Answer: \(A\left(\dfrac14,\dfrac{15}{4}\right)\), \(B(4,15)\); area \(=\dfrac{45}{16}\).

At the intersections,

\(1+x+5\sqrt{x}=3x+3.\)

Let \(u^2=x\), where \(u\geq0\). Then \(\sqrt{x}=u\), so

\(1+u^2+5u=3u^2+3.\)

Rearranging,

\(2u^2-5u+2=0.\)

Factorise:

\((2u-1)(u-2)=0.\)

Thus

\(u=\frac12\quad\text{or}\quad u=2.\)

Since \(x=u^2\),

\(x=\frac14\quad\text{or}\quad x=4.\)

Using the line \(y=3x+3\), the coordinates are

\(A\left(\frac14,\frac{15}{4}\right),\qquad B(4,15).\)

The shaded area is the area between the curve and the line from \(x=\frac14\) to \(x=4\):

\(\int_{1/4}^{4}\left[(1+x+5\sqrt{x})-(3x+3)\right]\,dx.\)

This simplifies to

\(\int_{1/4}^{4}\left(5\sqrt{x}-2x-2\right)\,dx.\)

Integrating,

\(\int \left(5x^{1/2}-2x-2\right)\,dx =\frac{10}{3}x^{3/2}-x^2-2x.\)

Therefore the area is

\(\left[\frac{10}{3}x^{3/2}-x^2-2x\right]_{1/4}^{4}.\)

At \(x=4\), the value is \(\frac83\). At \(x=\frac14\), the value is \(-\frac7{48}\). Hence

\(\text{area}=\frac83-\left(-\frac7{48}\right)=\frac{45}{16}.\)