8 Fig. 8 shows the graph of \(y = x \sqrt { 1 + x }\). The point P on the curve is on the \(x\)-axis.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{c7998a08-229a-40d2-ba34-b5f264139295-3_433_800_895_587}
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\caption{Fig. 8}
\end{figure}
- Write down the coordinates of P .
- Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 3 x + 2 } { 2 \sqrt { 1 + x } }\).
- Hence find the coordinates of the turning point on the curve.
What can you say about the gradient of the curve at P ?
- By using a suitable substitution, show that \(\int _ { - 1 } ^ { 0 } x \sqrt { 1 + x } \mathrm {~d} x = \int _ { 0 } ^ { 1 } \left( u ^ { \frac { 3 } { 2 } } - u ^ { \frac { 1 } { 2 } } \right) \mathrm { d } u\).
Evaluate this integral, giving your answer in an exact form.
What does this value represent? - Use your answer to part (ii) to differentiate \(y = x \sqrt { 1 + x } \sin 2 x\) with respect to \(x\).
(You need not simplify your result.)