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Naing Wai Yan Tun's Maths

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Naing Wai Yan Tun's Maths

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2024 Sample Question Set I

2024 SAMPLE QUESTION SET I
MATRICULATION EXAMINATION
DEPARTMENT OF MYANMAR EXAMINATIONS
MATHEMATICS Time: 3 hours

Answer ALL Questions. Write your answers in the answer booklet.

Section A (Each question carries 1 mark.)

Choose the correct or the most appropriate answer for each question. Write the letter of the correct or the most appropriate answer.

  1. $x+3+(y-2)i = 5+2i$, find the values of $x$ and $y$.
    A. $x=8$ and $y=4$ B. $x=2$ and $y=4$ C. $x=2$ and $y=0$ D. $x=8$ and $y=0$
  2. The Cartesian form of the equation of the plane $\vec{r}(2\hat{i}+3\hat{j}-\hat{k})=10$ is
    A. $2x+3y-z=10$ B. $2x+3y-z=\sqrt{14}$ C. $2x+3y-z=-\sqrt{14}$ D. $2x+3y+z+10=0$
  3. After a sport tournament, each player shakes hands with every other player once. If there are 36 handshakes in total, the number of players at the tournament is
    A. 18 B. 8 C. 10 D. 9
  4. For $y^2=-4px$, the directrix is
    A. $x=-p$ B. $x=p$ C. $y=-p$ D. $y=p$
  5. The period of the function $y = 3\cos \pi(x+1)+2$ is
    A. 1 B. 2 C. 3 D. $\pi$
  6. $x$-axis reflection for the graph of $y=\log_b x$ is
    A. $y=\log_b x$ B. $\log_{\frac{1}{b}} x$ C. $y=\log_{10}x$ D. $y=\log_b 2$
  7. The range of $y=-2e^{-x+1}+3$ is
    A. $\{3\}$ B. $\{y:y>3\}$ C. $\{y:y<3\}$ D. $\{y:y\ge3\}$
  8. How many inflection points are there in the graph of $x^4+2x^2+5$?
    A. 0 B. 1 C. 2 D. 3
  9. $\int 2^{3x+1}dx =$
    A. $\frac{1}{\ln 2} 2^{3x+1}+C$ B. $\frac{1}{3} 2^{3x+1}+C$ C. $\frac{1}{3 \ln 2} 2^{3x+1}+C$ D. $\frac{1}{6 \ln 2}2^{3x+1}+C$
  10. Area of the region bounded by the curve $y=\cos x$ between $x=0$ and $x=\pi$ is
    A. 2 sq. units B. 4 sq. units C. 3 sq. units D. 1 sq. units
Section B (Each question carries 2 marks.)

Write only the solution of each question. (There is no need to show your working.)

  1. Simplify $\frac{\overline{2+3i}}{\overline{-4-5i}}$.
  2. Find the unit vector of $-2\hat{i}+3\hat{j}-7\hat{k}$.
  3. In how many ways can a president, a treasurer and a secretary for a committee be selected from a group of 15 people?
  4. Find the center and radius of the circle $x^2-2x+y^2+4y-4=0$.
  5. If the point $(x,y)$ is on the graph of $y=\cos x$, then find the respective point on the graph of $y=f(3x-3)+2$.
  6. Points $(0,1)$ and $(1,b)$ are on the graph of $y=b^x$. Find the corresponding points on the graphs of $y=ab^x$ and $y=ab^{x-h}+k$.
  7. Determine the open intervals on which the graph $\frac{x}{x^2+1}$ is concave up or concave down.
  8. Evaluate $\int (2x+1)\cos x \,dx$.
  9. Find $\int \left(-\frac{2}{x}+3e^{x}\right) dx$.
  10. Find the shaded area.
    0 -2 2 4 x y f(x) = 4 - x2
Section C (Each question carries 3 marks.)
  1. Find the cube roots of $z=2-2i$.
  2. Use the mathematical induction principle to prove that $1+3+3^2+\dots+3^{n-1} = \frac{3^n-1}{2}$.
  3. Let $\vec{p} = \begin{pmatrix} 1 \\ -1 \\ 4 \end{pmatrix}$ and $\vec{q} = \begin{pmatrix} -2 \\ 0 \\ 2 \end{pmatrix}$. Find
    (a) $\frac{3}{2}\vec{p}-\vec{q}$ (b) $\vec{p}\cdot\vec{q}$ (c) $\vec{p}\times\vec{q}$
  4. A group of 15 friends go on a trip in 3 cars, which respectively can contain 6, 5 and 4 people (each a driver). The owners of cars are members of the group and want to drive their own car. In how many ways can the remaining 12 members be divided between 3 cars?
  5. Find the new coordinates of the point $(1,2)$ if the coordinates axes are rotated angle of $\theta=30^\circ$.
  6. Show that $y=a\sin(bx)$ is an odd function.
  7. Draw the graph of $y=2\log_2 x$.
  8. Find the range of $f(x)=x-e^x$.
  9. Evaluate $\int (xe^x+2\sin x)dx$.
  10. Find the area of the region enclosed by $y=x^2+1$, the x-axis, $x=1$ and $x=2$.
Section D (Each question carries 5 marks.)
  1. Let $z=-\sqrt{3}-i$. Using trigonometric form of $z$, find $z^{-1}$. Use your answer to show that $z^2(z^{-1})^2=1$.
  2. Use the mathematical induction principle to prove that $4n<2^n$ for all-natural numbers $n\ge5$.
  3. The points A(3, 1, 2), B$(-1, 1, 5)$ and C(7, 2, 3) are vertices of a parallelogram ABCD.
    1. Find the coordinates of D.
    2. Calculate the area of the parallelogram.
  4. From the letters of the word CELESTIAL, 7 letters are to be chosen and arrange them in a line. In how many ways can it be done
    1. if there is no other restriction?
    2. if there is at least one E in the arrangements?
    3. if there is at least one E and at most one L in the arrangements?
  5. Write the standard form and sketch the graph of $x^2-2x+8y-23=0$, showing the vertex, focus, directrix and end points of the latus rectum.
  6. Draw the graph of $y=3\sin 2(x-1)+4$.
  7. Differentiate the following functions with respect to $x$.
    (a) $\tan 3x+e^{7-2x^2}$ (b) $\frac{\ln 7x}{\sin (x^2+5)}$ (c) $\frac{x^2}{\log_{10} x}$
  8. Find the values of $\int_2^6 \left(\frac{1}{x} + \frac{4}{2x} + \frac{3}{3x-2}\right)dx$ and give your answer in the form $p \ln p + q \ln q$, where $p$ and $q$ are prime numbers to be formed.
*****************************************
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သင်္ချာမေးခွန်းလွှာဆိုင်ရာ သိကောင်းစရာများ

GRADE 12 MATHEMATICS

QUESTION FORMAT

သင်္ချာမေးခွန်းလွှာဆိုင်ရာသိကောင်းစရာများ

သင်္ချာမေးခွန်းလွှာသည် ဖြေဆိုချိန် (၃)နာရီဖြစ်ပြီး Section A, Section B, Section C နှင့် Section D အပိုင်း ၄ ပိုင်း ပါရှိပါသည်။

