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:

x	5	6	7	8	9
Frequency(f)	9	9	10	12	10

Step 1: Construct the calculation table

$x$	$f$	$f \cdot x$	$x^2$	$f \cdot x^2$
5	9	45	25	225
6	9	54	36	324
7	10	70	49	490
8	12	96	64	768
9	10	90	81	810
	$\sum f = 50$	$\sum fx = 355$		$\sum fx^2 = 2617$

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)	4	15	5	12	10

Step 1: Construct the calculation table with midpoints

Length	Midpoint (x)	f	$f \cdot x$	$x^2$	$f \cdot x^2$
(0, 10]	5	4	20	25	100
(10, 20]	15	15	225	225	3375
(20, 30]	25	5	125	625	3125
(30, 40]	35	12	420	1225	14700
(40, 50]	45	10	450	2025	20250
		$\sum f = 46$	$\sum fx = 1240$		$\sum fx^2 = 41150$

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)	0	1	2	3	4
Number of students (f)	15	10	20	8	2

Step 1: Construct the calculation table

$x$	$f$	$f \cdot x$	$x^2$	$f \cdot x^2$
0	15	0	0	0
1	10	10	1	10
2	20	40	4	80
3	8	24	9	72
4	2	8	16	32
	$\sum f = 55$	$\sum fx = 82$		$\sum fx^2 = 194$

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)	7	8	9	10	11
Frequency (f)	3	8	9	5	9

Step 1: Construct the calculation table

$x$	$f$	$f \cdot x$	$x^2$	$f \cdot x^2$
7	3	21	49	147
8	8	64	64	512
9	9	81	81	729
10	5	50	100	500
11	9	99	121	1089
	$ \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 batteries	12	8	15	30	15

Step 1: Construct the calculation table

Midpoint(x)	f	$f \cdot x$	$x^2$	$f \cdot x^2$
8	12	96	64	768
12.5	8	100	156.25	1250
17.5	15	262.5	306.25	4593.75
22.5	30	675	506.25	15187.5
27.5	15	412.5	756.25	11343.75
	$\sum f=80$	$\sum fx=1546$		$\sum fx^2=33143$

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 students	8	22	55	35	18	12

Step 1: Construct the calculation table

Distance (s)	Midpoint(x)	f	$f \cdot x$	$x^2$	$f \cdot x^2$
(0, 2]	1	8	8	1	8
(2, 4]	3	22	66	9	198
(4, 6]	5	55	275	25	1375
(6, 8]	7	35	245	49	1715
(8, 10]	9	18	162	81	1458
(10, 12]	11	12	132	121	1452
		$\sum f=150$	$\sum fx=888$		$\sum fx^2=6206$

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

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

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

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

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

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

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

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

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

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

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

Exercise 1.1

1. Solve the following equations.

(a) Solution

$$ x^2 - 6x + 10 = 0 $$ $$ x^2 - 6x + 9 = -1 $$ $$ (x - 3)^2 = i^2 $$ $$ x - 3 = \pm i $$ $$ x = 3 \pm i $$

So there are two solutions: \( x = 3 + i \) and \( x = 3 - i \).

(b) Solution

$$ -2x^2 + 4x - 3 = 0 $$ $$ x^2 - 2x + \frac{3}{2} = 0 $$ $$ x^2 - 2x + 1 = -\frac{1}{2} $$ $$ (x - 1)^2 = \frac{1}{2}i^2 $$ $$ x - 1 = \pm \sqrt{\frac{1}{2}}i $$ $$ x = 1 \pm \frac{\sqrt{2}}{2}i $$

So there are two solutions: \( x = 1 + \frac{\sqrt{2}}{2}i \) and \( x = 1 - \frac{\sqrt{2}}{2}i \).

(c) Solution

$$ 5x^2 - 2x + 1 = 0 $$ $$ x^2 - \frac{2}{5}x + \frac{1}{5} = 0 $$ $$ x^2 - \frac{2}{5}x + \frac{1}{25} = -\frac{4}{25} $$ $$ (x - \frac{1}{5})^2 = \frac{4}{25}i^2 $$ $$ x - \frac{1}{5} = \pm \frac{2}{5}i $$ $$ x = \frac{1}{5} \pm \frac{2}{5}i $$

So there are two solutions: \( x = \frac{1}{5} + \frac{2}{5}i \) and \( x = \frac{1}{5} - \frac{2}{5}i \).

