To solve the problem of finding the area enclosed by the curve \( y^2 = 4x \) and the line \( y = x \), we will follow these steps:
### Step 1: Sketch the Curve
The equation \( y^2 = 4x \) represents a parabola that opens to the right. To sketch it, we can find some key points:
- When \( x = 0 \):
\[
y^2 = 4 \cdot 0 \implies y = 0 \quad \text{(Point: (0, 0))}
\]
- When \( x = 1 \):
\[
y^2 = 4 \cdot 1 \implies y = \pm 2 \quad \text{(Points: (1, 2) and (1, -2))}
\]
- When \( x = 4 \):
\[
y^2 = 4 \cdot 4 \implies y = \pm 4 \quad \text{(Points: (4, 4) and (4, -4))}
\]
Now, we can sketch the parabola using these points. The parabola will pass through (0, 0), (1, 2), (1, -2), (4, 4), and (4, -4).
### Step 2: Sketch the Line
The line \( y = x \) is a diagonal line passing through the origin (0, 0) with a slope of 1.
### Step 3: Find Points of Intersection
To find the area enclosed by the curve and the line, we need to find their points of intersection. We set \( y = x \) in the parabola's equation:
\[
x^2 = 4x
\]
Rearranging gives:
\[
x^2 - 4x = 0
\]
Factoring out \( x \):
\[
x(x - 4) = 0
\]
Thus, \( x = 0 \) or \( x = 4 \). The corresponding \( y \) values are:
- For \( x = 0 \): \( y = 0 \) (Point: (0, 0))
- For \( x = 4 \): \( y = 4 \) (Point: (4, 4))
### Step 4: Set Up the Integral
The area \( A \) between the curves from \( x = 0 \) to \( x = 4 \) can be found using the integral:
\[
A = \int_{0}^{4} (y_{\text{upper}} - y_{\text{lower}}) \, dx
\]
Here, \( y_{\text{upper}} = \sqrt{4x} \) (from the parabola) and \( y_{\text{lower}} = x \) (from the line).
Thus, the area becomes:
\[
A = \int_{0}^{4} (\sqrt{4x} - x) \, dx
\]
### Step 5: Evaluate the Integral
First, simplify \( \sqrt{4x} \):
\[
\sqrt{4x} = 2\sqrt{x}
\]
So the integral becomes:
\[
A = \int_{0}^{4} (2\sqrt{x} - x) \, dx
\]
Now, we can integrate term by term:
1. The integral of \( 2\sqrt{x} \):
\[
\int 2\sqrt{x} \, dx = \frac{2}{\frac{3}{2}} x^{\frac{3}{2}} = \frac{4}{3} x^{\frac{3}{2}}
\]
2. The integral of \( x \):
\[
\int x \, dx = \frac{x^2}{2}
\]
Combining these, we have:
\[
A = \left[ \frac{4}{3} x^{\frac{3}{2}} - \frac{x^2}{2} \right]_{0}^{4}
\]
### Step 6: Calculate the Area
Now, evaluate at the limits:
\[
A = \left( \frac{4}{3} (4)^{\frac{3}{2}} - \frac{(4)^2}{2} \right) - \left( \frac{4}{3} (0)^{\frac{3}{2}} - \frac{(0)^2}{2} \right)
\]
Calculating \( (4)^{\frac{3}{2}} = 8 \):
\[
A = \left( \frac{4}{3} \cdot 8 - \frac{16}{2} \right) = \left( \frac{32}{3} - 8 \right) = \left( \frac{32}{3} - \frac{24}{3} \right) = \frac{8}{3}
\]
### Final Answer
The area of the region enclosed by the curve \( y^2 = 4x \) and the line \( y = x \) is:
\[
\boxed{\frac{8}{3}}
\]
