Multi-part questions combining substitution with curve/area analysis

10 \includegraphics[max width=\textwidth, alt={}, center]{0f081749-4fe0-46e3-96c2-466e69cf49d3-4_620_894_338_687} The function f is defined by \(\mathrm { f } ( x ) = ( \ln x ) ^ { 2 }\) for \(x > 0\). The diagram shows a sketch of the graph of \(y = \mathrm { f } ( x )\). The minimum point of the graph is \(A\). The point \(B\) has \(x\)-coordinate e .

1. State the \(x\)-coordinate of \(A\).
2. Show that \(\mathrm { f } ^ { \prime \prime } ( x ) = 0\) at \(B\).
3. Use the substitution \(x = \mathrm { e } ^ { u }\) to show that the area of the region bounded by the \(x\)-axis, the line \(x = \mathrm { e }\), and the part of the curve between \(A\) and \(B\) is given by $$\int _ { 0 } ^ { 1 } u ^ { 2 } \mathrm { e } ^ { u } \mathrm {~d} u .$$
4. Hence, or otherwise, find the exact value of this area.

9 \includegraphics[max width=\textwidth, alt={}, center]{b2136f5d-0d66-4524-bb76-fcc4cb59150c-3_639_387_1749_879} The diagram shows the curve \(y = \mathrm { e } ^ { 2 \sin x } \cos x\) for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\), and its maximum point \(M\).

1. Using the substitution \(u = \sin x\), find the exact value of the area of the shaded region bounded by the curve and the axes.
2. Find the \(x\)-coordinate of \(M\), giving your answer correct to 3 decimal places.

10 \includegraphics[max width=\textwidth, alt={}, center]{b00cefad-7c3c-4672-b309-f19aafab8b01-18_324_677_259_734} The diagram shows the curve \(y = \sin x \cos ^ { 2 } 2 x\) for \(0 \leqslant x \leqslant \frac { 1 } { 4 } \pi\) and its maximum point \(M\).

1. Using the substitution \(u = \cos x\), find by integration the exact area of the shaded region bounded by the curve and the \(x\)-axis.
2. Find the \(x\)-coordinate of \(M\). Give your answer correct to 2 decimal places.

10 \includegraphics[max width=\textwidth, alt={}, center]{dcfbe7af-c212-42b1-8a90-8e0418cf0ffd-16_330_689_264_726} The diagram shows the curve \(y = \sin ^ { 3 } x \sqrt { } ( \cos x )\) for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\), and its maximum point \(M\).

1. Using the substitution \(u = \cos x\), find by integration the exact area of the shaded region bounded by the curve and the \(x\)-axis.
2. Showing all your working, find the \(x\)-coordinate of \(M\), giving your answer correct to 3 decimal places.
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

7 \includegraphics[max width=\textwidth, alt={}, center]{adbef77f-e2ac-40ce-a56b-cf6776534ec1-3_543_1091_1402_529} The diagram shows part of the curve \(y = \sin ^ { 3 } 2 x \cos ^ { 3 } 2 x\). The shaded region shown is bounded by the curve and the \(x\)-axis and its exact area is denoted by \(A\).

1. Use the substitution \(u = \sin 2 x\) in a suitable integral to find the value of \(A\).
2. Given that \(\int _ { 0 } ^ { k \pi } \left| \sin ^ { 3 } 2 x \cos ^ { 3 } 2 x \right| \mathrm { d } x = 40 A\), find the value of the constant \(k\).

7 \includegraphics[max width=\textwidth, alt={}, center]{c861e691-66da-4269-9057-4a343be9835e-12_357_565_260_790} The diagram shows the curve \(y = 5 \sin ^ { 2 } x \cos ^ { 3 } x\) for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\), and its maximum point \(M\). The shaded region \(R\) is bounded by the curve and the \(x\)-axis.

