Given `(dy)/(dx)` is directly proportional to the square of x and `(dy)/(dx)=6` at x = 2. Then find the equation of the curve, when x = 2 and y = 4. Also find the area of the region bounded by curve between lines y = 1 and y = 3.

Answer

To solve the problem step by step, we will follow the instructions provided in the video transcript.

### Step 1: Establish the relationship
Given that \(\frac{dy}{dx}\) is directly proportional to \(x^2\), we can express this as:
\[ \frac{dy}{dx} = kx^2 \]
where \(k\) is a constant of proportionality.

### Step 2: Integrate to find \(y\)
To find the equation of the curve, we need to integrate both sides:
\[ dy = kx^2 dx \]
Integrating both sides gives:
\[ y = \frac{kx^3}{3} + C \]
where \(C\) is the constant of integration.

### Step 3: Use the given condition to find \(k\)
We know that \(\frac{dy}{dx} = 6\) when \(x = 2\):
\[ \frac{dy}{dx} = k(2^2) = 4k \]
Setting this equal to 6 gives:
\[ 4k = 6 \implies k = \frac{3}{2} \]

### Step 4: Substitute \(k\) back into the equation
Now substituting \(k\) back into our equation for \(y\):
\[ y = \frac{3}{2} \cdot \frac{x^3}{3} + C = \frac{x^3}{2} + C \]

### Step 5: Use the initial condition to find \(C\)
We are given that when \(x = 2\), \(y = 4\):
\[ 4 = \frac{(2)^3}{2} + C \implies 4 = \frac{8}{2} + C \implies 4 = 4 + C \]
Thus, \(C = 0\).

### Step 6: Final equation of the curve
The equation of the curve is:
\[ y = \frac{x^3}{2} \]

### Step 7: Find the area between the curves \(y = 1\) and \(y = 3\)
To find the area bounded by the curve between \(y = 1\) and \(y = 3\), we first express \(x\) in terms of \(y\):
\[ y = \frac{x^3}{2} \implies x^3 = 2y \implies x = (2y)^{1/3} \]

### Step 8: Set up the integral for the area
The area \(A\) can be calculated by integrating \(x\) with respect to \(y\) from \(y = 1\) to \(y = 3\):
\[ A = \int_{1}^{3} (2y)^{1/3} \, dy \]

### Step 9: Solve the integral
Calculating the integral:
\[ A = \int_{1}^{3} 2^{1/3} y^{1/3} \, dy = 2^{1/3} \int_{1}^{3} y^{1/3} \, dy \]
The integral of \(y^{1/3}\) is:
\[ \int y^{1/3} \, dy = \frac{y^{4/3}}{4/3} = \frac{3}{4} y^{4/3} \]
Thus,
\[ A = 2^{1/3} \left[ \frac{3}{4} y^{4/3} \right]_{1}^{3} = 2^{1/3} \left( \frac{3}{4} \left(3^{4/3} - 1^{4/3}\right) \right) \]

### Step 10: Calculate the final area
Calculating \(3^{4/3}\):
\[ 3^{4/3} = (3^{1/3})^4 = \sqrt[3]{81} \]
Thus,
\[ A = 2^{1/3} \cdot \frac{3}{4} \left( \sqrt[3]{81} - 1 \right) \]

### Final Answer
The equation of the curve is \(y = \frac{x^3}{2}\) and the area bounded by the curve between lines \(y = 1\) and \(y = 3\) is:
\[ A = 2^{1/3} \cdot \frac{3}{4} \left( \sqrt[3]{81} - 1 \right) \]