Figure 3 shows the design of a doorknob. The shape of the doorknob is formed by rotating the curve shown in Figure 4 through $$\(360^{\circ}\)$$ about the $$\(x\)$$-axis, where the units are centimetres. The equation of the curve is given by $$\[ \mathrm{f}(x)=\frac{1}{4}(4-x) \mathrm{e}^{x} \quad 0 \leqslant x \leqslant 4 \]$$ (a) Show that the volume, $$\(V \mathrm{~cm}^{3}\)$$, of the doorknob is given by $$\[ V=K \int_{0}^{4}\left(x^{2}-8 x+16\right) \mathrm{e}^{2 x} \mathrm{~d} x \]$$ where $$\(K\)$$ is a constant to be found. (3) (b) Hence, find the exact value of the volume of the doorknob. Give your answer in the form $$\(p \pi\left(\mathrm{e}^{q}+r\right) \mathrm{cm}^{3}\)$$ where $$\(p, q\)$$ and $$\(r\)$$ are simplified rational numbers to be found. (5)

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