Integration

Integration
Integration is used to find areas under curves.
Integration is the reversal of differentiation hence functions can be integrated
by indentifying the anti-derivative.
However, we will learn the process of integration as a set of rules rather than
identifying anti-derivatives.
Terminology

Indefinite and Definite integrals
There are two types of integrals: Indefinite and Definite.
Indefinite integrals are those with no limits and definite integrals have
limits.
When dealing with indefinite integrals you need to add a constant of
integration. For example, if integrating the function f(x) with respect to x:
( )
∫
f x dx = g(x) + C
where g(x) is the integrated function.

C is an arbitrary constant called the constant of integration.

dx indicates the variable with respect to which we are integrating, in this
case, x.

The function being integrated, f(x), is called the integrand.Integration- the basics
2
The rules

The Power Rule
∫
x dx
n
= C
n 1
x
n 1
+
+
+
provided that n ≠ -1
Examples:
∫
x dx
5
=
6
x
6
+ C

∫
x dx
-4
=
- 3
x
-3
+ C

When n = -1
∫
x dx
-1
=
∫
dx
x
1
= ln x + C

Constant rule
∫
k dx = kx + C where k is a constant
Example:
∫
2dx = 2x + C

Exponentials
∫
e dx
kx
= e C
k
1 kx
+
Example: e C
9
1
e dx
9x 9x
= +
∫
e dx e C
x x
= +
∫

Trig functions
- Cos
( ) =
∫
cos x dx sin(x) + C
( ) =
∫
cos kx dx sin( ) kx
k
1
+ C where k is a constant
Example: ( ) =
∫
cos 12x dx sin( ) 12x
12
1
+ C Integration- the basics
3
- Sin
( ) =
∫
sin x dx -cos(x) + C
( ) =
∫
sin kx dx cos( ) kx
k
1
− + C where k is a constant
Example: ( ) =
∫
sin 10x dx cos( ) 10x
10
1
− + C
( ) =
∫
sin - 5x dx cos( ) - 5x
5
1
+ C
Linearity
Suppose f(x) and g(x) are two functions in terms of x, then:
[ ( ) ( )] ( ) ( )
∫ ∫ ∫
f x ± g x dx = f x dx ± g x dx
Additionally, if A and B are constants, then
[ ( ) ( )] ( ) ( )
∫ ∫ ∫
Af x ±Bg x dx = A f x dx ± B g x dx
Examples:
( )
∫
2x + 3x dx
4 5
=
∫
2x dx
4
+
∫
3x dx
5
=
∫
2 x dx
4
+
∫
3 x dx
5
=




5
x
2
5
+




6
x
3
6
+ C
=
2
x
5
2x
5 6
+ + C

( ( ) )
∫
5cos 3x − 3e dx
7x
= ( )
∫ ∫
5cos 3x dx - 3e dx
7x
= ( )
∫ ∫
5 cos 3x dx - 3 e dx
7x
= ( ) 





 −




 7x
e
7
1
sin 3x 3
3
1
5
= ( )
7x
e
7
3
sin 3x
3
5
−Integration- the basics
4
Questions (General rules):
Integrate the following functions:
1. ( )
∫
x − x + 5 dx
2
3
x
6 1
2. ( )
∫
3x + x − 5 dx
8
3. ( )
∫
9x − 3x dx
2 -1
4. ( ( ) )
∫
sin 4x + e dx
3x
5. ( ( ) )
∫
cos 7x + 7x dx
2
(Solutions on page
Definite Integrals
Earlier we saw that
( )
∫
f x dx = g(x) + C
Suppose now we are given limits, i.e.
( )
∫
b
a
f x dx = g(x) + C
This can be interpreted as:
(value of g(x) + C at x = b) – (value of g(x) + C at x = a)
In other words, since C will cancel out:
( )
∫
b
a
f x dx = g(b) – g(a)
The full calculation of definite integrals is usually written out as:
( )
∫
b
a
f x dx = [ ( )]
b
a g x = g(b) – g(a)
i.e. integrate the function first (find g(x)) then substitute in the given limits
(always substitute the upper limit first).
(where a is the lower limit
and b is the upper limit) Integration- the basics
5
Examples
1.
∫
1
0
2
x dx =
1
0
3
x
3
1






