Consider two regions
`R_(1):"points P are nearer to (1,0) than to "x=-1.`
`R_(2): "Points P are nearer to (0,0) than to (8,0)"` Find the area of the region common to `R_(1) and R_(2).`

Explanation

To find the area of the region common to \( R_1 \) and \( R_2 \), we will follow these steps:

### Step 1: Define the regions mathematically.

**Region \( R_1 \)**: Points \( P \) are nearer to \( (1, 0) \) than to the line \( x = -1 \).

The distance from a point \( P(x, y) \) to \( (1, 0) \) is given by:
\[ d_1 = \sqrt{(x - 1)^2 + y^2} \]
The distance from \( P(x, y) \) to the line \( x = -1 \) is:
\[ d_2 = |x + 1| \]
Since we want points nearer to \( (1, 0) \) than to \( x = -1 \), we have:
\[ d_1 < d_2 \]
Squaring both sides (since both distances are non-negative):
\[ (x - 1)^2 + y^2 < (x + 1)^2 \]
Expanding both sides:
\[ x^2 - 2x + 1 + y^2 < x^2 + 2x + 1 \]
Simplifying:
\[ y^2 < 4x \]
This represents the area inside the parabola \( y^2 = 4x \).

**Region \( R_2 \)**: Points \( P \) are nearer to \( (0, 0) \) than to \( (8, 0) \).

The distance from \( P(x, y) \) to \( (0, 0) \) is:
\[ d_1 = \sqrt{x^2 + y^2} \]
The distance from \( P(x, y) \) to \( (8, 0) \) is:
\[ d_2 = \sqrt{(x - 8)^2 + y^2} \]
We want:
\[ d_1 < d_2 \]
Squaring both sides:
\[ x^2 + y^2 < (x - 8)^2 + y^2 \]
Simplifying:
\[ x^2 + y^2 < x^2 - 16x + 64 + y^2 \]
Cancelling \( y^2 \) and \( x^2 \):
\[ 0 < -16x + 64 \]
Rearranging gives:
\[ 16x < 64 \quad \Rightarrow \quad x < 4 \]
This represents the area to the left of the vertical line \( x = 4 \).

### Step 2: Find the area of the common region.

The common region is bounded by the parabola \( y^2 = 4x \) and the line \( x = 4 \).

To find the area, we will integrate from \( x = 0 \) to \( x = 4 \).

The upper half of the parabola is given by:
\[ y = \sqrt{4x} \]
The area \( A \) can be calculated as:
\[ A = 2 \int_0^4 \sqrt{4x} \, dx \]
Calculating the integral:
\[ A = 2 \int_0^4 2\sqrt{x} \, dx = 4 \int_0^4 x^{1/2} \, dx \]
Using the power rule for integration:
\[ = 4 \left[ \frac{x^{3/2}}{3/2} \right]_0^4 = 4 \left[ \frac{2}{3} x^{3/2} \right]_0^4 = \frac{8}{3} \left[ 4^{3/2} - 0 \right] \]
Calculating \( 4^{3/2} = 8 \):
\[ = \frac{8}{3} \cdot 8 = \frac{64}{3} \]

Thus, the area of the common region is:
\[ \text{Area} = \frac{64}{3} \text{ square units} \]