What is the area of the region enclosed between the curve `y^(2)=2x` and the straight line y=x?

Options

A. 2/3 sq unit
B. 4/3sq unit
C. 1/3 sq unit
D. 1 sq. unit

Answer

To find the area of the region enclosed between the curve \( y^2 = 2x \) and the straight line \( y = x \), we can follow these steps:

### Step 1: Find the Points of Intersection
To find the area between the two curves, we first need to determine where they intersect. We can do this by substituting \( y = x \) into the equation of the curve.

\[ y^2 = 2x \implies x^2 = 2x \]

Rearranging gives:

\[ x^2 - 2x = 0 \]

Factoring out \( x \):

\[ x(x - 2) = 0 \]

Thus, the solutions are:

\[ x = 0 \quad \text{and} \quad x = 2 \]

### Step 2: Find the Corresponding y-values
Now, we can find the corresponding \( y \)-values for these \( x \)-values:

- For \( x = 0 \):
\[ y = 0 \]

- For \( x = 2 \):
\[ y = 2 \]

So, the points of intersection are \( (0, 0) \) and \( (2, 2) \).

### Step 3: Set Up the Integral for Area
The area \( A \) between the curves from \( x = 0 \) to \( x = 2 \) can be found using the integral:

\[ A = \int_{0}^{2} (y_{\text{upper}} - y_{\text{lower}}) \, dx \]

Here, \( y_{\text{upper}} \) is the curve \( y = \sqrt{2x} \) (from \( y^2 = 2x \)) and \( y_{\text{lower}} \) is the line \( y = x \).

Thus, we have:

\[ A = \int_{0}^{2} (\sqrt{2x} - x) \, dx \]

### Step 4: Evaluate the Integral
Now, we can evaluate the integral:

\[ A = \int_{0}^{2} \sqrt{2x} \, dx - \int_{0}^{2} x \, dx \]

Calculating each integral separately:

1. For \( \int \sqrt{2x} \, dx \):
\[ \sqrt{2} \int x^{1/2} \, dx = \sqrt{2} \cdot \frac{x^{3/2}}{3/2} = \frac{2\sqrt{2}}{3} x^{3/2} \]

Evaluating from 0 to 2:
\[ \left[ \frac{2\sqrt{2}}{3} (2^{3/2}) \right] - 0 = \frac{2\sqrt{2}}{3} (2\sqrt{2}) = \frac{4}{3} \cdot 2 = \frac{8}{3} \]

2. For \( \int x \, dx \):
\[ \frac{x^2}{2} \]

Evaluating from 0 to 2:
\[ \left[ \frac{2^2}{2} \right] - 0 = 2 \]

### Step 5: Combine the Results
Now, we can combine the results of the two integrals:

\[ A = \frac{8}{3} - 2 = \frac{8}{3} - \frac{6}{3} = \frac{2}{3} \]

### Final Answer
Thus, the area of the region enclosed between the curve \( y^2 = 2x \) and the line \( y = x \) is:

\[ \boxed{\frac{2}{3}} \text{ square units} \]