1. Find the \(x\)-coordinate of \(M\), giving your answer correct to 3 decimal places.
2. Using the substitution \(u = \sin x\) and showing all necessary working, find the exact area of \(R\). [4]

10 \includegraphics[max width=\textwidth, alt={}, center]{5b5ed7d1-028e-4f9a-ae9e-26071d0df678-18_449_787_262_678} The diagram shows the graph of \(y = \mathrm { e } ^ { \cos x } \sin ^ { 3 } x\) for \(0 \leqslant x \leqslant \pi\), and its maximum point \(M\). The shaded region \(R\) is bounded by the curve and the \(x\)-axis.

1. Find the \(x\)-coordinate of \(M\). Show all necessary working and give your answer correct to 2 decimal places.
2. By first using the substitution \(u = \cos x\), find the exact value of the area of \(R\).
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

11 \includegraphics[max width=\textwidth, alt={}, center]{7cdf4db7-7217-4ef1-becf-359a70cfeb62-16_556_698_274_712} The diagram shows the curve \(y = \sin x \cos 2 x\) for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\), and its maximum point \(M\).

1. Find the \(x\)-coordinate of \(M\), giving your answer correct to 3 significant figures.
2. Using the substitution \(u = \cos x\), find the area of the shaded region enclosed by the curve and the \(x\)-axis in the first quadrant, giving your answer in a simplified exact form.
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

11 \includegraphics[max width=\textwidth, alt={}, center]{6280ab81-0bdb-47b4-8651-bff1261a0adf-18_565_634_260_717} The diagram shows the curve \(y = 2 \sin x \sqrt { 2 + \cos x }\), for \(0 \leqslant x \leqslant 2 \pi\), and its minimum point \(M\), where \(x = a\).

1. Find the value of \(a\) correct to 2 decimal places. \includegraphics[max width=\textwidth, alt={}, center]{6280ab81-0bdb-47b4-8651-bff1261a0adf-18_2716_38_109_2012} \includegraphics[max width=\textwidth, alt={}, center]{6280ab81-0bdb-47b4-8651-bff1261a0adf-19_2726_33_97_22}
2. Use the substitution \(u = 2 + \cos x\) to find the exact area of the shaded region \(R\).
   If you use the following page to complete the answer to any question, the question number must be clearly shown.

7 A curve is defined by the equation \(y = 2 x \ln ( 1 + x )\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and hence verify that the origin is a stationary point of the curve.
2. Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\), and use this to verify that the origin is a minimum point.
3. Using the substitution \(u = 1 + x\), show that \(\int \frac { x ^ { 2 } } { 1 + x } \mathrm {~d} x = \int \left( u - 2 + \frac { 1 } { u } \right) \mathrm { d } u\). Hence evaluate \(\int _ { 0 } ^ { 1 } \frac { x ^ { 2 } } { 1 + x } \mathrm {~d} x\), giving your answer in an exact form.
4. Using integration by parts and your answer to part (iii), evaluate \(\int _ { 0 } ^ { 1 } 2 x \ln ( 1 + x ) \mathrm { d } x\).

2 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 \(( 11,11 )\). \begin{figure}[h] \end{figure}

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

3

1. Use the substitution \(u = 1 + x\) to show that $$\int _ { 0 } ^ { 1 } \frac { x ^ { 3 } } { 1 + x } \mathrm {~d} x = \int _ { a } ^ { b } \left( u ^ { 2 } - 3 u + 3 - \frac { 1 } { u } \right) \mathrm { d } u$$ where \(a\) and \(b\) are to be found.
   Hence evaluate \(\int _ { 0 } ^ { 1 } \frac { x ^ { 3 } } { 1 + x } \mathrm {~d} x\), giving your answer in exact form. Fig. 8 shows the curve \(y = x ^ { 2 } \ln ( 1 + x )\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{d1206ce8-7716-4205-b98e-664e7ead8a25-3_830_806_907_706} \captionsetup{labelformat=empty} \caption{Fig. 8}
   \end{figure}
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\). Verify that the origin is a stationary point of the curve.
3. Using integration by parts, and the result of part (i), find the exact area enclosed by the curve \(y = x ^ { 2 } \ln ( 1 + x )\), the \(x\)-axis and the line \(x = 1\).