  • Section A
    မေးခွန်းနံပါတ် 1 မှ 10 အထိ ဖြစ်ပါသည်။ တစ်ပုဒ် ၁ မှတ် ဖြစ်ပါသည်။ တစ်ပုဒ်လျှင် A, B, C, D တို့မှ တစ်ခုရွေးချယ်ရမည့် ဓမ္မဓိဋ္ဌာန် ပုံစံဖြစ်ပါသည်။
  • Section B
    မေးခွန်းနံပါတ် 11 မှ 20 အထိ ဖြစ်ပါသည်။ တစ်ပုဒ် ၂ မှတ် ဖြစ်ပါသည်။ တွက်ချက်မှု အဆင့်ဆင့်ကို ရေးပြရန်မလိုအပ်ဘဲ အဖြေကိုသာရေးရမည့် ပုံစံဖြစ်ပါသည်။
  • Section C
    မေးခွန်းနံပါတ် 21 မှ 30 အထိ ဖြစ်ပါသည်။ တစ်ပုဒ် ၃ မှတ် ဖြစ်ပါသည်။ တွက်ချက်မှု အဆင့်ဆင့်ကို စနစ်တကျရေးပြရန် လိုအပ်ပါသည်။
  • Section D
    မေးခွန်းနံပါတ် 31 မှ 38 အထိ ဖြစ်ပါသည်။ တစ်ပုဒ် ၅ မှတ် ဖြစ်ပါသည်။ တွက်ချက်မှု အဆင့်ဆင့်ကို စနစ်တကျရေးပြရန် လိုအပ်ပါသည်။

မေးခွန်းလွှာတွင် အခန်းအားလုံးပါဝင်မည်။

  • အခန်း ၁၊ ၂၊ ၃၊ ၄၊ ၅၊ ၆ တို့မှ ၅၅ မှတ် ဖြစ်ပါသည်။
  • အခန်း ၇ (Section 7.1, Section 7.2, Section 7.3, Section 7.4) နှင့် အခန်း ၈ (Section 8.1, Section 8.3) တို့မှ ၁၅ မှတ် ဖြစ်ပါသည်။
  • အခန်း ၇ (Section 7.5)၊ အခန်း ၈ (Section 8.2, Section 8.4) နှင့် အခန်း ၉၊ အခန်း ၁၀၊ အခန်း ၁၁ တို့မှ အမှတ် ၃၀ မှတ် ဖြစ်ပါသည်။

မေးခွန်းအားလုံး ဖြေဆိုရမည့် Answer ALL Questions ဖြစ်ပြီး စဉ်းစားတွက်ချက် ဖြေဆိုရမည့် ပုစ္ဆာအမျိုးအစားများ ဖြစ်ပါသည်။ မေးခွန်းလွှာ၏ အခန်းအလိုက် အမှတ်ပေးမှုဇယားကို ဖော်ပြထားပါသည်။

Chapter Section A Section B Section C Section D Total
1123511
2--358
3+4123+3514
5123511
6123511
7-12-58
8-1+123-7
9+123511
1012+23-8
11123511
Total10203040100

သင်္ချာမေးခွန်းနံပါတ်အလိုက် အခန်းပါဝင်မှုများ

2025 MATRICULATION EXAMINATION
DEPARTMENT OF MYANMAR EXAMINATION
MATHEMATICS

Time: 3 hours
Answer ALL Questions. Write your answers in the answer booklet.

Section A (Each question carries 1 mark.)

Choose the correct or the most appropriate answer for each question. Write the letter of the correct or the most appropriate answer.

  • 1. Chapter 1
  • 2. Chapter 3, Chapter 4
  • 3. Chapter 5
  • 4. Chapter 6
  • 5. Chapter 7 Section 7.1, 7.2, 7.3, 7.4
  • 6. Chapter 8 Section 8.1
  • 7. Chapter 8 Section 8.3
  • 8. Chapter 9, Chapter 7 Section 7.5, Chapter 8 Section 8.2, 8.4
  • 9. Chapter 10
  • 10. Chapter 11

Section B (Each question carries 2 marks.)

Write only the solution of each question. (There is no need to show your working.)

  • 11. Chapter 1
  • 12. Chapter 3
  • 13. Chapter 4
  • 14. Chapter 5
  • 15. Chapter 6
  • 16. Chapter 7 Section 7.1, 7.2, 7.3, 7.4
  • 17. Chapter 8 Section 8.1, Section 8.3
  • 18. Chapter 9, Chapter 7 Section 7.5, Chapter 8 Section 8.2, 8.4
  • 19. Chapter 10
  • 20. Chapter 11

Section C (Each question carries 3 marks.)

  • 21. Chapter 1
  • 22. Chapter 2
  • 23. Chapter 3, Chapter 4
  • 24. Chapter 3, Chapter 4
  • 25. Grade 12 Chapter 5
  • 26. Grade 12 Chapter 6
  • 27. Grade 12 Chapter 8 Section 8.1, Section 8.3
  • 28. Grade 12 Chapter 9, Chapter 7 Section 7.5, Chapter 8 Section 8.2, 8.4
  • 29. Grade 12 Chapter 10
  • 30. Grade 12 Chapter 11

Section D (Each question carries 5 marks.)

  • 31. Grade 12 Chapter 1
  • 32. Grade 12 Chapter 2
  • 33. Chapter 3, Chapter 4
  • 34. Grade 12 Chapter 5
  • 35. Grade 12 Chapter 6
  • 36. Grade 12 Chapter 7 Section 7.1, 7.2, 7.3, 7.4
  • 37. Grade 12 Chapter 9, Chapter 7 Section 7.5, Chapter 8 Section 8.2, 8.4
  • 38. Grade 12 Chapter 11
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Measure of Variation

Chapter 6 Statistics - Section 6.1 Measure of Variation

6.1 Measure of Variation

Measures of centre of a data set

These measures include mean, median, and mode.

For a set of data, the mean is defined by the sum of the values, divided by the number of values.

$ \bar{x} = \frac{x_1 + x_2 + ... + x_n}{n} = \frac{1}{n}\sum_{i=1}^{n} x_i $

The median is the middle point of the data arranged in ascending order.

The mode is the value that occurs most often in a set of data.

The simplest measure of spread is the range R which is the difference between the largest value and the smallest value.

$ R = \text{largest value} - \text{smallest value} $

The data variation is on the difference of each data value from the mean. This difference is called a deviation, in symbol $x_i - \bar{x}$.

The variance $\sigma^2$ is the mean of squares of the deviations.

$ \sigma^2 = \frac{1}{n}\sum_{i=1}^{n} (x_i - \bar{x})^2 $

The standard deviation $\sigma$ is the square root of the variance.

$ \sigma = \sqrt{\text{variance}} $

We can compute the similar formula for the variance as follow:

$ \sigma^2 = \frac{\sum x^2}{n} - (\bar{x})^2 $

Quartiles $Q_1, Q_2$ and $Q_3$ divide the data arranged in ascending order into four equal groups. Here, $Q_2$ is the median of the whole data set. $Q_1$ and $Q_3$ are called the lower quartile and the upper quartile. The interquartile range is a measure of variability.