= [ ]
1
0
3
x
3
1
=
3
1
{(1)
3
– (0)
3
} =
3
1
(1 – 0) =
3
1
2. ( ) 2x 1 dx
3
∫1
+ =
3
1
2
x
2
2x






+ = [ ]
3
1
2
x + x = {(3
2
+ 3) – (1
2
+ 1)}
= {(9 + 3) – (1 + 1)} = 12 – 2 = 10
3. cos ( ) x dx
2
π
∫0
= [ ( )]
2
π
0
sin x = {(sin(
2
π
)) – (sin(0))}
= 1 – 0 = 1
Questions (Definite integrals):
Integrate the following functions:
1. (3x 2x 5)dx
2
1
2
∫
− +
2. e dx
1
0
7x
∫
3. ( )
∫
π
0
sin 2x dx
4. ( )
∫
+
4
1
4x
12e 4 x dx
(Solutions on page 9)
Integration that leads to log functions
We know that if we differentiate y = ln(x) we find
x
1
dx
dy
= .
We also know that if y = ln f(x), this differentiates as:
( )
f( ) x
'f x
dx
dy
=
If we can recognise that the function we are trying to integrate is the derivative
of another function, we can simply reverse the above process. So if the
function we are trying to integrate is a quotient, and if the numerator is the
derivative of the denominator, then the integral will involve a logarithm, i.e.
( )
( )
∫
dx
f x
'f x
= ln (f(x)) + C
Example: y = ln(2x
2
+ 5)
t = 2x
2
+ 5 y = ln t
dx
dt
= 4x
dt
dy
=
t
1
dx
dy
= 4x x
t
1
=
t
4x
=
+ 5
2
2x
4xIntegration- the basics
6
Example 1:
∫
+
dx
3 5x
5
The derivative of the denominator is 5 which is the same as the
numerator, hence
∫
+
dx
3 5x
5
= ln (3 + 5x) + C
Example 2:
∫
+
dx
1 x
x
2
The derivative of the denominator is 2x. This is not the same as the
numerator but we can make it the same by re-writing the function
2
1 x
x
+
as
2
1 x
2x
2
1
+
⋅ , therefore
∫
+
dx
1 x
x
2
=
∫
+
dx
1 x
2x
2
1
2
=
2
1
ln (1 + x
2
) + C
Example 3:
( )
∫
dx
xln x
1
The derivative of ln x is
x
1
, so we can rewrite the function as:
ln( ) x
x
1
. Hence
( )
∫
dx
xln x
1
=
( )
∫
dx
ln x
x
1
= ln(ln(x)) + C
Example 4:
∫






+
−
2
1
dx
x 1
3
x
3
∫






+
−
2
1
dx
x 1
3
x
3
=
∫






+
2
1
dx
x 1
1
-
x
1
3
= [ ( ) ( )]
2
3ln x − 3ln x +1 1
= {(3ln(2) – 3ln(3)) – (3ln(1) – 3ln(2))}
= 3ln(2) – 3ln(3) + 3ln(2) = 6ln(2) – 3ln(3)
= ln(2
6
) – ln(3
3
)
= ln(64) – ln(27) = 





27
64
lnIntegration- the basics
7
Questions (Integration that leads to log functions):
Integrate the following functions:
1.
∫
+
dx
2 3x
3
2.
∫
+
dx
1 2x
x
2
3.
∫
+
dx
e 1
e
2x
2x
4.
∫
+
dx
x 4
x
2-
-3
5.
∫






