To find the area of the shaded region, we need to integrate the difference between the two curves from \(x = 0\) to \(x = 4\).

The area \(A\) is given by:

\(A = \int_0^4 \left( 18 - \frac{3}{8}x^{\frac{5}{2}} - \left( \frac{9}{4}x^2 - 12x + 18 \right) \right) \, dx\)

This simplifies to:

\(A = \int_0^4 \left( 18 - \frac{3}{8}x^{\frac{5}{2}} - \frac{9}{4}x^2 + 12x - 18 \right) \, dx\)

\(A = \int_0^4 \left( -\frac{3}{8}x^{\frac{5}{2}} - \frac{9}{4}x^2 + 12x \right) \, dx\)

Integrating each term separately:

\(\int -\frac{3}{8}x^{\frac{5}{2}} \, dx = -\frac{3}{28}x^{\frac{7}{2}}\)

\(\int -\frac{9}{4}x^2 \, dx = -\frac{9}{12}x^3 = -\frac{3}{4}x^3\)

\(\int 12x \, dx = 6x^2\)

Evaluating from 0 to 4:

\(A = \left[ -\frac{3}{28}x^{\frac{7}{2}} - \frac{3}{4}x^3 + 6x^2 \right]_0^4\)

Substitute \(x = 4\):

\(A = -\frac{3}{28}(4)^{\frac{7}{2}} - \frac{3}{4}(4)^3 + 6(4)^2\)

\(A = -\frac{3}{28} \times 128 - \frac{3}{4} \times 64 + 6 \times 16\)

\(A = -\frac{384}{28} - 48 + 96\)

\(A = -13.7142857 + 48\)

\(A = \frac{240}{7} \approx 34.3\)