$ \text{interquartile range} = \text{upper quartile} - \text{lower quartile} = Q_3 - Q_1 $

From the frequency table, we can find the mean:

$ \bar{x} = \frac{\sum_{i=1}^{n} f_i x_i}{\sum_{i=1}^{n} f_i} $

We have the similar formula for the variance of group data as follow:

$ \sigma^2 = \frac{\sum f_i x_i^2}{\sum f_i} - (\bar{x})^2 $

Examples

Example 1

The marks scored in a test by seven students are 3, 4, 5, 2, 8, 8, 5. Find the mean and standard deviation.

Step 1: Calculate the Mean ($\bar{x}$)

$ \bar{x} = \frac{3+4+5+2+8+8+5}{7} = \frac{35}{7} = 5 $

Step 2: Calculate the Variance ($\sigma^2$) using Computational Formula

$\sum x^2 = 3^2+4^2+5^2+2^2+8^2+8^2+5^2 = 9+16+25+4+64+64+25 = 207$

$\sigma^2 = \frac{\sum x^2}{n} - (\bar{x})^2 = \frac{207}{7} - (5)^2 = 29.5714 - 25 = 4.5714$

Step 3: Calculate the Standard Deviation ($\sigma$)

$ \sigma = \sqrt{4.5714} = 2.1381 $

Example 2

Ten students are weighted (w kg). The summary data for the weights are $\sum w = 440$, $\sum w^2 = 19490$. Find the mean and standard deviation of the students' weight.

Step 1: Calculate the Mean ($\bar{w}$)

$ \bar{w} = \frac{\sum w}{n} = \frac{440}{10} = 44 $

Step 2: Calculate the Variance ($\sigma^2$)

$ \sigma^2 = \frac{\sum w^2}{n} - (\bar{w})^2 = \frac{19490}{10} - (44)^2 = 1949 - 1936 = 13 $

Step 3: Calculate the Standard Deviation ($\sigma$)

$ \sigma = \sqrt{13} = 3.6056 $

Example 3

Find the median and interquartile range for each of the set of data below.

(a) 7, 9, 4, 6, 3, 2, 8, 1, 10, 15, 11

Step 1: Arrange data in ascending order.

1, 2, 3, 4, 6, 7, 8, 9, 10, 11, 15

Step 2: Find the Median ($Q_2$).

Median ($Q_2$) = 7 (the 6th value)

Step 3: Find the Lower Quartile ($Q_1$) and Upper Quartile ($Q_3$).

Lower half: [1, 2, 3, 4, 6]. $Q_1 = 3$.

Upper half: [8, 9, 10, 11, 15]. $Q_3 = 10$.

Step 4: Calculate the Interquartile Range (IQR).

IQR = $Q_3 - Q_1 = 10 - 3 = 7$.

(b) 6, 5, 1, 14, 3, 7, 8, 2, 13

Step 1: Arrange data in ascending order.

1, 2, 3, 5, 6, 7, 8, 13, 14

Step 2: Find the Median ($Q_2$).

Median ($Q_2$) = 6 (the 5th value)

Step 3: Find $Q_1$ and $Q_3$.

Lower half: [1, 2, 3, 5]. $Q_1 = \frac{2+3}{2} = 2.5$.

Upper half: [7, 8, 13, 14]. $Q_3 = \frac{8+13}{2} = 10.5$.

Step 4: Calculate the IQR.

IQR = $Q_3 - Q_1 = 10.5 - 2.5 = 8$.

(c) 2, 15, 13, 6, 5, 12, 50, 22, 18, 52

Step 1: Arrange data in ascending order.

2, 5, 6, 12, 13, 15, 18, 22, 50, 52

Step 2: Find the Median ($Q_2$).

Median ($Q_2$) = $\frac{13+15}{2} = 14$.

Step 3: Find $Q_1$ and $Q_3$.

Lower half: [2, 5, 6, 12, 13]. $Q_1 = 6$.

Upper half: [15, 18, 22, 50, 52]. $Q_3 = 22$.

Step 4: Calculate the IQR.

IQR = $Q_3 - Q_1 = 22 - 6 = 16$.

(d) 5, 3, 5, 7, 8, 2, 4, 1, 2, 2, 3, 6

Step 1: Arrange data in ascending order.

1, 2, 2, 2, 3, 3, 4, 5, 5, 6, 7, 8

Step 2: Find the Median ($Q_2$).

Median ($Q_2$) = $\frac{3+4}{2} = 3.5$.

Step 3: Find $Q_1$ and $Q_3$.

Lower half: [1, 2, 2, 2, 3, 3]. $Q_1 = \frac{2+2}{2} = 2$.

Upper half: [4, 5, 5, 6, 7, 8]. $Q_3 = \frac{5+6}{2} = 5.5$.

Step 4: Calculate the IQR.

IQR = $Q_3 - Q_1 = 5.5 - 2 = 3.5$.

Example 4

Find the mean and standard deviation for the following frequency table:

x56789
Frequency(f)99101210

Step 1: Construct the calculation table

$x$$f$$f \cdot x$$x^2$$f \cdot x^2$
594525225
695436324
7107049490
8129664768
9109081810
$\sum f = 50$$\sum fx = 355$$\sum fx^2 = 2617$

Step 2: Calculate Mean, Variance, and Standard Deviation

Mean: $\bar{x} = \frac{\sum fx}{\sum f} = \frac{355}{50} = 7.1000$

Variance: $\sigma^2 = \frac{\sum fx^2}{\sum f} - (\bar{x})^2 = \frac{2617}{50} - (7.1)^2 = 52.34 - 50.41 = 1.9300$

Standard Deviation: $\sigma = \sqrt{1.93} = 1.3892$

Example 5

A man recorded the length, in minutes, of each telephone call he made for a month. Find the mean and standard deviation for the data.

Length (minute)(0, 10](10, 20](20, 30](30, 40](40, 50]
Frequency (f)41551210

Step 1: Construct the calculation table with midpoints

LengthMidpoint (x)f$f \cdot x$$x^2$$f \cdot x^2$
(0, 10]542025100
(10, 20]15152252253375
(20, 30]2551256253125
(30, 40]3512420122514700
(40, 50]4510450202520250
$\sum f = 46$$\sum fx = 1240$$\sum fx^2 = 41150$

Step 2: Calculate Mean, Variance, and Standard Deviation

Mean: $\bar{x} = \frac{\sum fx}{\sum f} = \frac{1240}{46} = 26.9565$

Variance: $\sigma^2 = \frac{\sum fx^2}{\sum f} - (\bar{x})^2 = \frac{41150}{46} - (\frac{1240}{46})^2 = 176.6068$

Standard Deviation: $\sigma = \sqrt{176.6068} = 13.2893$

Exercises 6.1

1. Find the median and the interquartile range for the following sets of data:

(a) 16, 18, 22, 19, 3, 21, 17, 20

Sorted Data: 3, 16, 17, 18, 19, 20, 21, 22

Median ($Q_2$) = $\frac{18+19}{2} = 18.5$.

