Answer: \(\frac32\leq x\leq7\); area \(\frac{1331}{24}\).

(a) Factorise:

\(2x^2-17x+21=(2x-3)(x-7).\)

So

\((2x-3)(x-7)\leq0.\)

The critical values are

\(x=\frac32,\qquad x=7.\)

Since the quadratic opens upwards, it is non-positive between the roots. Hence

\(\frac32\leq x\leq7.\)

(b) Between the roots, the curve is below the \(x\)-axis, so the enclosed area is

\(-\int_{\frac32}^{7}(2x^2-17x+21)\,dx.\)

Equivalently,

\(\int_{\frac32}^{7}(-2x^2+17x-21)\,dx.\)

Integrate:

This gives

\(\frac{1331}{24}.\)