Answer: \(a=3\), \(b=1\); area \(=\dfrac{76}{15}\).

(a) The graph \(y=(x-3)^2\) meets the \(x\)-axis when \(y=0\):

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

So

\(x=3,\)

and hence

\(a=3.\)

At \(B(b,4)\), the curve \(y=4x^{\frac23}\) has \(y=4\). Therefore

\(4b^{\frac23}=4.\)

So

\(b^{\frac23}=1.\)

The intersection shown is on the positive branch, so

\(b=1.\)

(b) The shaded area is the area under \(y=4x^{\frac23}\) from \(x=0\) to \(x=1\), plus the area under \(y=(x-3)^2\) from \(x=1\) to \(x=3\).

Thus

\(\text{area}=\int_0^1 4x^{\frac23}\,dx+\int_1^3 (x-3)^2\,dx.\)

First,

\(\int_0^1 4x^{\frac23}\,dx =\left[\frac{12}{5}x^{\frac53}\right]_0^1 =\frac{12}{5}.\)

Also,

\(\int_1^3 (x-3)^2\,dx =\left[\frac{(x-3)^3}{3}\right]_1^3 =0-\frac{(-2)^3}{3} =\frac83.\)

Therefore

\(\text{area}=\frac{12}{5}+\frac83 =\frac{36+40}{15} =\frac{76}{15}.\)