Lower half: [3, 16, 17, 18]. $Q_1 = \frac{16+17}{2} = 16.5000$.

Upper half: [19, 20, 21, 22]. $Q_3 = \frac{20+21}{2} = 20.5$.

IQR = $20.5 - 16.5 = 4$.

(b) 33, 38, 43, 30, 29, 40, 51

Sorted Data: 29, 30, 33, 38, 40, 43, 51

Median ($Q_2$) = 38 (the 4th value).

Lower half: [29, 30, 33]. $Q_1 = 30$.

Upper half: [40, 43, 51]. $Q_3 = 43$.

IQR = $43 - 30 = 13$.

(c) 24, 32, 54, 31, 16, 18, 19, 14, 17, 20

Sorted Data: 14, 16, 17, 18, 19, 20, 24, 31, 32, 54

Median ($Q_2$) = $\frac{19+20}{2} = 19.5$.

Lower half: [14, 16, 17, 18, 19]. $Q_1 = 17$.

Upper half: [20, 24, 31, 32, 54]. $Q_3 = 31$.

IQR = $31 - 17 = 14$.

(d) 14, 16, 27, 18, 13, 19, 36, 15, 20

Sorted Data: 13, 14, 15, 16, 18, 19, 20, 27, 36

Median ($Q_2$) = 18 (the 5th value).

Lower half: [13, 14, 15, 16]. $Q_1 = \frac{14+15}{2} = 14.5$.

Upper half: [19, 20, 27, 36]. $Q_3 = \frac{20+27}{2} = 23.5$.

IQR = $23.5 - 14.5 = 9$.

2. Fifteen students do a mathematics test. Their marks are as follows: 7, 4, 9, 7, 6, 10, 12, 11, 3, 8, 5, 9, 8, 7, 3. Find the median and the interquartile range.

Step 1: Arrange data in ascending order.

3, 3, 4, 5, 6, 7, 7, 7, 8, 8, 9, 9, 10, 11, 12

Step 2: Find the Median ($Q_2$).

Median ($Q_2$) = 7 (the 8th value)

Step 3: Find $Q_1$ and $Q_3$.

Lower half: [3, 3, 4, 5, 6, 7, 7]. $Q_1 = 5$.

Upper half: [8, 8, 9, 9, 10, 11, 12]. $Q_3 = 9$.

Step 4: Calculate the IQR.

IQR = $Q_3 - Q_1 = 9 - 5 = 4$.

3. Find the mean, variance and the standard deviation of the following data sets.

(a) 1, 2, 3, 4, 5, 6, 7

Mean: $\bar{x} = \frac{1+2+3+4+5+6+7}{7} = \frac{28}{7} = 4.0000$

$\sum x^2 = 1^2+2^2+3^2+4^2+5^2+6^2+7^2 = 1+4+9+16+25+36+49=140$

Variance: $\sigma^2 = \frac{140}{7} - (4)^2 = 20 - 16 = 4$

Standard Deviation: $\sigma = \sqrt{4} = 2$

(b) 5, 6, 7, 3, 2, 9, 11

Mean: $\bar{x} = \frac{5+6+7+3+2+9+11}{7} = \frac{43}{7} \approx 6.1429$

$\sum x^2 = 5^2+6^2+7^2+3^2+2^2+9^2+11^2 = 25+36+49+9+4+81+121=325$

Variance: $\sigma^2 = \frac{325}{7} - (\frac{43}{7})^2 \approx 46.4286 - 37.7347 = 8.6939$

Standard Deviation: $\sigma = \sqrt{8.6939} = 2.9485$

(c) 4, 3, 2, 7, 0, 9

Mean: $\bar{x} = \frac{4+3+2+7+0+9}{6} = \frac{25}{6} \approx 4.1667$

$\sum x^2 = 4^2+3^2+2^2+7^2+0^2+9^2 = 16+9+4+49+0+81=159$

Variance: $\sigma^2 = \frac{159}{6} - (\frac{25}{6})^2 \approx 26.5 - 17.3611 = 9.1389$

Standard Deviation: $\sigma = \sqrt{9.1389} = 3.0231$

(d) 5, 5, 5, 4, 2, 7, 7, 7, 7

Mean: $\bar{x} = \frac{5(3)+4+2+7(4)}{9} = \frac{15+4+2+28}{9} = \frac{49}{9} \approx 5.4444$

$\sum x^2 = 5^2(3)+4^2+2^2+7^2(4) = 25(3)+16+4+49(4) = 75+16+4+196 = 291$

Variance: $\sigma^2 = \frac{291}{9} - (\frac{49}{9})^2 \approx 32.3333 - 29.6420 = 2.6913$

Standard Deviation: $\sigma = \sqrt{2.6913} = 1.6405$

(e) 1, 3, 5, 5, 7, 9, 10, 15, 17

Mean: $\bar{x} = \frac{1+3+5+5+7+9+10+15+17}{9} = \frac{72}{9} = 8$

$\sum x^2 = 1^2+3^2+5^2+5^2+7^2+9^2+10^2+15^2+17^2 = 1+9+25+25+49+81+100+225+289=804$

Variance: $\sigma^2 = \frac{804}{9} - (8)^2 = 89.3333 - 64 = 25.3333$

Standard Deviation: $\sigma = \sqrt{25.3333} = 5.0332$

4. The numbers of errors, x, on each of 200 pages of typescript was monitored. The results when summarized showed that $\sum x = 1000, \sum x^2 = 5500$. Calculate the mean and the standard deviation of the number of errors per page.

Step 1: Calculate the Mean ($\bar{x}$)

$\bar{x} = \frac{\sum x}{n} = \frac{1000}{200} = 5$

Step 2: Calculate the Variance ($\sigma^2$)

$\sigma^2 = \frac{\sum x^2}{n} - (\bar{x})^2 = \frac{5500}{200} - (5)^2 = 27.5 - 25 = 2.5$

Step 3: Calculate the Standard Deviation ($\sigma$)

$\sigma = \sqrt{2.5} = 1.5811$

5. In a student group, a record was kept for the number of absent days each student had over one particular term. Calculate the mean and standard deviation for these data shown in the table below.

Number of absent days (x)01234
Number of students (f)15102082

Step 1: Construct the calculation table

$x$$f$$f \cdot x$$x^2$$f \cdot x^2$
015000
11010110
22040480
3824972
4281632
$\sum f = 55$$\sum fx = 82$$\sum fx^2 = 194$

Step 2: Calculate Mean, Variance, and Standard Deviation

Mean: $\bar{x} = \frac{\sum fx}{\sum f} = \frac{82}{55} = 1.4909$

Variance: $\sigma^2 = \frac{\sum fx^2}{\sum f} - (\bar{x})^2 = \frac{194}{55} - (1.4909)^2 = 3.5273 - 2.2228 = 1.3045$

Standard Deviation: $\sigma = \sqrt{1.3045} = 1.1421$

6. A moth trap was set every night for five weeks. The number of moths caught in the trap was recorded. Calculate the mean and standard deviation for these data shown in the table below.

