Using the substitution \(u = \sqrt{2x - 1}\), show that \(\displaystyle \displaystyle\int \frac{x}{\sqrt{2x - 1}} \, \mathrm{d}x = \displaystyle \displaystyle\int \frac{u^2 + 1}{2} \, \mathrm{d}u\).
The region \(R\) is bounded by the curve \(y = 2 - \frac{x}{\sqrt{2x - 1}}\) and the lines \(x = 1\), \(x = 3\) and \(y = \frac{1}{2}\). Find the exact area of \(R\).
The region \(S\) is bounded by the curves \(y = 2 - \frac{x}{\sqrt{2x - 1}}\), \(x = 2(y - 1)^2 + 1\), the lines \(x = 1\), \(x = 4\) and the \(x\)-axis.
Express \(x = 2(y - 1)^2 + 1\) in the form \(y = \mathrm{f}(x)\).
On the same diagram as in part (b)(i), sketch the graph of \(x = 2(y - 1)^2 + 1\).
Find the volume of the solid generated when \(S\) is rotated through \(2\pi\) radians about the \(x\)-axis. Give your answer correct to \(3\) decimal places.
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