Answer: \(\dfrac{40}{3}\).

We start with the main method. Use the measurements and relationships shown in the diagram, then translate them into algebraic or trigonometric equations. Find the relevant boundary equations and use integration to subtract the lower curve from the upper curve.

First find the point \(A\). Since \(A\) lies on the \(x\)-axis and on the curve, set \(y=0\):

\(32x-4x^2-48=0.\)

Divide by \(-4\):

\(x^2-8x+12=0.\)

So

\((x-2)(x-6)=0.\)

The point shown as \(A\) is the left intercept, so

\(A=(2,0).\)

The maximum point of the curve occurs halfway between the roots \(2\) and \(6\), so

\(x=\frac{2+6}{2}=4.\)

At \(x=4\),

\(y=32(4)-4(4)^2-48=128-64-48=16.\)

The line \(AB\) passes through \(A(2,0)\) and \(B(5,12)\). Its gradient is

\(\frac{12-0}{5-2}=4.\)

Thus the line has equation

\(y=4x-8.\)

At \(x=4\), this line has value

\(y=4(4)-8=8,\)

so \(D=(4,8)\).

The shaded area is the area between the curve and the line from \(x=2\) to \(x=4\):

\(\int_2^4 \left[(32x-4x^2-48)-(4x-8)\right]\,dx.\)

This is

\(\int_2^4 (28x-4x^2-40)\,dx.\)

Integrating gives

\(\left[14x^2-\frac43x^3-40x\right]_2^4.\)

Substitution gives

\(\left(224-\frac{256}{3}-160\right)-\left(56-\frac{32}{3}-80\right).\)

So the area is

\(-\frac{64}{3}+\frac{104}{3}=\frac{40}{3}.\)

This completes the solution and gives the required result.