Number of moths (x)7891011
Frequency (f)38959

Step 1: Construct the calculation table

$x$$f$$f \cdot x$$x^2$$f \cdot x^2$
732149147
886464512
998181729
10550100500
119991211089
$ \sum f = 34$$\sum fx = 315$$\sum fx^2=2977$

Step 2: Calculate Mean, Variance, and Standard Deviation

Mean: $\bar{x} = \frac{315}{34} = 9.2647$

Variance: $\sigma^2 = \frac{2977}{34} - (9.2647)^2 = 87.5588 - 85.8259 = 1.7329$

Standard Deviation: $\sigma = \sqrt{1.7329} = 1.3164$

7. The lifetimes of 80 batteries, to the nearest hour (h), is shown in the following table. Calculate the mean and standard deviation for these data.

Lifetime (h)(6, 10](10, 15](15, 20](20, 25](25, 30]
Number of batteries128153015

Step 1: Construct the calculation table

Midpoint(x)f$f \cdot x$$x^2$$f \cdot x^2$
8129664768
12.58100156.251250
17.515262.5306.254593.75
22.530675506.2515187.5
27.515412.5756.2511343.75
$\sum f=80$$\sum fx=1546$$\sum fx^2=33143$

Step 2: Calculate Mean, Variance, and Standard Deviation

Mean: $\bar{x} = \frac{1546}{80} = 19.3250$

Variance: $\sigma^2 = \frac{33143}{80} - (19.3250)^2 = 414.2875 - 373.4556 = 40.8319$

Standard Deviation: $\sigma = \sqrt{40.8319} = 6.39$

8. The table summarizes the distances travelled by 150 students to college each day. Calculate the mean and standard deviation for these data.

Distance (s)(0, 2](2, 4](4, 6](6, 8](8, 10](10, 12]
Number of students82255351812

Step 1: Construct the calculation table

Distance (s)Midpoint(x)f$f \cdot x$$x^2$$f \cdot x^2$
(0, 2]18818
(2, 4]322669198
(4, 6]555275251375
(6, 8]735245491715
(8, 10]918162811458
(10, 12]11121321211452
$\sum f=150$$\sum fx=888$$\sum fx^2=6206$

Step 2: Calculate Mean, Variance, and Standard Deviation

Mean: $\bar{x} = \frac{\sum fx}{\sum f} = \frac{888}{150} = 5.92$

Variance: $\sigma^2 = \frac{\sum fx^2}{\sum f} - (\bar{x})^2 = \frac{6206}{150} - (5.92)^2 = 41.3733 - 35.0464 = 6.3269$

Standard Deviation: $\sigma = \sqrt{6.3269} = 2.5153$

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Exercise 10.4

Calculus - Exercise 10.4

Exercise 10.4

1. Use the partial fractions method to evaluate the following integrals.

(a) $\int \frac{1}{2x^2+5x+3}dx$

Solution:

$$ \frac{1}{2x^2+5x+3} = \frac{1}{(2x+3)(x+1)} = \frac{A}{2x+3} + \frac{B}{x+1} $$

$$ 1 = A(x+1) + B(2x+3) = (A+2B)x + (A+3B) $$

By equating the coefficients of corresponding powers of x,

$$ A+2B=0 \tag{1} $$ $$ A+3B=1 \tag{2} $$

If eq(2)-eq(1), $B=1$.

If $B=1$ in eq(1), $A=-2$.

$$ \begin{aligned} \text{Therefore, } \int \frac{1}{(2x+3)(x+1)}dx &= \int (\frac{-2}{2x+3} + \frac{1}{x+1})dx \\ &= \ln|x+1| - \ln|2x+3| + C \\ &= \ln|\frac{x+1}{2x+3}| + C \end{aligned} $$

(b) $\int \frac{2x-1}{(x-3)^2}dx$

Solution:

$$ \frac{2x-1}{(x-3)^2} = \frac{A}{x-3} + \frac{B}{(x-3)^2} $$

$$ 2x-1 = A(x-3) + B = Ax - (3A-B) $$

By equating the coefficients of corresponding powers of x, $A=2$

$$ 3A-B=1 \tag{1} $$

If $A=2$ in eq (1), $B=5$.

$$ \begin{aligned} \text{Therefore, } \int \frac{2x-1}{(x-3)^2}dx &= \int (\frac{2}{x-3} + \frac{5}{(x-3)^2})dx \\ &= 2 \int \frac{1}{x-3}dx + 5 \int(x-3)^{-2}dx \\ &= 2\ln|x-3| - \frac{5}{x-3} + C \end{aligned} $$

(c) $\int \frac{x+1}{(2x+5)(x+4)}dx$

Solution:

$$ \frac{x+1}{(2x+5)(x+4)} = \frac{A}{2x+5} + \frac{B}{x+4} $$

$$ x+1 = A(x+4) + B(2x+5) = (A+2B)x + (4A+5B) $$

By equating coefficients:

$$ A+2B=1 \tag{1} $$ $$ 4A+5B=1 \tag{2} $$

If eq(2)-eq(1)x4, $B=-1$. If $B=-1$ in eq(1), $A=1$.

$$ \begin{aligned} \text{Therefore, } \int \frac{x+1}{(2x+5)(x+4)}dx &= \int (\frac{1}{2x+5} - \frac{1}{x+4})dx \\ &= \frac{1}{2}\ln|2x+5| - \ln|x+4| + C \\ &= \ln \frac{\sqrt{|2x+5|}}{|x+4|} + C \end{aligned} $$

(d) $\int \frac{2x^2-1}{x^2-1}dx$

Solution:

First, we write $$ \int \frac{2x^2-1}{x^2-1}dx = \int (2 + \frac{1}{x^2-1})dx $$

First, we write $$ \frac{1}{x^2-1} = \frac{1}{(x+1)(x-1)} = \frac{A}{x+1} + \frac{B}{x-1} $$

and so $$ 1 = A(x-1) + B(x+1) $$

When $x=1$, $1=B(2)$, so $B=1/2$.

