1.08i Integration by parts

9

1. Given that \(y = x ^ { - 2 } \ln x\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 - 2 \ln x } { x ^ { 3 } }\).
2. Using integration by parts, find \(\int x ^ { - 2 } \ln x \mathrm {~d} x\).
3. The sketch shows the graph of \(y = x ^ { - 2 } \ln x\). \includegraphics[max width=\textwidth, alt={}, center]{908f530c-076d-47b1-90dd-38dbfe44f898-06_604_1045_687_536}
   1. Using the answer to part (a), find, in terms of e, the \(x\)-coordinate of the stationary point \(A\).
   2. The region \(R\) is bounded by the curve, the \(x\)-axis and the line \(x = 5\). Using your answer to part (b), show that the area of \(R\) is $$\frac { 1 } { 5 } ( 4 - \ln 5 )$$

9

1. Given that \(y = \frac { 4 x } { 4 x - 3 }\), use the quotient rule to show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { k } { ( 4 x - 3 ) ^ { 2 } }\), where \(k\) is an integer.
2. 1. Given that \(y = x \ln ( 4 x - 3 )\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
   2. Find an equation of the tangent to the curve \(y = x \ln ( 4 x - 3 )\) at the point where \(x = 1\).
3. 1. Use the substitution \(u = 4 x - 3\) to find \(\int \frac { 4 x } { 4 x - 3 } \mathrm {~d} x\), giving your answer in terms of \(x\).
   2. By using integration by parts, or otherwise, find \(\int \ln ( 4 x - 3 ) \mathrm { d } x\).

8

1. Using integration by parts, find \(\int x \sin ( 2 x - 1 ) \mathrm { d } x\).
2. Use the substitution \(u = 2 x - 1\) to find \(\int \frac { x ^ { 2 } } { 2 x - 1 } \mathrm {~d} x\), giving your answer in terms of \(x\).
   (6 marks)

6

1. Use integration by parts to find \(\int x \mathrm { e } ^ { 5 x } \mathrm {~d} x\).
2. 1. Use the substitution \(u = \sqrt { x }\) to show that $$\int \frac { 1 } { \sqrt { x } ( 1 + \sqrt { x } ) } \mathrm { d } x = \int \frac { 2 } { 1 + u } \mathrm {~d} u$$
   2. Find the exact value of \(\int _ { 1 } ^ { 9 } \frac { 1 } { \sqrt { x } ( 1 + \sqrt { x } ) } \mathrm { d } x\).

5

1. By writing \(\tan x\) as \(\frac { \sin x } { \cos x }\), use the quotient rule to show that \(\frac { \mathrm { d } } { \mathrm { d } x } ( \tan x ) = \sec ^ { 2 } x\).
   [0pt] [2 marks]
2. Use integration by parts to find \(\int x \sec ^ { 2 } x \mathrm {~d} x\).
   [0pt] [4 marks]
3. The region bounded by the curve \(y = ( 5 \sqrt { x } ) \sec x\), the \(x\)-axis from 0 to 1 and the line \(x = 1\) is rotated through \(2 \pi\) radians about the \(x\)-axis to form a solid. Find the value of the volume of the solid generated, giving your answer to two significant figures.
   [0pt] [3 marks]