2 Fig. 7 shows part of the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = x \sqrt { 1 + x }\). The curve meets the \(x\)-axis at the origin and at the point P . \begin{figure}[h] \end{figure}

1. Verify that the point P has coordinates \(( - 1,0 )\). Hence state the domain of the function \(\mathrm { f } ( x )\).
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 + 3 x } { 2 \sqrt { 1 + x } }\).
3. Find the exact coordinates of the turning point of the curve. Hence write down the range of the function.
4. Use the substitution \(u = 1 + x\) to show that $$\int _ { - 1 } ^ { 0 } x \sqrt { 1 + x } \mathrm {~d} x = \int _ { 0 } ^ { 1 } \left( \begin{array} { l l } u ^ { \frac { 3 } { 2 } } & u ^ { \frac { 1 } { 2 } } \end{array} \right) \mathrm { d } u .$$ Hence find the area of the region enclosed by the curve and the \(x\)-axis.

5 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 \(( 11,11 )\). \begin{figure}[h] \end{figure}

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

2

1. Use the substitution \(u = 1 + x\) to show that $$\int _ { 0 } ^ { 1 } \frac { x ^ { 3 } } { 1 + x } \mathrm {~d} x = \int _ { a } ^ { b } \left( u ^ { 2 } - 3 u + 3 - \frac { 1 } { u } \right) \mathrm { d } u$$ where \(a\) and \(b\) are to be found.
   Hence evaluate \(\int _ { 0 } ^ { 1 } \frac { x ^ { 3 } } { 1 + x } \mathrm {~d} x\), giving your answer in exact form. Fig. 8 shows the curve \(y = x ^ { 2 } \ln ( 1 + x )\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{35646966-3747-4f1d-bf94-60e9e3130afe-2_829_806_944_706} \captionsetup{labelformat=empty} \caption{Fig. 8}
   \end{figure}
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\). Verify that the origin is a stationary point of the curve.
3. Using integration by parts, and the result of part (i), find the exact area enclosed by the curve \(y = x ^ { 2 } \ln ( 1 + x )\), the \(x\)-axis and the line \(x = 1\).

8

1. The curve \(C _ { 1 }\) has equation \(y = - \ln ( \cos x )\). Show that the length of the arc of \(C _ { 1 }\) from the point where \(x = 0\) to the point where \(x = \frac { 1 } { 3 } \pi\) is \(\ln ( 2 + \sqrt { 3 } )\).
2. The curve \(C _ { 2 }\) has equation \(y = 2 \sqrt { } ( x + 3 )\). The arc of \(C _ { 2 }\) joining the point where \(x = 0\) to the point where \(x = 1\) is rotated through one complete revolution about the \(x\)-axis. Show that the area of the surface generated is $$\frac { 8 } { 3 } \pi ( 5 \sqrt { } 5 - 8 )$$

8

1. Show that $$\int _ { 0 } ^ { \ln 2 } \mathrm { e } ^ { 1 - 2 x } \mathrm {~d} x = \frac { 3 } { 8 } \mathrm { e }$$
2. Use the substitution \(u = \tan x\) to find the exact value of $$\int _ { 0 } ^ { \frac { \pi } { 4 } } \sec ^ { 4 } x \sqrt { \tan x } d x$$ (8 marks)

10

1. Show that \(\int _ { 0 } ^ { 2 } \frac { x } { x ^ { 2 } + 5 } \mathrm {~d} x = \ln \left( \frac { 3 } { \sqrt { 5 } } \right)\).
2. Find \(\int x \sqrt { x - 2 } \mathrm {~d} x\).

\includegraphics{figure_7} The diagram shows the curve \(y = 5\sin^2 x \cos^3 x\) for \(0 \leqslant x \leqslant \frac{1}{2}\pi\), and its maximum point \(M\). The shaded region \(R\) is bounded by the curve and the \(x\)-axis.

1. Find the \(x\)-coordinate of \(M\), giving your answer correct to 3 decimal places. [5]
2. Using the substitution \(u = \sin x\) and showing all necessary working, find the exact area of \(R\). [4]