When $x=-1$, $1=A(-2)$, so $A=-1/2$.

$$ \begin{aligned} \text{Therefore, } \int \frac{2x^2-1}{x^2-1}dx &= \int 2 dx - \int \frac{\frac{1}{2}}{x+1}dx + \int \frac{\frac{1}{2}}{x-1}dx \\ &= 2x - \frac{1}{2}\ln|x+1| + \frac{1}{2}\ln|x-1| + C \\ &= 2x + \frac{1}{2}\ln|\frac{x-1}{x+1}| + C \end{aligned} $$

2. Find the function $f(x)$ that satisfying the equation $f'(x) = \sin 4x \cos 2x$ with $f(\frac{\pi}{2}) = 0$.

Solution:

By integrating, $$ f(x) = \int \sin 4x \cos 2x dx = \frac{1}{2} \int (\sin 6x + \sin 2x) dx $$

$$ = \frac{1}{2} (-\frac{1}{6}\cos 6x - \frac{1}{2}\cos 2x) + C $$

Since $f(\frac{\pi}{2}) = 0$:

$$ \frac{1}{2} (-\frac{1}{6}\cos 6(\frac{\pi}{2}) - \frac{1}{2}\cos 2(\frac{\pi}{2})) + C = 0 $$

$$ -\frac{1}{12}\cos 3\pi - \frac{1}{4}\cos\pi + C = 0 $$

$$ C = \frac{1}{12}\cos 3\pi + \frac{1}{4}\cos\pi = \frac{1}{12}(-1) + \frac{1}{4}(-1) = -\frac{1}{3} $$

Therefore, $$ f(x) = -\frac{1}{12}\cos 6x - \frac{1}{4}\cos 2x - \frac{1}{3} $$

3. Find the function $g(x)$ that satisfying the equation $g'(x) = x^2 e^{x^3}$ with $g(0) = -\frac{2}{3}$.

Solution:

By integrating, $$ g(x) = \int x^2 e^{x^3} dx $$

Let $u = x^3$, so $\frac{1}{3}du = x^2 dx$.

$$ g(x) = \int \frac{1}{3}e^u du = \frac{1}{3} e^u + C = \frac{1}{3} e^{x^3} + C $$

Since $g(0) = -\frac{2}{3}$:

$$ \frac{1}{3} e^{0} + C = -\frac{2}{3} \text{, so } \frac{1}{3} + C = -\frac{2}{3} \text{, so } C = -1 $$

Therefore, $$ g(x) = \frac{1}{3}e^{x^3} - 1 $$

4. Find the function $h(x)$, that satisfying the equation $h'(x) = \frac{x}{x^2-1}$ with $h(2) = \frac{1}{2}$.

Solution:

By integrating, $$ h(x) = \int \frac{x}{x^2-1} dx $$

Let $u = x^2-1$, so $\frac{1}{2}du = x dx$.

$$ h(x) = \int \frac{1}{2u} du = \frac{1}{2}\ln|u| + C = \frac{1}{2}\ln|x^2-1| + C $$

Since $h(2) = \frac{1}{2}$:

$$ \frac{1}{2}\ln|2^2-1| + C = \frac{1}{2} \text{, so } \frac{1}{2}\ln|3| + C = \frac{1}{2} $$

$$ \ln|3| + 2C = 1 \text{, so } C = \frac{1 - \ln 3}{2} $$

Therefore, $$ h(x) = \frac{1}{2}\ln|x^2-1| + \frac{1 - \ln 3}{2} $$

5. Find the function $f(x)$ satisfying the equation $f''(x) = 2x - 1$ with $f(0) = -1$ and $f'(1) = 2$.

Solution:

By integrating, $$ f'(x) = \int (2x-1)dx = x^2 - x + C $$

Since $f'(1) = 2$: $1^2 - 1 + C = 2$, so $C = 2$.

Therefore $f'(x) = x^2 - x + 2$.

By integrating, we get $$ f(x) = \int (x^2 - x + 2)dx = \frac{x^3}{3} - \frac{x^2}{2} + 2x + C $$

Since $f(0) = -1$: $\frac{0}{3} - \frac{0}{2} + 2(0) + C = -1$, so $C = -1$.

Therefore, $$ f(x) = \frac{x^3}{3} - \frac{x^2}{2} + 2x - 1 $$

6. Find the function $g(x)$ that satisfying the equation $g''(x) = x \sin x$ with $g(\frac{\pi}{2}) = 0$ and $g'(0) = 0$.

Solution:

By integrating, $$ g'(x) = \int x \sin x dx $$

Let $u=x$, $dv = \sin x dx$, then $du=dx$, $v=-\cos x$.

$$ g'(x) = -x \cos x - \int (-\cos x) dx = -x \cos x + \sin x + C $$

Since $g'(0) = 0$: $-(0) \cos 0 + \sin 0 + C = 0$, so $C = 0$.

Therefore, $g'(x) = -x \cos x + \sin x$.

By integrating, $$ g(x) = \int (-x \cos x + \sin x)dx = - \int x \cos x dx + \int \sin x dx $$

Let $u=x$, $dv = \cos x dx$, then $du=dx$, $v=\sin x$.

$$ \therefore g(x) = -(x \sin x - \int \sin x dx) + \int \sin x dx = -x \sin x + 2 \int \sin x dx $$

$$ = -x \sin x - 2\cos x + C $$

Since $g(\frac{\pi}{2}) = 0$:

$$ -\frac{\pi}{2}\sin(\frac{\pi}{2}) - 2\cos(\frac{\pi}{2}) + C = 0 \text{, so } -\frac{\pi}{2}(1) - 2(0) + C = 0 \text{, so } C = \frac{\pi}{2} $$

Therefore, $$ g(x) = -x \sin x - 2 \cos x + \frac{\pi}{2} $$

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Exercise 10.3

Exercise 10.3

Use the integration by parts to evaluate the following integrals.

(a) $\int s e^{-2s} ds$

Solution:

Let $u = s, \quad dv = e^{-2s} ds$

$du = ds, \quad v = -\frac{1}{2} e^{-2s}$

$$ \begin{aligned} \int s e^{-2s} ds &= s(-\frac{1}{2} e^{-2s}) - \int (-\frac{1}{2} e^{-2s}) ds \\ &= -\frac{1}{2} s e^{-2s} + \frac{1}{2} \int e^{-2s} ds \\ &= -\frac{1}{2} s e^{-2s} + \frac{1}{2} (-\frac{1}{2}) e^{-2s} + C \\ &= -\frac{1}{2} s e^{-2s} - \frac{1}{4} e^{-2s} + C \end{aligned} $$

(b) $\int \ln(x+1) dx$

Solution:

Let $u = \ln(x+1), \quad dv = dx$

$du = \frac{1}{x+1} dx, \quad v = x$

$$ \begin{aligned} \int \ln(x+1) dx &= \ln(x+1) \cdot x - \int x \cdot \frac{1}{x+1} dx \\ &= x \ln(x+1) - \int \frac{x+1-1}{x+1} dx \\ &= x \ln(x+1) - \int (1 - \frac{1}{x+1}) dx \\ &= x \ln(x+1) - x + \ln|x+1| + C \end{aligned} $$

(c) $\int t \sin 2t dt$

Solution:

Let $u = t, \quad dv = \sin 2t dt$

$du = dt, \quad v = -\frac{1}{2} \cos 2t$

$$ \begin{aligned} \int t \sin 2t dt &= t(-\frac{1}{2} \cos 2t) - \int (-\frac{1}{2} \cos 2t) dt \\ &= -\frac{1}{2} t \cos 2t + \frac{1}{2} \int \cos 2t dt \\ &= -\frac{1}{2} t \cos 2t + \frac{1}{2} \cdot \frac{1}{2} \sin 2t + C \\ &= -\frac{1}{2} t \cos 2t + \frac{1}{4} \sin 2t + C \end{aligned} $$

(d) $\int x 2^x dx$

Solution:

