Step 1: Understanding the Concept:
The problem involves calculating the geometric area enclosed between two intersecting curves in a two-dimensional Cartesian plane. This is a fundamental application of integral calculus, specifically the calculation of the area under a curve.
The region is bounded by a parabola, \(y^{2} = 4x\), which is a rightward-opening curve with its vertex at the origin \((0,0)\) and its axis of symmetry being the x-axis. The second boundary is a straight line, \(y = x\), which passes through the origin at a \(45^{\circ}\) angle relative to the positive x-axis.
To find the area between these two curves, we must determine the range over which they enclose a space. This is done by finding their points of intersection. Once the boundaries (limits of integration) are established, the area is found by integrating the difference between the "upper" function and the "lower" function with respect to \(x\). The upper function is the one that has a greater y-value for a given x-value within the interval.
Geometrically, we are summing the areas of infinitely thin vertical strips, each having a width \(dx\) and a height equal to the difference between the y-values of the two curves.
Key Formula or Approach:
The general formula for the area \(A\) between two curves \(y_{upper} = f(x)\) and \(y_{lower} = g(x)\) from \(x = a\) to \(x = b\) is given by:
\[ \text{Area} = \int_{a}^{b} [f(x) - g(x)] \, dx \]
In this problem, the parabola equation \(y^{2} = 4x\) can be rewritten in terms of \(x\) as \(y = \sqrt{4x} = 2\sqrt{x}\) (we consider the upper branch as it intersects with \(y = x\) in the first quadrant). The line is already given as \(y = x\).
Step 2: Detailed Explanation:
First, we calculate the intersection points by solving the equations simultaneously:
Substitute \(y = x\) into \(y^{2} = 4x\):
\[ x^{2} = 4x \]
\[ x^{2} - 4x = 0 \]
Factoring the quadratic expression:
\[ x(x - 4) = 0 \]
This yields two solutions: \(x = 0\) and \(x = 4\).
For \(x = 0\), \(y = 0\), giving the intersection point \((0, 0)\).
For \(x = 4\), \(y = 4\), giving the intersection point \((4, 4)\).
These values, \(x = 0\) and \(x = 4\), serve as our lower and upper limits of integration, respectively.
Next, we determine which curve is higher in the interval \([0, 4]\). If we test \(x = 1\), the parabola gives \(y = 2\sqrt{1} = 2\), and the line gives \(y = 1\). Since \(2>1\), the parabola is the upper curve.
Now, we set up and solve the definite integral:
\[ \text{Area} = \int_{0}^{4} (2\sqrt{x} - x) \, dx \]
\[ \text{Area} = \int_{0}^{4} (2x^{1/2} - x) \, dx \]
Using the fundamental theorem of calculus and the power rule for integration \(\int x^{n} dx = \frac{x^{n+1}}{n+1}\):
\[ \text{Area} = \left[ 2 \cdot \frac{x^{3/2}}{3/2} - \frac{x^{2}}{2} \right]_{0}^{4} \]
\[ \text{Area} = \left[ \frac{4}{3}x^{3/2} - \frac{x^{2}}{2} \right]_{0}^{4} \]
Substituting the upper limit \(4\):
\[ \text{Area} = \left( \frac{4}{3}(4)^{3/2} - \frac{4^{2}}{2} \right) - \left( \frac{4}{3}(0)^{3/2} - \frac{0^{2}}{2} \right) \]
Since \(4^{3/2} = (\sqrt{4})^{3} = 2^{3} = 8\):
\[ \text{Area} = \left( \frac{4}{3} \cdot 8 - \frac{16}{2} \right) \]
\[ \text{Area} = \frac{32}{3} - 8 \]
Converting \(8\) to a fraction with a denominator of \(3\):
\[ \text{Area} = \frac{32}{3} - \frac{24}{3} = \frac{8}{3} \]
Step 3: Final Answer:
The points of intersection are \((0,0)\) and \((4,4)\). Integrating the difference \(2\sqrt{x} - x\) over the interval \([0, 4]\) yields a total area of \(\frac{8}{3}\) square units.