Answer: (i) \(y=x+1\). (ii) Area \(=15.5\) square units.

(i) Differentiate the curve:

\(y=x+e^{5-2x}, \qquad \frac{dy}{dx}=1-2e^{5-2x}.\)

At \(x=2.5\),

\(e^{5-2(2.5)}=e^0=1,\)

so

\(\frac{dy}{dx}=1-2=-1.\)

The gradient of the normal is therefore \(1\). The \(y\)-coordinate of \(A\) is

\(y=2.5+e^0=3.5.\)

So \(A=(2.5,3.5)\). The normal has gradient \(1\), so

\(y-3.5=1(x-2.5).\)

Thus

\(y=x+1.\)

(ii) The shaded area is the area under the line \(y=x+1\) from \(x=0\) to \(x=2.5\), plus the area under the curve from \(x=2.5\) to \(x=5\).

The area under the line is a trapezium with parallel sides \(1\) and \(3.5\), and width \(2.5\):

\(\frac12(1+3.5)(2.5)=5.625.\)

For the curve,

\(\int (x+e^{5-2x})\,dx = \frac{x^2}{2}-\frac12e^{5-2x}.\)

So the area under the curve from \(2.5\) to \(5\) is

\(\left[\frac{x^2}{2}-\frac12e^{5-2x}\right]_{2.5}^{5}.\)

Evaluating this gives approximately

\(9.87.\)

Therefore the shaded area is

\(5.625+9.87=15.5\)

square units, to 3 significant figures.