Area with parametric/substitution

9. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation \(y = \mathrm { e } ^ { \sqrt { x } } , x > 0\) The finite region \(R\), shown shaded in Figure 1, is bounded by the curve, the \(x\)-axis and the lines \(x = 4\) and \(x = 9\)

1. Use the trapezium rule, with 5 strips of equal width, to obtain an estimate for the area of \(R\), giving your answer to 2 decimal places.
2. Use the substitution \(u = \sqrt { x }\) to find, by integrating, the exact value for the area of \(R\).

5. \includegraphics[max width=\textwidth, alt={}, center]{5840974b-b08a-4818-9a59-97b2d3ce9890-1_469_809_1777_484} The diagram shows the curve with equation \(y = x \sqrt { 1 - x } , 0 \leq x \leq 1\).
Use the substitution \(u ^ { 2 } = 1 - x\) to show that the area of the region bounded by the curve and the \(x\)-axis is \(\frac { 4 } { 15 }\).

Fig. 8 shows the curve \(y = \frac{x}{\sqrt{x-2}}\), together with the lines \(y = x\) and \(x = 11\). The curve meets these lines at P and Q respectively. R is the point \((11, 11)\). \includegraphics{figure_8}

1. Verify that the \(x\)-coordinate of P is 3. [2]
2. Show that, for the curve, \(\frac{dy}{dx} = \frac{x-4}{2(x-2)^{\frac{3}{2}}}\). Hence find the gradient of the curve at P. Use the result to show that the curve is not symmetrical about \(y = x\). [7]
3. Using the substitution \(u = x - 2\), show that \(\int_3^{11} \frac{x}{\sqrt{x-2}} \, dx = 25\frac{1}{3}\). Hence find the area of the region PQR bounded by the curve and the lines \(y = x\) and \(x = 11\). [9]