(d) Solution

$$ 3x^2 + 7x + 5 = 0 $$ $$ x^2 + \frac{7}{3}x + \frac{5}{3} = 0 $$ $$ x^2 + \frac{7}{3}x + \frac{49}{36} = -\frac{11}{36} $$ $$ (x + \frac{7}{6})^2 = \frac{11}{36}i^2 $$ $$ x + \frac{7}{6} = \pm \frac{\sqrt{11}}{6}i $$ $$ x = -\frac{7}{6} \pm \frac{\sqrt{11}}{6}i $$

So there are two solutions: \( x = -\frac{7}{6} + \frac{\sqrt{11}}{6}i \) and \( x = -\frac{7}{6} - \frac{\sqrt{11}}{6}i \).

2. Solve the following equations and check your answers.

(a) Solution

$$x^2 - 2x + 4 = 0$$ $$x^2 - 2x + 1 = -3$$ $$(x - 1)^2 = 3i^2$$ $$x - 1 = \pm \sqrt{3}i$$ $$x = 1 \pm \sqrt{3}i$$

So the two solutions are \( x = 1 + \sqrt{3}i \) and \( x = 1 - \sqrt{3}i \).

For \(x = 1 + \sqrt{3}i\),

\((1 + \sqrt{3}i)^2 - 2(1 + \sqrt{3}i) + 4\)

\(= (1 + 2\sqrt{3}i - 3) - 2 - 2\sqrt{3}i + 4\)

\(= -2 - 2 + 4 = 0\)

For \(x = 1 - \sqrt{3}i\),

\((1 - \sqrt{3}i)^2 - 2(1 - \sqrt{3}i) + 4\)

\(= (1 - 2\sqrt{3}i - 3) - 2 + 2\sqrt{3}i + 4\)

\(= -2 - 2 + 4 = 0\)

(b) Solution

$$x^2 - 4x + 5 = 0$$ $$x^2 - 4x + 4 = -1$$ $$(x - 2)^2 = i^2$$ $$x - 2 = \pm i$$ $$x = 2 \pm i$$

So the two solutions are \( x = 2 + i \) and \( x = 2 - i \).

For \(x = 2 + i\),

\((2+i)^2 - 4(2+i) + 5\)

\(= (4 + 4i - 1) - 8 - 4i + 5\)

\(= 3 - 8 + 5 = 0\)

For \(x = 2 - i\),

\((2-i)^2 - 4(2-i) + 5\)

\(= (4 - 4i - 1) - 8 + 4i + 5\)

\(= 3 - 8 + 5 = 0\)

3. Find the value of \(i^n\) for every positive integer \(n\).

Solution (Method 1)

The powers of \(i\) follow a repeating cycle. We can find this pattern by calculating the first few powers step-by-step, using the definition \(i^2 = -1\).

• Step 1: Calculate \(i^2\).
  \(i^2 = -1\) (by definition)
• Step 2: Calculate \(i^3\).
  \(i^3 = i^2 \cdot i = (-1) \cdot i = -i\)
• Step 3: Calculate \(i^4\).
  \(i^4 = i^2 \cdot i^2 = (-1)(-1) = 1\)
• Step 4: Calculate \(i^5\).
  \(i^5 = i^4 \cdot i = (1) \cdot i = i\)

The pattern \(i, -1, -i, 1\) begins to repeat. This means the value of \(i^n\) depends on which category the exponent \(n\) falls into. We can express this using an integer \(k\).

• If \(n\) is of the form \(4k+1\) (e.g., 1, 5, 9,...), then \(i^n = i\). (for \(k \ge 0\))
• If \(n\) is of the form \(4k+2\) (e.g., 2, 6, 10,...), then \(i^n = -1\). (for \(k \ge 0\))
• If \(n\) is of the form \(4k+3\) (e.g., 3, 7, 11,...), then \(i^n = -i\). (for \(k \ge 0\))
• If \(n\) is of the form \(4k\) (e.g., 4, 8, 12,...), then \(i^n = 1\). (for \(k \ge 1\))

This can be written as a piecewise-defined function where \(k\) is an integer satisfying the conditions above:

$$ i^n = \begin{cases} i & \text{if } n = 4k+1 \\ -1 & \text{if } n = 4k+2 \\ -i & \text{if } n = 4k+3 \\ 1 & \text{if } n = 4k \end{cases} $$

Solution (Method 2: )

An alternative method is to consider whether the exponent \(n\) is even or odd.

If \(n\) is an even integer, we can write \(n=2k\). Then \(i^n = i^{2k} = (i^2)^k = (-1)^k\). Since \(k=\frac{n}{2}\), this gives \(i^n = (-1)^{\frac{n}{2}}\).

