0 - 0 \right) \]
Since \( 4^{3/2} = (4^{1/2})^3 = 2^3 = 8 \):
\[ A = \frac{4}{3} \cdot 8 - \frac{16}{2} \]
\[ A = \frac{32}{3} - 8 \]
\[ A = \frac{32 - 24}{3} = \frac{8}{3} \text{ sq. units} \]
Step 4: Final Answer:
The enclosed area is \( \frac{8}{3} \) sq. units.