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 \(x = 4\):

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

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

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

\(A = -13.7142857 + 48\)

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

(i) The curve is given by \(y = 2\sqrt{x}\). To find the equation of the normal, we first find the derivative \(\frac{dy}{dx}\). For \(y = 2\sqrt{x}\), \(\frac{dy}{dx} = x^{-\frac{1}{2}}\).

At \(x = 4\), \(\frac{dy}{dx} = \frac{1}{2}\). The slope of the normal is the negative reciprocal, so \(m = -2\).

The equation of the normal line at point B (4, 4) is \(y - 4 = -2(x - 4)\), which simplifies to \(y = -2x + 12\).

(ii) To find the area of the shaded region, we integrate the curve from \(x = 1\) to \(x = 4\). The integral is \(\int 2\sqrt{x} \, dx = \int 2x^{0.5} \, dx = \frac{2}{1.5}x^{1.5}\).

Evaluating from 1 to 4 gives \(\left[ \frac{2}{1.5}x^{1.5} \right]_1^4 = \frac{2}{1.5}(4^{1.5} - 1^{1.5}) = \frac{2}{1.5}(8 - 1) = \frac{2}{1.5} \times 7 = \frac{28}{3}\).

(i) To find the coordinates of \(P\), set \(y = 3\sqrt{x} = x\). Solving for \(x\), we have:

\(3\sqrt{x} = x\)

\(3 = \sqrt{x}\)

\(x = 9\)

Thus, \(y = x = 9\), so \(P\) is (9, 9).

(ii) To find the area of the shaded region, calculate the area under the curve \(y = 3\sqrt{x}\) from \(x = 0\) to \(x = 9\), and subtract the area under the line \(y = x\) over the same interval.

The area under the curve \(y = 3\sqrt{x}\) is given by:

\(\int_0^9 3\sqrt{x} \, dx = \int_0^9 3x^{1/2} \, dx\)

\(= \left[ 3 \cdot \frac{2}{3} x^{3/2} \right]_0^9\)

\(= \left[ 2x^{3/2} \right]_0^9\)

\(= 2(9^{3/2}) - 2(0^{3/2})\)

\(= 2(27)\)

\(= 54\)

The area under the line \(y = x\) is the area of a right triangle with base and height of 9:

\(\frac{1}{2} \times 9 \times 9 = 40.5\)

Thus, the area of the shaded region is:

\(54 - 40.5 = 13.5\)

(a) To find the area between the curves, integrate the difference of the functions from \(x = 0\) to \(x = 4\):

\(\int (2x^{\frac{1}{2}} + 1) - \left(\frac{1}{2}x^2 - x + 1\right) \, dx\)

Calculate the integrals:

\(\int 2x^{\frac{1}{2}} \, dx = \frac{4}{3}x^{\frac{3}{2}}\)

\(\int \left(\frac{1}{2}x^2 - x + 1\right) \, dx = \frac{x^3}{6} - \frac{x^2}{2} + x\)

Evaluate from 0 to 4:

\(\left(\frac{32}{3} + 32 - 0\right) - \left(\frac{44}{3} - 20 + 0\right) = 8\)

(b) Differentiate both curves to find the gradients at \(x = 4\):

Upper curve: \(\frac{dy}{dx} = x^{-\frac{1}{2}}\), at \(x = 4\), gradient = \(\frac{1}{2}\).

Lower curve: \(\frac{dy}{dx} = x - 1\), at \(x = 4\), gradient = 3.

Find \(\alpha\) using the inverse tangent:

\(\alpha = \arctan(3) - \arctan\left(\frac{1}{2}\right)\)

\(\alpha = 45^\circ\)

(a) To find the equation of the tangent, first differentiate the curve equation \(y = x^{\frac{1}{2}} + 4x^{-\frac{1}{2}}\).

The derivative is \(\frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}} - 2x^{-\frac{3}{2}}\).

At \(x = 1\), \(\frac{dy}{dx} = \frac{1}{2} \times 1^{-\frac{1}{2}} - 2 \times 1^{-\frac{3}{2}} = \frac{1}{2} - 2 = -\frac{3}{2}\).

The equation of the tangent at \(A(1, 5)\) is \(y - 5 = -\frac{3}{2}(x - 1)\).

Simplifying gives \(y = -\frac{3}{2}x + \frac{13}{2}\).

(b) To find the area of the shaded region, integrate the curve equation from \(x = 1\) to \(x = 16\).

The integral of \(y = x^{\frac{1}{2}} + 4x^{-\frac{1}{2}}\) is \(\frac{x^{\frac{3}{2}}}{\frac{3}{2}} + 8x^{\frac{1}{2}}\).

Evaluating from 1 to 16 gives \(\left[ \frac{128}{3} + 32 \right] - \left[ \frac{2}{3} + 8 \right] = 66\).

The area under the line \(y = 5\) from \(x = 1\) to \(x = 16\) is \(15 \times 5 = 75\).

The required area is \(75 - 66 = 9\).