Let $u = x, \quad dv = 2^x dx$

$du = dx, \quad v = \frac{2^x}{\ln 2}$

$$ \begin{aligned} \int x 2^x dx &= x \cdot \frac{2^x}{\ln 2} - \int \frac{2^x}{\ln 2} dx \\ &= \frac{x \cdot 2^x}{\ln 2} - \frac{1}{\ln 2} \int 2^x dx \\ &= \frac{x \cdot 2^x}{\ln 2} - \frac{1}{\ln 2} \cdot \frac{2^x}{\ln 2} + C \\ &= \frac{x \cdot 2^x}{\ln 2} - \frac{2^x}{(\ln 2)^2} + C \end{aligned} $$

(e) $\int x \cos 5x dx$

Solution:

Let $u = x, \quad dv = \cos 5x dx$

$du = dx, \quad v = \frac{1}{5} \sin 5x$

$$ \begin{aligned} \int x \cos 5x dx &= x \cdot \frac{1}{5} \sin 5x - \int \frac{1}{5} \sin 5x dx \\ &= \frac{1}{5} x \sin 5x - \frac{1}{5} \int \sin 5x dx \\ &= \frac{1}{5} x \sin 5x - \frac{1}{5} (-\frac{1}{5} \cos 5x) + C \\ &= \frac{1}{5} x \sin 5x + \frac{1}{25} \cos 5x + C \end{aligned} $$

(f) $\int e^x \cos x dx$

Solution:

Let $u = \cos x, \quad dv = e^x dx$

$du = -\sin x dx, \quad v = e^x$

$$ \int e^x \cos x dx = \cos x \cdot e^x - \int e^x (-\sin x dx) $$ $$ \int e^x \cos x dx = \cos x \cdot e^x + \int e^x \sin x dx \quad (1) $$

For $\int e^x \sin x dx$,

Let $u = \sin x, \quad dv = e^x dx$

$du = \cos x dx, \quad v = e^x$

$$ \int e^x \sin x dx = \sin x \cdot e^x - \int e^x (\cos x dx) $$

By substituting $\int e^x \sin x dx = \sin x \cdot e^x - \int e^x \cos x dx$ in (1), we get

$$ \int e^x \cos x dx = \cos x \cdot e^x + \sin x \cdot e^x - \int e^x \cos x dx $$ $$ 2 \int e^x \cos x dx = \cos x \cdot e^x + \sin x \cdot e^x + C_1 $$ $$ \therefore \int e^x \cos x dx = \frac{1}{2} (\cos x \cdot e^x + \sin x \cdot e^x) + C, \text{ where } C = \frac{1}{2}C_1 $$
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Exercise 10.2

Exercise 10.2: Integration by Substitution

1. Integrate the following functions using the given substitutions.

(a) \(\int 4x^3 \sqrt{x^4-1} \,dx; \quad u = x^4 - 1\)

Solution

\( u = x^4 - 1 \)

Then \( du = 4x^3 \,dx \)

\(\int 4x^3 \sqrt{x^4-1} \,dx = \int \sqrt{u} \,du = \int u^{\frac{1}{2}} \,du \)

\(= \frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1} + C \)

\(= \frac{2}{3} u^{\frac{3}{2}} + C \)

\(= \frac{2}{3} (x^4 - 1)^{\frac{3}{2}} + C \)

(b) \(\int \cos^3 x \sin x \,dx; \quad u = \cos x\)

Solution

\( u = \cos x \)

Then \( du = -\sin x \,dx \implies -du = \sin x \,dx \)

\(\int \cos^3 x \sin x \,dx = \int u^3 (-du) \)

\(= -\int u^3 \,du \)

\(= -\frac{u^{3+1}}{3+1} + C \)

\(= -\frac{1}{4} u^4 + C \)

\(= -\frac{1}{4} \cos^4 x + C \)

(c) \(\int \frac{1}{x \ln|x|} \,dx; \quad u = \ln|x|\)

Solution

\( u = \ln|x| \)

Then \( du = \frac{1}{x} \,dx \)

\(\int \frac{1}{x \ln|x|} \,dx = \int \frac{1}{u} \,du \)

\(= \ln|u| + C \)

\(= \ln|\ln|x|| + C \)

(d) \(\int \sin^5 x \cos x \,dx; \quad u = \sin x\)

Solution

\( u = \sin x \)

Then \( du = \cos x \,dx \)

\(\int \sin^5 x \cos x \,dx = \int u^5 \,du \)

\(= \frac{u^{5+1}}{5+1} + C \)

\(= \frac{u^6}{6} + C \)

\(= \frac{\sin^6 x}{6} + C \)

(e) \(\int \frac{\ln x}{x} \,dx, x > 0; \quad u = \ln x\)

Solution

\( u = \ln x \)

Then \( du = \frac{1}{x} \,dx \)

\(\int \frac{\ln x}{x} \,dx = \int u \,du \)

\(= \frac{u^{1+1}}{1+1} + C \)

\(= \frac{1}{2} u^2 + C \)

\(= \frac{1}{2} (\ln x)^2 + C \)

(f) \(\int x^3 e^{x^4} \,dx; \quad u = x^4\)

Solution

\( u = x^4 \)

Then \( du = 4x^3 \,dx \implies \frac{1}{4} du = x^3 \,dx \)

\(\int x^3 e^{x^4} \,dx = \int e^u (\frac{1}{4} du) \)

\(= \frac{1}{4} \int e^u \,du \)

\(= \frac{1}{4} e^u + C \)

\(= \frac{1}{4} e^{x^4} + C \)

2. Use the substitution method to evaluate the following integrals.

(a) \(\int x \sqrt{1-x} \,dx\)

Solution

Let \( u = 1 - x \implies x = 1 - u \)

Then \( dx = -du \)

\(\int x \sqrt{1-x} \,dx = \int (1-u)\sqrt{u}(-du) \)

\(= \int (u-1)u^{\frac{1}{2}} \,du = \int (u^{\frac{3}{2}} - u^{\frac{1}{2}}) \,du \)

\(= \frac{2}{5}u^{\frac{5}{2}} - \frac{2}{3}u^{\frac{3}{2}} + C \)

\(= \frac{2}{5}(1-x)^{\frac{5}{2}} - \frac{2}{3}(1-x)^{\frac{3}{2}} + C \)

(b) \(\int (2x+1)(x^2+x)^7 \,dx\)

Solution

Let \( u = x^2+x \)

Then \( du = (2x+1) \,dx \)

\(\int (2x+1)(x^2+x)^7 \,dx = \int u^7 \,du \)

\(= \frac{1}{8} u^8 + C \)

\(= \frac{1}{8} (x^2+x)^8 + C \)

(c) \(\int \sin^3 x \,dx\)

Solution

\(\int \sin^3 x \,dx = \int \sin^2 x \cdot \sin x \,dx \)

\(= \int (1 - \cos^2 x) \sin x \,dx \)

Let \( u = \cos x \implies du = -\sin x \,dx \)

\(= \int (1-u^2) (-du) = \int (u^2-1) \,du \)

\(= \frac{1}{3}u^3 - u + C \)

\(= \frac{1}{3}\cos^3 x - \cos x + C \)