+
−
+
1
0
dx
x 2
1
x 1
1
(Solutions on page 10) Integration- the basics
8
Solutions (General rules):
1. ( )
∫
x − x + 5 dx
2
3
x
6 1
=
∫ ∫ ∫
x dx - x dx + 5 dx
2
3
x
6 1
=
∫ ∫ ∫
−
x dx - x dx + x dx
6 2 5
3
=
( ) 4
x x
7
x
4
2
5
7 2
5
−
− +
−
+ C
=
4
7
4x
1
5
2x
7
x
2
5
− − + C
2. ( )
∫
3x + x − 5 dx
8
=
∫
3x dx
8
+
∫
x dx -
∫
5dx
=
∫
3 x dx
8
+
∫
x dx -
∫
5dx
= 5x
2
x
9
3x
9 2
+ − + C
= 5x
2
x
3
x
9 2
+ − + C
3. ( )
∫
9x − 3x dx
2 -1
=
∫
9x dx
2
-
∫
3x dx
-1
=
∫
9 x dx
2
-
∫
3 x dx
-1
=
3
9x
3
- 3ln + C (x)
=
3
3x - 3ln + C (x)
4. ( ( ) )
∫
sin 4x + e dx
3x
= ( )
∫
sin 4x dx +
∫
e dx
3x
= ( )
3x
e
3
1
cos 4x
4
1
− + + C
5. ( ( ) )
∫
cos 7x + 7x dx
2
= ( )
∫
cos 7x dx +
∫
7 x dx
2
= ( )
3
x
3
7
sin 7x
7
1
+ + C Integration- the basics
9
Solutions (Definite integrals):
1. (3x 2x 5)dx
2
1
2
∫
− + =
2
1
3 2
5x
2
2x
3
3x






− +
= [ ]
2
1
3 2
x − x + 5x
= {(2
3
– 2
2
+ 5(2)) – (1
3
– 1
2
+ 5(1))}
= {(8 – 4 + 10) – (1 – 1 + 5)}
= 14 – 5
= 9
2. [ ]
1
0
7x
1
0
7x
1
0
7x
e
7
1
e
7
1
e dx =






=
∫
=
7
1
{e
7
– e
0
} =
7
1
(e
7
– 1)
3. ( )
∫
π
0
sin 2x dx = ( )
π
0
cos 2x
2
1






− = [ ( )]
π
2 0
1
− cos 2x
= -
2
1
{cos(2π) – cos (0)}
= -
2
1
{1 – 1}
= 0
4. ( )
∫
+
4
1
4x
12e 4 x dx = ( )
∫
+
4
1
4x
12e 4x dx
2
1
=
4
1
2
3
4x 2
3
4x
4
12e




+ =
4
1
4x
3
8x
3e
2
3




+
=
( )
















− +




+
3
8
3e
3
8 4
3e
16 4
2
3
=
( )












 − +





+
3
8
3e
3
8 8
3e
16 4
=
3
8
3e
3
64
3e
16 4
+ − −
=
3
56
3e 3e
16 4
− +Integration- the basics
10
Solutions (Integration that leads to log functions):
1.
∫
+
dx
2 3x
3
= ln (2 + 3x) + C
2.
∫
+
dx
1 2x
x
2
Differentiating the denominator gives 4x
Therefore rewrite the function:
2
1 2x
x
+
=
2
1 2x
4x
4
1
+
⋅
Hence,
∫
+
dx
1 2x
x
2
=
∫
+
⋅ dx
1 2x
4x
4
1
2
=
∫
+
dx
1 2x
4x
4 2
1
=
4
1
ln (1 + 2x
2
) + C
3.
∫
+
dx
e 1
e
2x
2x
Differentiating the denominator gives 2e
2x
hence we can rewrite the
function as:
e 1
e
2x
2x
+
=
e 1
2e
2
1
2x
2x
+
⋅
∫
+
⋅ dx
e 1
e
2
1
2x
2x
=
2
1
ln (e
2x
+ 1) + C
4.
∫
+
dx
x 4
x
2-
-3
Differentiating the denominator gives -2x
-3
, hence the function can be
rewritten as:
x 4
x
2-
-3
+
=
x 4
2x
2
1
2-
-3
+
− ⋅
∫
+
dx
x 4
x
2-
-3
=
∫
+
− dx
x 4
2x
2
1
2-
-3
= -
2
1
ln(x
-2
+ 4) + C