Answer: \(\dfrac25+\ln9\).

At \(A\), the line and curve intersect. Since \(y=5x-3\) and \(xy=2\), substitute \(y=5x-3\) into the curve:

\(x(5x-3)=2.\)

So

\(5x^2-3x-2=0.\)

Factorising,

\((5x+2)(x-1)=0.\)

The point \(A\) shown on the diagram is in the first quadrant, so \(x=1\). Then

\(y=5(1)-3=2.\)

Thus

\(A=(1,2).\)

The point \(B\) is where the line meets the \(x\)-axis, so \(y=0\):

\(0=5x-3.\)

Hence

\(x=\frac35,\)

so

\(B=\left(\frac35,0\right).\)

The shaded region is made from the triangle under the line from \(x=\frac35\) to \(x=1\), plus the area under the curve from \(x=1\) to \(x=3\).

The triangle has base

\(1-\frac35=\frac25\)

and height \(2\), so its area is

\(\frac12\cdot\frac25\cdot2=\frac25.\)

For the curve \(xy=2\),

\(y=\frac2x.\)

The area under the curve from \(x=1\) to \(x=3\) is

\(\int_1^3 \frac2x\,dx =\left[2\ln x\right]_1^3.\)

So this area is

\(2\ln3=\ln9.\)

Therefore the exact shaded area is

\(\frac25+\ln9.\)