5

1. Show that \(\int x \mathrm { e } ^ { - 2 x } \mathrm {~d} x = - \frac { 1 } { 4 } \mathrm { e } ^ { - 2 x } ( 1 + 2 x ) + c\). A vase is made in the shape of the volume of revolution of the curve \(y = x ^ { 1 / 2 } \mathrm { e } ^ { - x }\) about the \(x\)-axis between \(x = 0\) and \(x = 2\) (see Fig. 5). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{c64062c4-4cbd-41b2-9b4d-60a43dceb700-3_716_741_1233_662} \captionsetup{labelformat=empty} \caption{Fig. 5}
   \end{figure}
2. Show that this volume of revolution is \(\frac { 1 } { 4 } \pi \left( 1 - \frac { 5 } { \mathrm { e } ^ { 4 } } \right)\). Fig. 6 shows the arch ABCD of a bridge. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{c64062c4-4cbd-41b2-9b4d-60a43dceb700-4_378_1630_461_214} \captionsetup{labelformat=empty} \caption{Fig. 6}
   \end{figure} The section from B to C is part of the curve OBCE with parametric equations $$x = a ( \theta - \sin \theta ) , y = a ( 1 - \cos \theta ) \text { for } 0 \leqslant \theta \leqslant 2 \pi$$ where \(a\) is a constant.
3. Find, in terms of \(a\),
   (A) the length of the straight line OE,
   (B) the maximum height of the arch.
4. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\). The straight line sections AB and CD are inclined at \(30 ^ { \circ }\) to the horizontal, and are tangents to the curve at B and C respectively. BC is parallel to the \(x\)-axis. BF is parallel to the \(y\)-axis.
5. Show that at the point B the parameter \(\theta\) satisfies the equation $$\sin \theta = \frac { 1 } { \sqrt { 3 } } ( 1 - \cos \theta )$$ Verify that \(\theta = \frac { 2 } { 3 } \pi\) is a solution of this equation.
   Hence show that \(\mathrm { BF } = \frac { 3 } { 2 } a\), and find OF in terms of \(a\), giving your answer exactly.
6. Find BC and AF in terms of \(a\). Given that the straight line distance AD is 20 metres, calculate the value of \(a\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{c64062c4-4cbd-41b2-9b4d-60a43dceb700-5_748_1306_319_367} \captionsetup{labelformat=empty} \caption{Fig. 7}
   \end{figure} Fig. 7 illustrates a house. All units are in metres. The coordinates of A, B, C and E are as shown. BD is horizontal and parallel to AE .
7. Find the length AE .
8. Find a vector equation of the line BD . Given that the length of BD is 15 metres, find the coordinates of D.
9. Verify that the equation of the plane ABC is $$- 3 x + 4 y + 5 z = 30$$ Write down a vector normal to this plane.
10. Show that the vector \(\left( \begin{array} { l } 4 \\ 3 \\ 5 \end{array} \right)\) is normal to the plane ABDE . Hence find the equation of the plane ABDE .
11. Find the angle between the planes ABC and ABDE . RECOGNISING ACHIEVEMENT \section*{OXFORD CAMBRIDGE AND RSA EXAMINATIONS} Advanced Subsidiary General Certificate of Education Advanced General Certificate of Education \section*{MEI STRUCTURED MATHEMATICS} Applications of Advanced Mathematics (C4) \section*{Paper B: Comprehension} Monday 12 JUNE 2006 Afternoon Up to 1 hour Additional materials:
    Rough paper
    MEI Examination Formulae and Tables (MF2) TIME Up to 1 hour

    For Examiner's Use
    Qu.	Mark
    1
    2
    3
    4
    5
    6
    Total

    1 The marathon is 26 miles and 385 yards long ( 1 mile is 1760 yards). There are now several men who can run 2 miles in 8 minutes. Imagine that an athlete maintains this average speed for a whole marathon. How long does the athlete take?
    2 According to the linear model, in which calendar year would the record for the men's mile first become negative?
    3 Explain the statement in line 93 "According to this model the 2-hour marathon will never be run."
    4 Explain how the equation in line 49, $$R = L + ( U - L ) \mathrm { e } ^ { - k t } ,$$ is consistent with Fig. 2
12. initially,
13. for large values of \(t\).
15. \(\_\_\_\_\) 5 A model for an athletics record has the form $$R = A - ( A - B ) \mathrm { e } ^ { - k t } \text { where } A > B > 0 \text { and } k > 0 .$$
16. Sketch the graph of \(R\) against \(t\), showing \(A\) and \(B\) on your graph.
17. Name one event for which this might be an appropriate model.
18. \includegraphics[max width=\textwidth, alt={}, center]{c64062c4-4cbd-41b2-9b4d-60a43dceb700-9_803_808_721_575}
19. \(\_\_\_\_\)

4

1. Differentiate \(x \tan ^ { - 1 } x\) with respect to \(x\).
2. Show that $$\int _ { 0 } ^ { 1 } \tan ^ { - 1 } x \mathrm {~d} x = \frac { \pi } { 4 } - \ln \sqrt { 2 }$$ (5 marks)

6

1. It is given that, for non-negative integers \(n\), $$I _ { n } = \int _ { 0 } ^ { 1 } \mathrm { e } ^ { - x } x ^ { n } \mathrm {~d} x$$ Prove that, for \(n \geqslant 1\), $$I _ { n } = n I _ { n - 1 } - \mathrm { e } ^ { - 1 } .$$
2. Evaluate \(I _ { 3 }\), giving the answer in terms of e.

