The curve y  kx2  mx  n passes through the points (1, 0), (2, 0) and its gradient at the point (2, 0) is 2. Find value of the area included between the curve and the axis of x.

Let's analyze the problem step by step.

### Step 1: Use the given points to find the coefficients \( k \), \( m \), and \( n \).
The curve \( y = kx^2 + mx + n \) passes through the points \( (1, 0) \) and \( (2, 0) \). This means the curve has roots at \( x = 1 \) and \( x = 2 \). Therefore, we can write the equation as:
\[
y = k(x - 1)(x - 2)
\]
Expanding this, we get:
\[
y = k(x^2 - 3x + 2)
\]
So, the equation of the curve becomes:
\[
y = kx^2 - 3kx + 2k
\]
Comparing this with the general form \( y = kx^2 + mx + n \), we can identify:
\[
m = -3k, \quad n = 2k
\]

### Step 2: Use the gradient condition.
We are also given that the gradient (derivative of \( y \)) at the point \( (2, 0) \) is 2. The gradient function is given by:
\[
\frac{dy}{dx} = 2kx - 3k
\]
Substituting \( x = 2 \) and setting the gradient equal to 2:
\[
2k(2) - 3k = 2
\]
\[
4k - 3k = 2
\]
\[
k = 2
\]

### Step 3: Find \( m \) and \( n \).
Now that \( k = 2 \), we can find \( m \) and \( n \):
\[
m = -3k = -6, \quad n = 2k = 4
\]
So, the equation of the curve is:
\[
y = 2x^2 - 6x + 4
\]

### Step 4: Find the area under the curve.
The area \( A \) between the curve and the \( x \)-axis from \( x = 1 \) to \( x = 2 \) is given by the definite integral:
\[
A = \int_{1}^{2} (2x^2 - 6x + 4) \, dx
\]
Now, let's compute this integral:

1. Integrate each term:
   \[
   \int 2x^2 \, dx = \frac{2x^3}{3}, \quad \int -6x \, dx = -3x^2, \quad \int 4 \, dx = 4x
   \]
   
2. Evaluate the definite integral from 1 to 2:
   \[
   A = \left[\frac{2x^3}{3} - 3x^2 + 4x\right]_{1}^{2}
   \]
   Substitute \( x = 2 \) and \( x = 1 \) into the expression:
   \[
   A = \left(\frac{2(2)^3}{3} - 3(2)^2 + 4(2)\right) - \left(\frac{2(1)^3}{3} - 3(1)^2 + 4(1)\right)
   \]
   \[
   A = \left(\frac{16}{3} - 12 + 8\right) - \left(\frac{2}{3} - 3 + 4\right)
   \]
   \[
   A = \left(\frac{16}{3} - 4\right) - \left(\frac{2}{3} + 1\right)
   \]
   \[
   A = \left(\frac{16}{3} - \frac{12}{3}\right) - \left(\frac{2}{3} + \frac{3}{3}\right)
   \]
   \[
   A = \frac{4}{3} - \frac{5}{3} = -\frac{1}{3}
   \]
   Since the area cannot be negative, we take the absolute value:
   \[
   A = \frac{1}{3}
   \]

### Final Answer:
The area enclosed between the curve and the \( x \)-axis is \( \frac{1}{3} \) square units.

Would you like further details on this solution or have any other questions?

Here are some related questions:
1. What is the significance of the roots of a quadratic equation in determining the area under the curve?
2. How do changes in the coefficients \( k \), \( m \), and \( n \) affect the shape of a parabola?
3. What is the geometric interpretation of the gradient of a curve?
4. How does the integral of a function relate to the area under its curve?
5. What happens to the area calculation if the curve crosses the \( x \)-axis multiple times?

**Tip:** Always pay attention to the sign of the area when integrating. If the curve is below the \( x \)-axis, the area might be negative, so take the absolute value to find the actual enclosed area.

Learn how to find the area under the quadratic curve y = 2x^2 - 6x + 4 using definite integration. This problem involves solving for coefficients k, m, and n, and applying the gradient condition at a given point. Suitable for advanced high school students in Grades 11-12.

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