The area bounded by the curve `y = x^2+ 2x + 1,` the tangent at `(1, 4)` and the y-axis is

Options

A. `2/3` sq.unit
B. `1/3` sq.unit
C. 2 sq.unit
D. None of these

Explanation

To find the area bounded by the curve \( y = x^2 + 2x + 1 \), the tangent at the point \( (1, 4) \), and the y-axis, we will follow these steps:

### Step 1: Identify the curve and the point of tangency
The given curve is:
\[ y = x^2 + 2x + 1 \]
This can be rewritten as:
\[ y = (x + 1)^2 \]
The point of tangency is \( (1, 4) \).

### Step 2: Differentiate the curve to find the slope of the tangent
To find the slope of the tangent at \( x = 1 \), we differentiate the curve:
\[ \frac{dy}{dx} = 2x + 2 \]
Now, substituting \( x = 1 \):
\[ \frac{dy}{dx} \bigg|_{x=1} = 2(1) + 2 = 4 \]
So, the slope of the tangent line at the point \( (1, 4) \) is 4.

### Step 3: Write the equation of the tangent line
Using the point-slope form of the equation of a line, \( y - y_1 = m(x - x_1) \):
\[ y - 4 = 4(x - 1) \]
Simplifying this gives:
\[ y - 4 = 4x - 4 \implies y = 4x \]
Thus, the equation of the tangent line is \( y = 4x \).

### Step 4: Find the area bounded by the curve, tangent, and y-axis
We need to find the area between the curve \( y = x^2 + 2x + 1 \) and the tangent line \( y = 4x \) from \( x = 0 \) to \( x = 1 \).

#### Step 4.1: Calculate the area under the curve from \( x = 0 \) to \( x = 1 \)
The area under the curve can be calculated using the integral:
\[ \text{Area}_{\text{curve}} = \int_0^1 (x^2 + 2x + 1) \, dx \]
Calculating the integral:
\[ \int (x^2 + 2x + 1) \, dx = \frac{x^3}{3} + x^2 + x \]
Evaluating from 0 to 1:
\[ \left[ \frac{1^3}{3} + 1^2 + 1 \right] - \left[ \frac{0^3}{3} + 0^2 + 0 \right] = \left( \frac{1}{3} + 1 + 1 \right) - 0 = \frac{1}{3} + 2 = \frac{7}{3} \]

#### Step 4.2: Calculate the area under the tangent line from \( x = 0 \) to \( x = 1 \)
The area under the tangent line can be calculated as:
\[ \text{Area}_{\text{tangent}} = \int_0^1 4x \, dx \]
Calculating the integral:
\[ \int 4x \, dx = 2x^2 \]
Evaluating from 0 to 1:
\[ \left[ 2(1^2) \right] - \left[ 2(0^2) \right] = 2 - 0 = 2 \]

### Step 5: Calculate the bounded area
The area bounded by the curve, the tangent, and the y-axis is given by:
\[ \text{Area} = \text{Area}_{\text{tangent}} - \text{Area}_{\text{curve}} = 2 - \frac{7}{3} \]
Finding a common denominator:
\[ 2 = \frac{6}{3} \]
Thus:
\[ \text{Area} = \frac{6}{3} - \frac{7}{3} = -\frac{1}{3} \]
Since area cannot be negative, we take the absolute value:
\[ \text{Area} = \frac{1}{3} \]

### Final Answer
The area bounded by the curve, the tangent line, and the y-axis is:
\[ \frac{1}{3} \text{ square units} \]