Find the area between x - axis, curve `x = y^(2)` and its normal at the point (1, 1).

Answer

To find the area between the x-axis, the curve \( x = y^2 \), and its normal at the point \( (1, 1) \), we will follow these steps:

### Step 1: Find the equation of the normal at the point (1, 1)

1. The given curve is \( x = y^2 \). We can rewrite it as \( y^2 = x \).
2. To find the slope of the tangent, we differentiate \( y^2 = x \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{1}{2y} \]
3. At the point \( (1, 1) \), substituting \( y = 1 \):
\[ \frac{dy}{dx} = \frac{1}{2 \cdot 1} = \frac{1}{2} \]
4. The slope of the normal is the negative reciprocal of the slope of the tangent:
\[ m' = -\frac{1}{\frac{1}{2}} = -2 \]
5. Using the point-slope form of the line, the equation of the normal at the point \( (1, 1) \) is:
\[ y - 1 = -2(x - 1) \]
Simplifying this gives:
\[ y = -2x + 3 \]

### Step 2: Find the point where the normal intersects the x-axis

1. To find the intersection with the x-axis, set \( y = 0 \):
\[ 0 = -2x + 3 \]
Solving for \( x \):
\[ 2x = 3 \implies x = \frac{3}{2} \]
2. Thus, the point of intersection is \( \left(\frac{3}{2}, 0\right) \).

### Step 3: Set up the area to be calculated

1. The area we need to find is bounded by the normal line, the curve, and the x-axis, from \( x = 0 \) to \( x = 1 \) (the point of tangency).
2. The area can be divided into two parts:
- The area under the curve from \( x = 0 \) to \( x = 1 \).
- The area of the triangle formed by the normal line and the x-axis.

### Step 4: Calculate the area under the curve

1. The equation of the curve is \( y = \sqrt{x} \).
2. The area under the curve from \( x = 0 \) to \( x = 1 \) is given by:
\[ A_1 = \int_0^1 \sqrt{x} \, dx \]
3. Evaluating the integral:
\[ A_1 = \int_0^1 x^{1/2} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_0^1 = \frac{2}{3} \]

### Step 5: Calculate the area of the triangle

1. The base of the triangle is the distance from \( x = 1 \) to \( x = \frac{3}{2} \):
\[ \text{Base} = \frac{3}{2} - 1 = \frac{1}{2} \]
2. The height of the triangle is the y-coordinate of the normal at \( x = 1 \):
\[ y = -2(1) + 3 = 1 \]
3. The area of the triangle is:
\[ A_2 = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times \frac{1}{2} \times 1 = \frac{1}{4} \]

### Step 6: Total area

1. The total area \( A \) between the x-axis, the curve, and the normal is:
\[ A = A_1 + A_2 = \frac{2}{3} + \frac{1}{4} \]
2. To add these fractions, find a common denominator:
\[ A = \frac{8}{12} + \frac{3}{12} = \frac{11}{12} \]

### Final Answer
The area between the x-axis, the curve \( x = y^2 \), and its normal at the point \( (1, 1) \) is \( \frac{11}{12} \) square units.