CHAPTER 1: INTEGRATION

HELLO AND WELCOME TO THE SECOND SEMESTER!

Integration is connected to function. Function F(x) is an antideravitive of f(x) if for all x in the domain of f(x), dF/dx = f

RULES OF INTEGRATION

1. Integrating [a . dx] = ax + c (where c is a constant)
2. Integrating [xⁿ . dx] = xⁿ⁺¹/n+1 + c
3. Integrating [a f(x) . dx] = a Integrate [f(x) . dx] (where a is factorised)
4. Integrating [f(x) + g(x)] . dx = Integrate [f(x) . dx] + Integrate [g(x) .dx]
5. Integrating [a xⁿ . dx] = a xⁿ⁺¹/n+1 + c
6. Integrating [(ax)ⁿ . dx] = (ax)ⁿ⁺¹/ (n+1)a +c

DEFINITE INTEGRAL
Suppose that f(x) is continuos on the interval [a,b]. If F(x) is any antideravative of f(x), then Integrating from a to b [f(x) . dx] = F(b) - F(a)


THE INTEGRATION PRODUCING LOGARITHM FUNCTIONS

1. Integrating [1/x . fx] = ln x + c
2. Integrating [1/ax + b . dx] = 1/a ln (ax + b) + c
3. Integrating [f'(x)/f(x) . fx] = ln |f(x)| + c

Eg; Integrating [-2x/x² - 1 . dx]
When the denominator is differentiated, dy/dx = 2x
Thus, the numenator should be written such that it equals to 2x
Integrating [-2x/x² - 1 . dx] can be written as - {Integrating [2x/x² - 1 . dx]}, where the negative sign is factorised. Now that the numenator is equals to the differentiation of the denominator, the function can be integrated.

Integrating [-2x/x² - 1 . dx]
=  - {Integrating [2x/x² - 1 . dx]}
= - ln |x² - 1| + c


INTEGRATION  OF EXPONENTIAL FUNCTIONS

REFRESH YOUR MIND! :D

1. d/dx e^x = e^x
2. d/dx ax = a^x ln a
3. d/dx e^{ax + b} = ae^{ax + b}
4. d/dx e^f(x) = f'(x) e^f(x)

Therefore,

1. Integrating [e^x . dx] = e^x + c
2. Integrating [a^x . dx] = a^x/ln a + c

Eg; Integrating [e^2x . dx]
Step 1. When integrating e^2x, e^2x is written again
Step 2. d/dx (2x) is find, which is 2
Step 3. The exponential is divided by the d/dx 2x,2 : e^2x/2
Step 4. Don't forget to add the constant, c! :D
Therefore, Integrating [e^2x . dx] = e^2x/2 + c


INTEGRATION OF TRIGONOMETRIC FUNCTIONS

REFRESH YOUR MIND! :D

1. d/dx (sin ax) = a cos ax
2. d/dx (cos ax) = -a sin ax
3. d/dx (tan ax) = a sec2 ax

Therefore,

1. Integrating [cos ax . dx] = 1/a sin ax + c
2. Integrating [sin ax . dx] = -1/a cos ax + c
3. Integrating [sec² ax . dx] = 1/a tan ax + c

Eg; d/dx (sin 7x) = 7 cos 7x + c
Integrating [d/dx (sin 7x) . dx] = Integrating [7 cos 7x . dx]
sin 7x =7 {Integrating [cos 7x . dx]} (where 7 is factorised)
Therefore, Integrating [cos 7x . dx] = 1/7 sin 7x + c


TECHNIQUES OF INTEGRATION
For some complicated/ complex functions, techniques is used to solve the integration of the functions
1. Substitution Method
eg; find Integrating [(2x + 1)⁵ . dx]
Let u = 2x + 1
du/dx = 2
dx = du/2

Therefore, Integrating [(2x + 1)⁵ . dx]
= Integrating [u⁵ . du/2]
= 1/2 Integrating  [u⁵ . du] (where 1/2 from du/2 is factorised)
= 1/2 (u⁶/6) + c
= u⁶/12 + c
= (2x + 1)⁶/12 + c (where u is substituted with 2x + 1)

2. Alternative Method
Integrating [(ax + b)ⁿ . dx] = (ax + b)^{n + 1}/a(n + 1) + c
Eg; Integrating  [x (1 + x²)^3/2 .dx]
Let u = 1 + x²
du/dx = 2x
x dx = du/2
Integrating [x (1 + x²)^3/2 .dx]
= Integrating [u^3/2 (du/2)]
= 1/2 {Integrating [u^3/2 . du]}
= 1/2 (u^5/2)/(5/2) + c
= 1/5 u^5/2 + c
= 1/5 (1 + x²)^5/2 + c


INTEGRATION BY PARTS
d(uv)/dx = u dv/dx + v du/dx
Integrating [d(uv)/dx . dx] = Integrating [u dv/dx . dx] + Integrating [v du/dx . dx]
uv = Integrating [u dv] + Integrating [v du]
Therefore, Integrating [u dv] = uv - Integrating  [v du]


INTEGRATION OF RATIONAL FUNCTIONS BY USING PARTIAL FRACTIONS
Polynomial is involved here!
Eg; Integrating [5x - 3/(x² - 2x - 3) . dx] = Integrating  [5x - 3/(x + 1)(x - 3) . dx
5x - 3/x² - 2x - 3 = A/x + 1 + B/x - 3
Using polynomial, A = 2 and B = 3

Therefore, Integrating [5x - 3/x² - 2x - 3 . dx]
= Integrating [2/(x + 1) . dx] + Integrating [3/(x - 3) . dx]
= 2 ln |x + 1| + 3 ln |x - 3| + c

*Note that this integration involves the integration producing logarithm functions
If the funtion involved is an improper fraction, long division is used to change it into a proper function
PLEASE REFER FIRST SEMESTER NOTES, TOPIC 6: POLYNOMIAL :)


INTEGRATION CAN BE USED TO FIND THE AREA AND VOLUME OF A GIVEN GRAPH

1. Area = Integrating from a to b [f(x) . dx]
2. Volume when rotated through x-axis = pi {Integrating from a to b [y² . dx]}
3. Volume when rotated through y-axis = pi {Integrating from c to d [x² . dy]}

As you can see, this integration topic involves differentiation and polynomial. Please refer those notes from Semester 1 to refresh your memory and help you to solve the questions! Besides, differentiation can be used to make sure the answers integrated are correct. Lastly, we're sorry for the low presentation quality as we don't have the program for Mathematics' symbols. Hope you can understand the notes well! :D

From Maths unit of JPP Academic Biro

Enjoy Maths :)