Area enclosed by the curves `y = x ^(2) + 1 ` and a normal drawn to it with gradient -1, is equal to:

Options

A. `2/3`
B. `1/3`
C. `3/4`
D. `4/3`

Answer

To find the area enclosed by the curve \( y = x^2 + 1 \) and a normal drawn to it with a gradient of -1, we will follow these steps:

### Step 1: Find the derivative of the curve
The first step is to differentiate the curve \( y = x^2 + 1 \) to find the slope of the tangent line.

\[ \frac{dy}{dx} = 2x \]

### Step 2: Find the slope of the normal
The slope of the normal line is the negative reciprocal of the slope of the tangent line. Given that the slope of the normal is -1, we can set up the equation:

\[ -1 = -\frac{1}{2x} \]

### Step 3: Solve for \( x \)
Rearranging the equation gives:

\[ 1 = \frac{1}{2x} \implies 2x = 1 \implies x = \frac{1}{2} \]

### Step 4: Find the corresponding \( y \) value
Now we substitute \( x = \frac{1}{2} \) back into the equation of the curve to find the corresponding \( y \) value.

\[ y = \left(\frac{1}{2}\right)^2 + 1 = \frac{1}{4} + 1 = \frac{5}{4} \]

So, the point of tangency is \( \left(\frac{1}{2}, \frac{5}{4}\right) \).

### Step 5: Write the equation of the normal line
Using the point-slope form of the equation of a line, we can write the equation of the normal line:

\[ y - \frac{5}{4} = -1\left(x - \frac{1}{2}\right) \]

This simplifies to:

\[ y = -x + \frac{1}{2} + \frac{5}{4} = -x + \frac{7}{4} \]

### Step 6: Set up the area calculation
Now we need to find the area between the curve \( y = x^2 + 1 \) and the normal line \( y = -x + \frac{7}{4} \). We first need to find the points of intersection by equating the two equations:

\[ x^2 + 1 = -x + \frac{7}{4} \]

Rearranging gives:

\[ x^2 + x + 1 - \frac{7}{4} = 0 \implies 4x^2 + 4x + 4 - 7 = 0 \implies 4x^2 + 4x - 3 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 4 \cdot (-3)}}{2 \cdot 4} = \frac{-4 \pm \sqrt{16 + 48}}{8} = \frac{-4 \pm \sqrt{64}}{8} = \frac{-4 \pm 8}{8} \]

This gives us:

\[ x = \frac{4}{8} = \frac{1}{2} \quad \text{and} \quad x = \frac{-12}{8} = -\frac{3}{2} \]

### Step 8: Calculate the area
The area \( A \) between the curves from \( x = -\frac{3}{2} \) to \( x = \frac{1}{2} \) is given by:

\[ A = \int_{-\frac{3}{2}}^{\frac{1}{2}} \left( \left(-x + \frac{7}{4}\right) - (x^2 + 1) \right) dx \]

Simplifying the integrand:

\[ A = \int_{-\frac{3}{2}}^{\frac{1}{2}} \left(-x + \frac{7}{4} - x^2 - 1\right) dx = \int_{-\frac{3}{2}}^{\frac{1}{2}} \left(-x - x^2 + \frac{3}{4}\right) dx \]

### Step 9: Evaluate the integral
Calculating the integral:

\[ A = \left[-\frac{x^2}{2} - \frac{x^3}{3} + \frac{3}{4}x\right]_{-\frac{3}{2}}^{\frac{1}{2}} \]

Calculating the definite integral gives:

\[ A = \left[-\frac{(\frac{1}{2})^2}{2} - \frac{(\frac{1}{2})^3}{3} + \frac{3}{4}(\frac{1}{2})\right] - \left[-\frac{(-\frac{3}{2})^2}{2} - \frac{(-\frac{3}{2})^3}{3} + \frac{3}{4}(-\frac{3}{2})\right] \]

After evaluating the above expression, we find that the area enclosed is:

\[ A = \frac{4}{3} \text{ square units} \]

### Final Answer
The area enclosed by the curves is \( \frac{4}{3} \) square units.