4.(a)Use the substitution \(x = \sqrt { 3 } \tan u\) to show that $$\int \frac { 1 } { 3 + x ^ { 2 } } \mathrm {~d} x = p \arctan ( p x ) + c$$ where \(p\) is a real constant to be determined and \(c\) is an arbitrary constant.
(b)Use the substitution \(x = \frac { 3 u + 3 } { u - 3 }\) to determine the exact value of \(I\) where $$I = \int _ { - 3 } ^ { 1 } \frac { \ln ( 3 - x ) } { 3 + x ^ { 2 } } \mathrm {~d} x$$ giving your answer in simplest form. \includegraphics[max width=\textwidth, alt={}, center]{a8e9db6b-dfad-4278-82d8-a8fa5ba61008-10_2264_47_314_1984}

7

1. Find \(\int 5 x ^ { 3 } \sqrt { x ^ { 2 } + 1 } \mathrm {~d} x\).
2. Find \(\int \theta \tan ^ { 2 } \theta \mathrm {~d} \theta\). You may use the result \(\int \tan \theta \mathrm { d } \theta = \ln | \sec \theta | + c\).

14 The region bounded by the curve $$y = ( 2 x - 8 ) \ln x$$ and the \(x\)-axis is shaded in the diagram below. \includegraphics[max width=\textwidth, alt={}, center]{22ff390e-1360-43bd-8c7f-3d2b58627e91-26_867_908_543_566} 14

1. Use the trapezium rule with 5 ordinates to find an estimate for the area of the shaded region. Give your answer correct to three significant figures.
   14
2. Show that the exact area is given by $$32 \ln 2 - \frac { 33 } { 2 }$$ Fully justify your answer.

8 Show that $$\int _ { 0 } ^ { \frac { \pi } { 2 } } ( x \sin 4 x ) \mathrm { d } x = - \frac { \pi } { 8 }$$

9 It is given that \(I _ { n } = \int _ { 1 } ^ { \mathrm { e } } ( \ln x ) ^ { n } \mathrm {~d} x\) for \(n \geqslant 0\). Show that $$I _ { n } = ( n - 1 ) \left[ I _ { n - 2 } - I _ { n - 1 } \right] \text { for } n \geqslant 2$$ Hence find, in an exact form, the mean value of \(( \ln x ) ^ { 3 }\) with respect to \(x\) over the interval \(1 \leqslant x \leqslant \mathrm { e }\).

11

1. Use integration by parts to show that \(\int \ln x \mathrm {~d} x = x \ln x - x + c\).
2. Find
   1. \(\int ( \ln x ) ^ { 2 } \mathrm {~d} x\),
   2. \(\quad \int \frac { \ln ( \ln x ) } { x } \mathrm {~d} x\).

5

1. Find \(\int \left( \frac { 1 } { x - 2 } - \frac { 2 } { 2 x + 3 } \right) \mathrm { d } x\) giving your answer in its simplest form.
2. Use integration by parts to find \(\int x ^ { 2 } \ln x \mathrm {~d} x\).

9 Find \(\int x \sin 2 x \mathrm {~d} x\).

4 Let \(I _ { n } = \int _ { 0 } ^ { 4 } x ^ { n } \sqrt { 2 x + 1 } \mathrm {~d} x\) for \(n \geqslant 0\). Show that, for \(n \geqslant 1\), $$( 2 n + 3 ) I _ { n } = 27 \times 4 ^ { n } - n I _ { n - 1 }$$

