Definite integral with logarithmic form

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1. Find \(\int _ { 4 } ^ { 10 } \frac { 4 } { 2 x - 5 } \mathrm {~d} x\), giving your answer in the form \(\ln a\), where \(a\) is an integer.
2. Find the exact value of \(\int _ { 4 } ^ { 10 } \mathrm { e } ^ { 2 x - 5 } \mathrm {~d} x\).

2 Show that \(\int _ { 0 } ^ { 6 } \frac { 1 } { x + 2 } \mathrm {~d} x = 2 \ln 2\).

6

1. Find \(\int \frac { 4 + \mathrm { e } ^ { x } } { 2 \mathrm { e } ^ { 2 x } } \mathrm {~d} x\).
2. Without using a calculator, find \(\int _ { 2 } ^ { 10 } \frac { 1 } { 2 x + 5 } \mathrm {~d} x\), giving your answer in the form \(\ln k\).
3. 
   \includegraphics[max width=\textwidth, alt={}, center]{a07e6d2f-ded1-4c62-957b-41fb94b46a2d-3_446_755_580_735} The diagram shows the curve \(y = \log _ { 10 } ( x + 2 )\) for \(0 \leqslant x \leqslant 6\). The region bounded by the curve and the lines \(x = 0 , x = 6\) and \(y = 0\) is denoted by \(R\). Use the trapezium rule with 2 strips to find an estimate of the area of \(R\), giving your answer correct to 1 decimal place.

6

1. Find \(\int \frac { 4 + \mathrm { e } ^ { x } } { 2 \mathrm { e } ^ { 2 x } } \mathrm {~d} x\).
2. Without using a calculator, find \(\int _ { 2 } ^ { 10 } \frac { 1 } { 2 x + 5 } \mathrm {~d} x\), giving your answer in the form \(\ln k\).
3. 
   \includegraphics[max width=\textwidth, alt={}, center]{f85c4010-17b1-441c-ae8a-e77573d1b0c3-3_446_755_580_735} The diagram shows the curve \(y = \log _ { 10 } ( x + 2 )\) for \(0 \leqslant x \leqslant 6\). The region bounded by the curve and the lines \(x = 0 , x = 6\) and \(y = 0\) is denoted by \(R\). Use the trapezium rule with 2 strips to find an estimate of the area of \(R\), giving your answer correct to 1 decimal place.

7

1. Find \(\int ( 2 \cos \theta - 3 ) ( \cos \theta + 1 ) \mathrm { d } \theta\).
2. 1. Find \(\int \left( \frac { 4 } { 2 x + 1 } + \frac { 1 } { 2 x } \right) \mathrm { d } x\).
   2. Hence find \(\int _ { 1 } ^ { 4 } \left( \frac { 4 } { 2 x + 1 } + \frac { 1 } { 2 x } \right) \mathrm { d } x\), giving your answer in the form \(\ln k\).

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1. Show that \(\int _ { 1 } ^ { 4 } \left( \frac { 2 } { x } + \frac { 2 } { 2 x + 1 } \right) \mathrm { d } x = \ln 48\).
2. Find \(\int \sin 2 x ( \cot x + 2 \operatorname { cosec } x ) \mathrm { d } x\).

1 Show that $$\int _ { 1 } ^ { 4 } \frac { 1 } { 2 x + 1 } \mathrm {~d} x = \frac { 1 } { 2 } \ln 3$$

5 Show that \(\int _ { 1 } ^ { 2 } \left( \frac { 1 } { x } - \frac { 4 } { 2 x + 1 } \right) \mathrm { d } x = \ln \frac { 18 } { 25 }\).

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1. Find the exact value of \(\int _ { 0 } ^ { \frac { 1 } { 3 } \pi } \left( \sin 2 x + \sec ^ { 2 } x \right) \mathrm { d } x\).
2. Show that \(\int _ { 1 } ^ { 4 } \left( \frac { 1 } { 2 x } + \frac { 1 } { x + 1 } \right) \mathrm { d } x = \ln 5\).

2 Show that \(\int _ { 2 } ^ { 6 } \frac { 2 } { 4 x + 1 } \mathrm {~d} x = \ln \frac { 5 } { 3 }\).

6

1. Find \(\int 4 \mathrm { e } ^ { - \frac { 1 } { 2 } x } \mathrm {~d} x\).
2. Show that \(\int _ { 1 } ^ { 3 } \frac { 6 } { 3 x - 1 } \mathrm {~d} x = \ln 16\).

