Answer: \(A=(0,20)\); \(y_B=31.7\); integral \(2e^{2x}-8e^{-2x}+C\); shaded area \(\approx6.16\).

At \(A\), \(x=0\). Therefore

\(y=4e^0+16e^0=20.\)

So

\(A=(0,20).\)

At \(B\), \(x=1\). Hence

\(y=4e^2+16e^{-2}=31.7\ldots\)

So the \(y\)-coordinate of \(B\) is approximately \(31.7\).

Integrating,

\(\int(4e^{2x}+16e^{-2x})\,dx=2e^{2x}-8e^{-2x}+C.\)

The straight line \(AB\) forms a trapezium with the \(x\)-axis between \(x=0\) and \(x=1\). Its area is

\(\dfrac12(20+31.7)(1)=25.85\ldots.\)

The area under the curve from \(x=0\) to \(x=1\) is

\(\left[2e^{2x}-8e^{-2x}\right]_0^1.\)

This is

\((2e^2-8e^{-2})-(2-8)=19.69\ldots.\)

Therefore the shaded area is

\(25.85\ldots-19.69\ldots=6.16\ldots.\)

So the area is

\(\boxed{6.16}\) square units, to 3 significant figures.