Answer: \(A\) has \(x=-2\), \(B\) has \(x=-\dfrac49\); \(C=(0,4)\), \(D=\left(\dfrac13,0\right)\); area \(=\dfrac{241}{324}\).

First expand the curve:

\(y=(x+2)^2(1-3x)=(x^2+4x+4)(1-3x).\)

So

\(y=-3x^3-11x^2-8x+4.\)

Differentiate:

\(\dfrac{dy}{dx}=-9x^2-22x-8.\)

Factorise:

\(-9x^2-22x-8=-(9x^2+22x+8)=-(x+2)(9x+4).\)

For stationary points, set \(\dfrac{dy}{dx}=0\). Therefore

\(x=-2\quad\text{or}\quad x=-\dfrac49.\)

From the diagram, \(A\) is at \(x=-2\) and \(B\) is at \(x=-\dfrac49\).

For \(C\), put \(x=0\):

\(y=(2)^2(1)=4,\)

so

\(C=(0,4).\)

For \(D\), put \(y=0\):

\((x+2)^2(1-3x)=0.\)

The positive \(x\)-intercept shown is \(1-3x=0\), so

\(x=\dfrac13.\)

Hence

\(D=\left(\dfrac13,0\right).\)

The shaded region is under the curve from \(x=0\) to \(x=\dfrac13\). Its area is

\(\displaystyle \int_0^{1/3}(-3x^3-11x^2-8x+4)\,dx.\)

Integrating gives

\(\left[-\dfrac34x^4-\dfrac{11}{3}x^3-4x^2+4x\right]_0^{1/3}.\)

Substitute \(x=\dfrac13\):

\(-\dfrac34\cdot\dfrac1{81}-\dfrac{11}{3}\cdot\dfrac1{27}-4\cdot\dfrac19+\dfrac43.\)

This simplifies to

\(\boxed{\dfrac{241}{324}}.\)