(i) Use integration by parts to evaluate \(\int x \ln x \, dx\).

Let \(u = \ln x\) and \(dv = x \, dx\). Then \(du = \frac{1}{x} \, dx\) and \(v = \frac{x^2}{2}\).

Integration by parts gives:

\(\int x \ln x \, dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx = \frac{x^2}{2} \ln x - \frac{1}{2} \int x \, dx\).

\(= \frac{x^2}{2} \ln x - \frac{1}{4} x^2\).

Substitute the limits from 1 to \(a\) and equate to 22:

\(\left[ \frac{a^2}{2} \ln a - \frac{1}{4} a^2 \right] - \left[ \frac{1}{2} \ln 1 - \frac{1}{4} \right] = 22\).

\(\frac{a^2}{2} \ln a - \frac{1}{4} a^2 + \frac{1}{4} = 22\).

Rearrange to find \(a\):

\(a = \sqrt{\frac{87}{2 \ln a - 1}}\).

(ii) Use the iterative formula \(a_{n+1} = \sqrt{\frac{87}{2 \ln a_n - 1}}\) with \(a_1 = 6\).

Calculate each iteration to 4 decimal places:

\(a_1 = 6\)

\(a_2 = 5.8030\)

\(a_3 = 5.8795\)

\(a_4 = 5.8491\)

\(a_5 = 5.8611\)

\(a_6 = 5.8564\)

The value of \(a\) correct to 2 decimal places is 5.86.