The diagram shows part of the curve $y = (1 - x)(x - 4)$ and two lines $l_1$ and $l_2$. The lines $l_1$ and $l_2$ are tangents to the curve at points $P$ and $Q$ and have gradients $2$ and $-2$ respectively.

1. Expand the equation of the curve and find its derivative.

$y = (1 - x)(x - 4) = x - 4 - x^2 + 4x = -x^2 + 5x - 4$
$\frac{dy}{dx} = -2x + 5$

2. Equate the derivative to the given gradients to find the x-coordinates of P and Q.

For $P$, gradient is $2$:
$-2x + 5 = 2$
$2x = 3 \implies x = \frac{3}{2}$

For $Q$, gradient is $-2$:
$-2x + 5 = -2$
$2x = 7 \implies x = \frac{7}{2}$

3. Substitute the x-coordinates back into the curve equation to find the y-coordinates.

When $x = \frac{3}{2}$:
$y = \left(1 - \frac{3}{2}\right)\left(\frac{3}{2} - 4\right) = \left(-\frac{1}{2}\right)\left(-\frac{5}{2}\right) = \frac{5}{4}$
So $P$ is $\left(\frac{3}{2}, \frac{5}{4}\right)$.

When $x = \frac{7}{2}$:
$y = \left(1 - \frac{7}{2}\right)\left(\frac{7}{2} - 4\right) = \left(-\frac{5}{2}\right)\left(-\frac{1}{2}\right) = \frac{5}{4}$
So $Q$ is $\left(\frac{7}{2}, \frac{5}{4}\right)$.

1. Find the equation of tangent $l_1$ at $P$

Gradient of $l_1$ is $2$. Point $P$ is $\left(\frac{3}{2}, \frac{5}{4}\right)$.
Equation of $l_1$:
$y - \frac{5}{4} = 2\left(x - \frac{3}{2}\right)$
$y = 2x - 3 + \frac{5}{4}$
$y = 2x - \frac{7}{4}$

2. Find the equation of tangent $l_2$ at $Q$

Gradient of $l_2$ is $-2$. Point $Q$ is $\left(\frac{7}{2}, \frac{5}{4}\right)$.
Equation of $l_2$:
$y - \frac{5}{4} = -2\left(x - \frac{7}{2}\right)$
$y = -2x + 7 + \frac{5}{4}$
$y = -2x + \frac{33}{4}$

3. Solve the simultaneous equations to find the intersection point $R$

Equating the two equations:
$2x - \frac{7}{4} = -2x + \frac{33}{4}$
$4x = \frac{40}{4} = 10$
$x = \frac{5}{2}$
Substitute $x = \frac{5}{2}$ into $l_1$:
$y = 2\left(\frac{5}{2}\right) - \frac{7}{4} = 5 - \frac{7}{4} = \frac{13}{4}$
Coordinates of $R$ are $\left(\frac{5}{2}, \frac{13}{4}\right)$.

1. Find the coordinates of $R$ by finding the equations of $l_1$ and $l_2$ (from part b).

Gradient of $l_1$ is $2$, passing through $P\left(\frac{3}{2}, \frac{5}{4}\right)$:
$y - \frac{5}{4} = 2\left(x - \frac{3}{2}\right) \implies y = 2x - \frac{7}{4}$

Gradient of $l_2$ is $-2$, passing through $Q\left(\frac{7}{2}, \frac{5}{4}\right)$:
$y - \frac{5}{4} = -2\left(x - \frac{7}{2}\right) \implies y = -2x + \frac{33}{4}$

Intersection $R$:
$2x - \frac{7}{4} = -2x + \frac{33}{4}$
$4x = 10 \implies x = \frac{5}{2}$
$y = 2\left(\frac{5}{2}\right) - \frac{7}{4} = \frac{13}{4}$
$R = \left(\frac{5}{2}, \frac{13}{4}\right)$

2. Calculate the area under the tangents (which forms a triangle with the line $PQ$ plus a rectangle below it).

The area under the tangents from $x = \frac{3}{2}$ to $x = \frac{7}{2}$ can be found by calculating the area of triangle $PQR$ and the rectangle below $PQ$.
Base of triangle $PQR = \frac{7}{2} - \frac{3}{2} = 2$
Height of triangle $PQR = \frac{13}{4} - \frac{5}{4} = 2$
Area of triangle $PQR = \frac{1}{2} \times 2 \times 2 = 2$

Area of rectangle below $PQ$ (from $y=0$ to $y=\frac{5}{4}$) = $2 \times \frac{5}{4} = \frac{5}{2}$
Total area under tangents = $2 + \frac{5}{2} = \frac{9}{2}$

3. Calculate the area under the curve from $x = \frac{3}{2}$ to $x = \frac{7}{2}$.

Area under curve = $\int_{\frac{3}{2}}^{\frac{7}{2}} (1 - x)(x - 4) \, dx$
$= \int_{\frac{3}{2}}^{\frac{7}{2}} (-x^2 + 5x - 4) \, dx$
$= \left[ -\frac{x^3}{3} + \frac{5x^2}{2} - 4x \right]_{\frac{3}{2}}^{\frac{7}{2}}$
$= \left( -\frac{343}{24} + \frac{245}{8} - 14 \right) - \left( -\frac{27}{24} + \frac{45}{8} - 6 \right)$
$= \frac{56}{24} - \left( -\frac{36}{24} \right) = \frac{92}{24} = \frac{23}{6}$

4. Find the shaded area by subtracting the area under the curve from the area under the tangents.

Shaded area = Area under tangents - Area under curve
$= \frac{9}{2} - \frac{23}{6}$
$= \frac{27}{6} - \frac{23}{6} = \frac{4}{6} = \frac{2}{3}$