The diagram shows part of the curve of $y = x(x-2)^2$, which passes through the point $P(1, 1)$.

1. Calculate the gradient of the line $OP$

The coordinates of $O$ are $(0, 0)$ and $P$ are $(1, 1)$.
Gradient of $OP = \frac{1 - 0}{1 - 0} = 1$

2. Form the equation of the line $OP$

Since the line passes through the origin $(0, 0)$, the $y$-intercept $c = 0$.
Equation of $OP$ is $y = 1x + 0$
$y = x$ (Shown)

1. Find the derivative of the curve's equation.

$y = x(x-2)^2 = x(x^2 - 4x + 4) = x^3 - 4x^2 + 4x$
$\frac{dy}{dx} = 3x^2 - 8x + 4$

2. Calculate the gradient of the tangent at $P(1, 1)$.

At $x = 1$,
$\frac{dy}{dx} = 3(1)^2 - 8(1) + 4 = -1$
The gradient of the tangent at $P$ is $-1$.

3. Show that $OP$ is perpendicular to the tangent at $P$.

From part (a), the equation of $OP$ is $y = x$, so the gradient of $OP$ is $1$.
Product of gradients = $(-1) \times 1 = -1$.
Since the product of the gradients is $-1$, $OP$ is perpendicular to the tangent at $P$. Therefore, $OP$ is the normal to the curve at $P$.

1. Set up the integral for the area between the curve and the line

$\text{Area} = \int_{0}^{1} \left[ x(x-2)^2 - x \right] \, dx$

2. Expand and simplify the integrand

$x(x-2)^2 - x = x(x^2 - 4x + 4) - x$
$= x^3 - 4x^2 + 4x - x$
$= x^3 - 4x^2 + 3x$

3. Integrate the polynomial

$\int_{0}^{1} (x^3 - 4x^2 + 3x) \, dx = \left[ \frac{x^4}{4} - \frac{4x^3}{3} + \frac{3x^2}{2} \right]_{0}^{1}$

4. Evaluate the definite integral using the limits

$= \left( \frac{1}{4} - \frac{4}{3} + \frac{3}{2} \right) - (0)$
$= \frac{3}{12} - \frac{16}{12} + \frac{18}{12}$
$= \frac{5}{12}$