1

1. Find \(\int \frac { 2 } { 4 x - 1 } \mathrm {~d} x\).
2. Hence find \(\int _ { 1 } ^ { 7 } \frac { 2 } { 4 x - 1 } \mathrm {~d} x\), expressing your answer in the form \(\ln a\), where \(a\) is an integer.

1 Find the exact value of \(\int _ { - 1 } ^ { 35 } \frac { 3 } { 2 x + 5 } \mathrm {~d} x\), giving the answer in the form \(\ln k\).

2 Show that \(\int _ { 1 } ^ { 7 } \frac { 6 } { 2 x + 1 } \mathrm {~d} x = \ln 125\).

1 Show that \(\int _ { 2 } ^ { 8 } \frac { 3 } { x } \mathrm {~d} x = \ln 64\).

1. Show that

$$\int _ { 1 } ^ { 7 } \frac { 2 } { 4 x - 1 } \mathrm {~d} x = \ln 3$$

2 Fig. 7 shows the curve \(y = \frac { x ^ { 2 } } { 1 + 2 x ^ { 3 } }\). It is undefined at \(x = a\); the line \(x = a\) is a vertical asymptote. \begin{figure}[h] \end{figure}

1. Calculate the value of \(a\), giving your answer correct to 3 significant figures.
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 x - 2 x ^ { 4 } } { \left( 1 + 2 x ^ { 3 } \right) ^ { 2 } }\). Hence determine the coordinates of the turning points of the curve.
3. Show that the area of the region between the curve and the \(x\)-axis from \(x = 0\) to \(x = 1\) is \(\frac { 1 } { 6 } \ln 3\).

1 Show that \(\int _ { \sqrt { 2 } } ^ { \sqrt { 6 } } \frac { 2 } { x } \mathrm {~d} x = \ln 3\).

7. Given that \(k \in \mathbb { Z } ^ { + }\)
a. show that \(\int _ { 2 k } ^ { 3 k } \frac { 6 } { ( 7 k - 2 x ) } \mathrm { d } x\) is independent of \(k\),
b. show that \(\int _ { k } ^ { 2 k } \frac { 2 } { 3 ( 2 x - k ) ^ { 2 } } \mathrm {~d} x\) is inversely proportional to \(k\).

1. (a) Express \(\lim _ { \delta x \rightarrow 0 } \sum _ { x = 2.1 } ^ { 6.3 } \frac { 2 } { x } \delta x\) as an integral.
   (b) Hence show that

$$\lim _ { \delta x \rightarrow 0 } \sum _ { x = 2.1 } ^ { 6.3 } \frac { 2 } { x } \delta x = \ln k$$ where \(k\) is a constant to be found.

11 Given that \(k\) is a positive constant, show that \(\int _ { k } ^ { 2 k } \frac { 2 } { ( 2 x + k ) ^ { 2 } } d x\) is inversely proportional to \(k\).

10 In this question you must show detailed reasoning.
Show that \(\int _ { \mathrm { e } } ^ { \pi } \frac { 1 } { x } \mathrm {~d} x = \ln \pi - 1\) as given in line 37.

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- 1 \end{array} \right) + \mu \left( \begin{array} { c } 1
- 2
4 \end{array} \right) .$$ (ii) Find the point of intersection of \(l _ { 1 }\) and \(l _ { 2 }\).
(iii) Find the acute angle between \(l _ { 1 }\) and \(l _ { 2 }\). 2 In this question you must show detailed reasoning.
(i) Find \(\int _ { \frac { 1 } { 4 } \pi } ^ { \frac { 1 } { 3 } \pi } 2 \tan x \mathrm {~d} x\) giving your answer in the form \(\ln p\).
(ii) Show that \(\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } 2 \tan x \mathrm {~d} x\) is undefined explaining your reasoning.

6 Four students, Tom, Josh, Floella and Georgia are attempting to complete the indefinite integral $$\int \frac { 1 } { x } \mathrm {~d} x \quad \text { for } x > 0$$ Each of the students' solutions is shown below: $$\begin{array} { l l } \text { Tom } & \int \frac { 1 } { x } \mathrm {~d} x = \ln x \\ \text { Josh } & \int \frac { 1 } { x } \mathrm {~d} x = k \ln x \\ \text { Floella } & \int \frac { 1 } { x } \mathrm {~d} x = \ln A x \\ \text { Georgia } & \int \frac { 1 } { x } \mathrm {~d} x = \ln x + c \end{array}$$ 6

1. 1. Explain what is wrong with Tom's answer. 6
2. (ii) Explain what is wrong with Josh's answer.
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3. Explain why Floella and Georgia's answers are equivalent.