If \(n\) is an odd integer, we can write \(n=2k+1\). Then \(i^n = i^{2k+1} = i^{2k} \cdot i = (-1)^k \cdot i\). Since \(k=\frac{n-1}{2}\), this gives \(i^n = (-1)^{\frac{n-1}{2}}i\).

This can be summarized in the following piecewise-defined function:

$$ i^n = \begin{cases} (-1)^{\frac{n}{2}} & \text{if } n \text{ is even} \\ (-1)^{\frac{n-1}{2}}i & \text{if } n \text{ is odd}\end{cases}$$

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Axis of Symmetry နှင့် ဖြတ်မှတ်နှစ်မှတ် ပေးထားသော Parabola ပုစ္ဆာများ

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၁။ Axis of symmetry ကို ကြည့်၍ ညီမျှခြင်းကို $y^2$ သို့မဟုတ် $x^2$ ဟု ခွဲခြားနိုင်ပါသည်။

$y = k$ ဖြစ်လျှင်

$(y - k)² = \pm4p (x - h)$ ဖြစ်မည်။

$x= h$ ဖြစ်လျှင်

$(x - h)² = \pm4p (y - k)$ ဖြစ်မည်။

၂။ ညီမျှခြင်း မှ $\pm$ ကို ဆုံးဖြတ်ရန်အတွက် ပေးရင်း အမှတ် နှစ်မှတ်ကို အသုံးပြုပါမည်။

ဝင်ရိုး (Axis of Symmetry) မှ ဝေးရာသို့ ရွေ့လျားလေလေ၊ Parabola သည် ၎င်း၏ ဖွင့်သည့်ဘက်သို့ ပို၍ ရွေ့လျားသွားလေလေ ဖြစ်သည်။

$(y - k)² = \pm4p (x - h)$ အတွက်

ဝင်ရိုး Axis of symmetry မှ အဝေးဆုံးအမှတ် တွင် $x$ တန်ဖိုး ပိုငယ်နေလျှင် Parabola opens to the left ,

Equation: $(y - k)² = -4p (x - h)$.

ဝင်ရိုး Axis of symmetry မှ အဝေးဆုံးအမှတ် တွင် $x$ တန်ဖိုး ပိုကြီးနေလျှင် Parabola opens to the right ,

Equation: $(y - k)² = 4p (x - h)$.

$(x - h)² = \pm 4p (y - k)$ အတွက်

ဝင်ရိုး Axis of symmetry မှ အဝေးဆုံးအမှတ် တွင် $y$ တန်ဖိုး ပိုငယ်နေလျှင် Parabola opens down ,

Equation: $(x - h)² = -4p (y - k)$.

ဝင်ရိုး Axis of symmetry မှ အဝေးဆုံးအမှတ် တွင် $y$ တန်ဖိုး ပိုကြီးနေလျှင် Parabola opens up ,

Equation: $(x - h)² = 4p (y - k)$.

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Examples

Find an equation of the parabola satisfying the given conditions.

(1) Axis : $y = 0$ passes through $(3, 2)$ and $(2, -3)$

Thinking

-3 သည် Axis of Symmetry: $y = 0$ မှ အဝေးဆုံးအမှတ်ဖြစ်ပြီး $x$ တန်ဖိုးမှာ ငယ်သည်။ ထို့ကြောင့် Parabola opens to the left ဖြစ်သည်။

Equation: $(y - k)² = -4p (x - h)$ ကို အသုံးပြုရမည်။

(2) Axis : $x = 0$ passes through $(2, -1)$ and $(-4, 5)$

Thinking

-4 သည် Axis of Symmetry: $x = 0$ မှ အဝေးဆုံးအမှတ်ဖြစ်ပြီး $y$ တန်ဖိုးမှာ ကြီးသည်။ ထို့ကြောင့် Parabola opens up ဖြစ်သည်။

Equation: $(x - h)² = 4p (y - k)$. ကို အသုံးပြုရမည်။

(3) Axis : $y = 1$ passes through $(-4, -2)$ and $(-3, -4)$

Thinking

-4 သည် Axis of Symmetry: $y = 1$ မှ အဝေးဆုံးအမှတ်ဖြစ်ပြီး $x$ တန်ဖိုးမှာ ကြီးသည်။ ထို့ကြောင့် Parabola opens to the right ဖြစ်သည်။

Equation: $(y - k)² = 4p (x - h)$ ကို အသုံးပြုရမည်။