The area bounded by the curve `y=x^(2)`, the normal at (1, 1) and the x-axis is:

Options

A. `(4)/(3)`
B. `(2)/(3)`
C. `(1)/(3)`
D. None

Answer

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

### Step 1: Find the equation of the normal at the point (1, 1)
1. **Find the slope of the tangent** to the curve \( y = x^2 \) at the point \( (1, 1) \).
- The derivative \( \frac{dy}{dx} = 2x \).
- At \( x = 1 \), \( \frac{dy}{dx} = 2(1) = 2 \).

2. **Determine the slope of the normal**.
- The slope of the normal \( m \) is the negative reciprocal of the slope of the tangent.
- Therefore, \( m = -\frac{1}{2} \).

3. **Use the point-slope form to find the normal equation**.
- The point-slope form is given by \( y - y_1 = m(x - x_1) \).
- Substituting \( (x_1, y_1) = (1, 1) \) and \( m = -\frac{1}{2} \):
\[ y - 1 = -\frac{1}{2}(x - 1) \]
- Rearranging gives:
\[ y = -\frac{1}{2}x + \frac{3}{2} \]

### Step 2: Find the x-intercept of the normal
1. **Set \( y = 0 \) to find the x-intercept**:
\[ 0 = -\frac{1}{2}x + \frac{3}{2} \]
- Solving for \( x \):
\[ \frac{1}{2}x = \frac{3}{2} \implies x = 3 \]

### Step 3: Set up the area to be calculated
1. **The area is bounded by**:
- The curve \( y = x^2 \) from \( x = 0 \) to \( x = 1 \).
- The normal line from \( x = 1 \) to \( x = 3 \).
- The x-axis.

### Step 4: Calculate the area under the curve from \( x = 0 \) to \( x = 1 \)
1. **Integrate \( y = x^2 \)**:
\[ \text{Area}_1 = \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1^3}{3} - 0 = \frac{1}{3} \]

### Step 5: Calculate the area under the normal from \( x = 1 \) to \( x = 3 \)
1. **The equation of the normal is \( y = -\frac{1}{2}x + \frac{3}{2} \)**.
2. **Integrate from \( x = 1 \) to \( x = 3 \)**:
\[ \text{Area}_2 = \int_1^3 \left(-\frac{1}{2}x + \frac{3}{2}\right) \, dx \]
- This can be split into two integrals:
\[ = \int_1^3 -\frac{1}{2}x \, dx + \int_1^3 \frac{3}{2} \, dx \]

3. **Calculate each integral**:
- For \( \int_1^3 -\frac{1}{2}x \, dx \):
\[ = -\frac{1}{2} \left[ \frac{x^2}{2} \right]_1^3 = -\frac{1}{2} \left( \frac{9}{2} - \frac{1}{2} \right) = -\frac{1}{2} \cdot 4 = -2 \]
- For \( \int_1^3 \frac{3}{2} \, dx \):
\[ = \frac{3}{2} \left[ x \right]_1^3 = \frac{3}{2} (3 - 1) = \frac{3}{2} \cdot 2 = 3 \]

4. **Combine the areas**:
\[ \text{Area}_2 = -2 + 3 = 1 \]

### Step 6: Total area bounded by the curve, normal, and x-axis
1. **Total area**:
\[ \text{Total Area} = \text{Area}_1 + \text{Area}_2 = \frac{1}{3} + 1 = \frac{1}{3} + \frac{3}{3} = \frac{4}{3} \]

### Final Answer
The area bounded by the curve \( y = x^2 \), the normal at \( (1, 1) \), and the x-axis is \( \frac{4}{3} \).

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