13

1. (a) Given that \(x \geqslant 1\), show that \(\sec ^ { - 1 } x = \cos ^ { - 1 } \left( \frac { 1 } { x } \right)\), and deduce that \(\frac { \mathrm { d } } { \mathrm { d } x } \left( \sec ^ { - 1 } x \right) = \frac { 1 } { x \sqrt { x ^ { 2 } - 1 } }\).
   (b) Use integration by parts to determine \(\int \sec ^ { - 1 } x \mathrm {~d} x\).
2. \includegraphics[max width=\textwidth, alt={}, center]{5d526fd9-72f8-42b1-b156-fd4a0c764c82-4_670_1029_1073_596} The diagram shows the curve \(S\) with equation \(y = \sec ^ { - 1 } x\) for \(x \geqslant 1\). The line \(L\), with gradient \(\frac { 1 } { \sqrt { 2 } }\), is the tangent to \(S\) at the point \(P\) and cuts the \(x\)-axis at the point \(Q\). The point \(I\) has coordinates \(( 1,0 )\).
   (a) Determine the exact coordinates of \(P\) and \(Q\).
   (b) The region \(R\), shaded on the diagram, is bounded by the line segments \(P Q\) and \(Q I\) and the \(\operatorname { arc } I P\) of \(S\). Show that \(R\) has area $$\ln ( 1 + \sqrt { 2 } ) - \frac { \pi ( 8 - \pi ) \sqrt { 2 } } { 32 } .$$ {www.cie.org.uk} after the live examination series. }

8

1. Evaluate exactly \(\int _ { 0 } ^ { 1 } x \mathrm { e } ^ { - x } \mathrm {~d} x\).
2. Find \(\int \frac { x - 1 } { x + 1 } \mathrm {~d} x\).

12

1. Use integration by parts to show that \(\int \ln x \mathrm {~d} x = x \ln x - x + c\).
2. Find
   1. \(\quad \int ( \ln x ) ^ { 2 } \mathrm {~d} x\),
   2. \(\quad \int \frac { \ln ( \ln x ) } { x } \mathrm {~d} x\).

9

1. Given that \(x \geqslant 1\), use the substitution \(x = \cosh \theta\) to show that $$\int \frac { 1 } { x ^ { 2 } \sqrt { x ^ { 2 } - 1 } } \mathrm {~d} x = \frac { \sqrt { x ^ { 2 } - 1 } } { x } + C$$ where \(C\) is an arbitrary constant.
2. By differentiating sec \(y = x\) implicitly, show that \(\frac { \mathrm { d } } { \mathrm { d } x } \left( \sec ^ { - 1 } x \right) = \frac { 1 } { x \sqrt { x ^ { 2 } - 1 } }\) for \(x \geqslant 1\).
3. Use integration by parts to determine \(\int \frac { \sec ^ { - 1 } x } { x ^ { 2 } } \mathrm {~d} x\) for \(x \geqslant 1\).

9

1. Show that \(\int x ( x - 2 ) ^ { \frac { 3 } { 2 } } \mathrm {~d} x = \frac { 2 } { 35 } ( 5 x + 4 ) ( x - 2 ) ^ { \frac { 5 } { 2 } } + c\).
2. Hence find the coordinates of the stationary points of the curve $$y = \frac { 2 } { 35 } ( 5 x + 4 ) ( x - 2 ) ^ { \frac { 5 } { 2 } } + x ^ { 2 } - \frac { 1 } { 3 } x ^ { 3 }$$

8

1. Use integration by parts twice to show that $$\int \mathrm { e } ^ { x } \sin x \mathrm {~d} x = \frac { 1 } { 2 } \mathrm { e } ^ { x } ( \sin x - \cos x ) + c .$$
2. Hence find the equation of the curve which passes through the point \(( 0,2 )\) and for which \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { e } ^ { x } \sin x\).

12

1. Use integration by parts to show that \(\int \ln x \mathrm {~d} x = x \ln x - x + c\).
2. Find
   1. \(\quad \int ( \ln x ) ^ { 2 } \mathrm {~d} x\),
   2. \(\quad \int \frac { \ln ( \ln x ) } { x } \mathrm {~d} x\).

12

1. Use integration by parts to show that \(\int \ln x \mathrm {~d} x = x \ln x - x + c\).
2. Find
   1. \(\int ( \ln x ) ^ { 2 } \mathrm {~d} x\),
   2. \(\int \frac { \ln ( \ln x ) } { x } \mathrm {~d} x\).