Using integration find the area bounded by the tangent to the curve `y = 3x^(2)` at the point (1, 3), and the Lines whose equations are `y = (x)/(3)` and `x+y=4`.

Answer

To find the area bounded by the tangent to the curve \( y = 3x^2 \) at the point \( (1, 3) \) and the lines \( y = \frac{x}{3} \) and \( x + y = 4 \), we will follow these steps:

### Step 1: Find the equation of the tangent line
1. **Differentiate the curve**: The derivative of \( y = 3x^2 \) is given by:
\[ \frac{dy}{dx} = 6x \]
2. **Evaluate the derivative at \( x = 1 \)**:
\[ \frac{dy}{dx} \bigg|_{x=1} = 6 \cdot 1 = 6 \]
3. **Use the point-slope form of the line**: The equation of the tangent line at the point \( (1, 3) \) is:
\[ y - 3 = 6(x - 1) \]
Simplifying this gives:
\[ y = 6x - 3 \]

### Step 2: Find the intersection points of the lines
1. **Intersection of the tangent line \( y = 6x - 3 \) and \( y = \frac{x}{3} \)**:
\[ 6x - 3 = \frac{x}{3} \]
Multiply through by 3 to eliminate the fraction:
\[ 18x - 9 = x \]
Rearranging gives:
\[ 17x = 9 \implies x = \frac{9}{17} \]
Substitute \( x \) back to find \( y \):
\[ y = \frac{9}{17} \cdot \frac{1}{3} = \frac{3}{17} \]
So, the intersection point \( C \) is \( \left(\frac{9}{17}, \frac{3}{17}\right) \).

2. **Intersection of \( y = \frac{x}{3} \) and \( x + y = 4 \)**:
Substitute \( y \):
\[ x + \frac{x}{3} = 4 \implies \frac{4x}{3} = 4 \implies x = 3 \]
Substitute back to find \( y \):
\[ y = \frac{3}{3} = 1 \]
So, the intersection point \( B \) is \( (3, 1) \).

3. **Intersection of the tangent line \( y = 6x - 3 \) and \( x + y = 4 \)**:
Substitute \( y \):
\[ x + (6x - 3) = 4 \implies 7x - 3 = 4 \implies 7x = 7 \implies x = 1 \]
Substitute back to find \( y \):
\[ y = 4 - 1 = 3 \]
So, the intersection point \( A \) is \( (1, 3) \).

### Step 3: Calculate the area of triangle formed by points A, B, and C
The vertices of the triangle are:
- \( A(1, 3) \)
- \( B(3, 1) \)
- \( C\left(\frac{9}{17}, \frac{3}{17}\right) \)

The area \( A \) of the triangle can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the points:
\[ \text{Area} = \frac{1}{2} \left| 1(1 - \frac{3}{17}) + 3\left(\frac{3}{17} - 3\right) + \frac{9}{17}(3 - 1) \right| \]
Calculating each term:
1. \( 1(1 - \frac{3}{17}) = 1 \cdot \frac{14}{17} = \frac{14}{17} \)
2. \( 3\left(\frac{3}{17} - 3\right) = 3\left(\frac{3 - 51}{17}\right) = 3\left(\frac{-48}{17}\right) = -\frac{144}{17} \)
3. \( \frac{9}{17}(3 - 1) = \frac{9}{17} \cdot 2 = \frac{18}{17} \)

Now substituting back:
\[ \text{Area} = \frac{1}{2} \left| \frac{14}{17} - \frac{144}{17} + \frac{18}{17} \right| = \frac{1}{2} \left| \frac{14 - 144 + 18}{17} \right| = \frac{1}{2} \left| \frac{-112}{17} \right| = \frac{56}{17} \]

### Final Answer:
The area bounded by the tangent line and the two lines is:
\[ \text{Area} = \frac{56}{17} \text{ square units} \]