(d) \(\int x^2 \sqrt{x^3 - 2} \,dx\)

Solution

Let \( u = x^3 - 2 \)

Then \( du = 3x^2 \,dx \implies \frac{1}{3}du = x^2 \,dx \)

\(\int x^2 \sqrt{x^3 - 2} \,dx = \int \sqrt{u} (\frac{1}{3} du) \)

\(= \frac{1}{3} \int u^{\frac{1}{2}} \,du = \frac{1}{3} \cdot \frac{2}{3} u^{\frac{3}{2}} + C \)

\(= \frac{2}{9} u^{\frac{3}{2}} + C \)

\(= \frac{2}{9} (x^3-2)^{\frac{3}{2}} + C \)

(e) \(\int \frac{\sec^2 x}{\tan x} \,dx\)

Solution

Let \( u = \tan x \)

Then \( du = \sec^2 x \,dx \)

\(\int \frac{\sec^2 x}{\tan x} \,dx = \int \frac{1}{u} \,du \)

\(= \ln|u| + C \)

\(= \ln|\tan x| + C \)

(f) \(\int \frac{x}{\sqrt{x+1}} \,dx\)

Solution

Let \( u = x+1 \implies x = u-1 \)

Then \( dx = du \)

\(\int \frac{x}{\sqrt{x+1}} \,dx = \int \frac{u-1}{\sqrt{u}} \,du \)

\(= \int (u^{\frac{1}{2}} - u^{-\frac{1}{2}}) \,du \)

\(= \frac{2}{3}u^{\frac{3}{2}} - 2u^{\frac{1}{2}} + C \)

\(= \frac{2}{3}(x+1)^{\frac{3}{2}} - 2(x+1)^{\frac{1}{2}} + C \)

3. Evaluate the integral \(\int \frac{x}{(x^2+1)\ln(x^2+1)} \,dx\).

Solution

Let \( u = \ln(x^2+1) \)

Then \( du = \frac{1}{x^2+1} \cdot 2x \,dx \implies \frac{1}{2}du = \frac{x}{x^2+1} \,dx \)

\(\int \frac{x}{(x^2+1)\ln(x^2+1)} \,dx = \int \frac{1}{u} (\frac{1}{2} du) \)

\(= \frac{1}{2} \int \frac{1}{u} \,du \)

\(= \frac{1}{2} \ln|u| + C \)

\(= \frac{1}{2} \ln|\ln(x^2+1)| + C \)

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Exercise 10.1

Calculus: Exercises

Exercise 10.1

Evaluate the following integrals.

(a) $\int 4x^8 dx$

$$ = 4\int x^8 dx = 4\frac{x^{8+1}}{8+1} + C = \frac{4}{9}x^9 + C $$

(b) $\int \frac{3}{2}x^{\frac{3}{2}}\sqrt[3]{x} dx$

$$ = \frac{3}{2}\int x^{\frac{7}{3}} dx = \frac{3}{2}\frac{x^{\frac{7}{3}+1}}{\frac{7}{3}+1} + C = \frac{9}{20}x^{\frac{10}{3}} + C $$

(c) $\int (5^x + 2) dx$

$$ = \int 5^x dx + \int 2 dx = \frac{5^x}{\ln 5} + 2x + C $$

(d) $\int \sin^2 x dx$

$$ = \int\frac{1-\cos 2x}{2} dx = \frac{1}{2}\int(1-\cos 2x) dx = \frac{1}{2}(x - \frac{\sin 2x}{2}) + C $$

(e) $\int \frac{x+3}{\sqrt{x}} dx$

$$ = \int (x^{\frac{1}{2}} + 3x^{-\frac{1}{2}}) dx = \frac{x^{\frac{3}{2}}}{\frac{3}{2}} + 3\frac{x^{\frac{1}{2}}}{\frac{1}{2}} + C = \frac{2}{3}x^{\frac{3}{2}} + 6x^{\frac{1}{2}} + C $$

(f) $\int (\frac{1}{x} + 5) dx$

$$ = \int \frac{1}{x} dx + 5\int dx = \ln|x| + 5x + C $$

(g) $\int (e^x + \frac{2}{x}) dx$

$$ = \int e^x dx + 2\int\frac{1}{x} dx = e^x + 2\ln|x| + C $$

(h) $\int (\frac{1}{x^5} + 4e^x) dx$

$$ = \int x^{-5} dx + 4\int e^x dx = \frac{x^{-5+1}}{-5+1} + 4e^x + C = -\frac{1}{4x^4} + 4e^x + C $$

(i) $\int (\frac{3}{x} + e^x + 10) dx$

$$ = 3\int\frac{1}{x} dx + \int e^x dx + 10\int dx = 3\ln|x| + e^x + 10x + C $$

(j) $\int \sin^2 3x dx$

$$ = \int\frac{1-\cos(6x)}{2}dx = \frac{1}{2}\int(1-\cos 6x)dx = \frac{1}{2}(x - \frac{1}{6}\sin 6x) + C = \frac{1}{2}x - \frac{1}{12}\sin 6x + C $$

(k) $\int \sin 5x \sin 2x dx$

$$ = -\frac{1}{2}\int(\cos 7x - \cos 3x)dx = -\frac{1}{2}(\frac{\sin 7x}{7} - \frac{\sin 3x}{3}) + C = \frac{\sin 3x}{6} - \frac{\sin 7x}{14} + C $$

(l) $\int \cos 7x \cos 4x dx$

$$ = \frac{1}{2}\int(\cos 11x + \cos 3x)dx = \frac{1}{2}(\frac{\sin 11x}{11} + \frac{\sin 3x}{3}) + C = \frac{\sin 11x}{22} + \frac{\sin 3x}{6} + C $$

Problem 2

Evaluate the following integrals.

(a) $\int (1-2x)^3 dx$

$$ = -\frac{1}{2}\frac{(1-2x)^{3+1}}{3+1} + C = -\frac{1}{8}(1-2x)^4 + C $$

(b) $\int \sin(2\pi x + 7) dx$

$$ = \frac{1}{2\pi}(-\cos(2\pi x + 7)) + C = -\frac{1}{2\pi}\cos(2\pi x + 7) + C $$

(c) $\int \cos(3x-7) dx$

$$ = \frac{1}{3}\sin(3x-7) + C $$

(d) $\int 3^{5x-2} dx$

$$ = \frac{3^{5x-2}}{5\ln 3} + C $$

(e) $\int \frac{1}{7x-6} dx$

$$ = \frac{1}{7}\ln|7x-6| + C $$

(f) $\int \frac{\sin 2x}{\sin x} dx$

$$ = \int \frac{2\sin x \cos x}{\sin x} dx = 2\int\cos x dx = 2\sin x + C $$

(g) $\int \sec^2(2x+3) dx$

$$ = \frac{1}{2}\tan(2x+3) + C $$

(h) $\int e^{7x-3} dx$

$$ = \frac{1}{7}e^{7x-3} + C $$

(i) $\int (1 + \tan^2 2\pi x) dx$

$$ = \int\sec^2 2\pi x dx = \frac{1}{2\pi}\tan 2\pi x + C $$

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