Integration for the beginners | Mathematical Guide
Integration and the definite integral: Supportive lecture for the
beginners.
The symbol for integration is ∫
Basic Integration Rules
1. ∫ k dx = kx + C ; k is a constant
2. ∫ k f (x) dx = k ∫ f(x) dx
3. ∫ [ f(x) + g(x) ] dx = ∫ f (x) dx + ∫ g(x) dx
4. ∫ [ f(x) - g(x) ] dx = ∫ f (x) dx - ∫ g(x) dx
5. ∫ xn dx = [xn+1 / n+1] + C ; n≠ (-1)
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In every integration operation without limits, we added C as a
constant after the integration. Below will explain it simply;
As an example:
d/dx (X2 + 5) is equal to -> d/dx (X2) +
d/dx (5)
After the differentiation;
d/dx (X2 + 5) is equal to -> 2X
If it is difficult to understand: Let’s visit the simple steps of
differentiation
Then, let we integrate 2X, in here the answer should be [X2
+ 5]
∫ 2X1 is equal to -> 2 ∫ X1
The answer is equal to
= 2. [ X(1+1)/(1+1)
]
= 2.[X2/2]
= X2
But the answer should be [X2 + 5]
Therefore we have to add a constant value at the end of the result,
consuming the constant value is equal to any value.
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Rule 01:
∫ k dx =
kx + C ; k is a constant
Ex:
1. Find ∫ 2 dx
∫ 2 dx = ∫ 2.1 dx
Then, 1 is equal to X0; (Power zero of any number is
equal to one)
∫ 2 dx = ∫ 2. X0 dx
= 2 ∫ X0
dx
= 2. [X0+1/
0+1] ; According to the Rule Number 05
= 2. [X / 1]
= 2.X + C ; C is a constant
2. Find ∫
(-3) dx
∫ (-3) dx = -3X
3. Find ∫ 1 dx
∫ 1 dx = 1X
= X + C ; C is a constant
4. Find ∫ 10 dt
∫ 10 dt = 10t + C ; C is a
constant
Rule 02:
∫ k f (x) dx = k ∫ f(x) dx
Ex:
1. Find ∫ 4.X dx
In here, X is equal to f(X) and 04 is equal to k;
∫ 4X dx = 4 ∫ X dx
= 4 ∫ X1 dx
= 4 [X (1+1)/ (1+1)] + C
= 4 [X2/2] + C
= 4/2 X2 + C
= 2X2
+ C ; C is a constant
2. Find ∫ 2/X3 dx
∫ 2/X3 dx =
2 ∫ 1/X3 dx
=
2 ∫ X-3 dx
=
2. X(-3+1)/ (-3+1) + C
= 2.X-2/(-2)
+ C
=
-X-2 + C ; C is a
constant
Rule 03.
∫ [f(x) + g(x)] dx = ∫
f (x) dx + ∫ g(x) dx
Ex:
Find ∫ (X + 5) dx
In here, X is equal to f(X) and 5 is equal to g(X);
∫ (X + 5) dx = ∫ (X)
dx + ∫ (5) dx
=
∫ X1 dx + 5 ∫ X0 dx
=
[X(1+1)/(1+1)] + 5.[X(0+1)/(0+1)] ; According to the Rule 05
=
[X2/2] + 5.[X1/1]
=
[X2/2] + 5X + C ;
C is a constant
Rule 04.
∫ [f(x) - g(x)] dx = ∫
f (x) dx - ∫ g(x) dx
Ex:
Find ∫ (X3 - 2) dx
In here, f(x) is equal to X3 and g(x) is equal
to 2;
∫ (X3 - 2) dx =
∫ X3 dx - ∫ 2 dx
=
∫ X3 dx –∫ 2.X0 dx
=
∫ X3 dx – 2 ∫ X0 dx
=
[X(3+1) / (3+1)] – 2. [X(0+1) / (0+1)] ;
According to the Rule 05
=
[X(4) / 4] – 2. [X(1) / 1]
=
[X(4) / 4] – 2. [X]
=
X(4) / 4 – 2X + C ;
C is a constant
Rule 05.
∫ xn dx = [xn+1 / n+1] +
C ; n≠ (-1)
Ex:
Find ∫ (x4) dx
In here 4 is equal to n;
∫ (x4) dx =
X(4+1) / (4+1)
=
X5 / 5 + C ;C
is a constant
Find ∫ 4X3 dx
∫ 4X3 dx = 4. ∫ X3
dx
=
4. [ X(3+1) / (3+1) ]
=
4. [X4 / 4]
=
X4 + C ;C is a
constant
Definite Integration
The expression of
∫ab f(X)
Is called the definite integral from a to b.
The lowest limit of the integration is a & the upper
limit of the integration is b.
Simply the function, f(X) is a graph
of a coordinate plane. We can define the area of that graph by integrating the
function, f(X) with using the limits. The area that covered by
the function is limited by those limits, a & b.
Ex:
Find f(X) = X3 – 2, 2 <= X <= 4
In here, f(X) is equal to (X3
– 2) and a & b is equal to 2 & 4.
Let’s find the area;
· First we have to integrate the function
· Then we have to define the limitations
Area =
∫42 (X3 – 2) dx
=
[X4/4 – 2X] 42
Then we have to add limits values to
the X.
=
[44 / 4 - 2(4) ] – [24 / 4 – 2(2)]
=
[256/4 – 8] – [16/4 - 4]
= [64 - 8] – 0
= 56
Find f(X) = (X2 - 5) 1 <= X <= 3
In here, f(X) is equal to (X2
– 5) and a & b is equal to 1 & 3.
Let’s find the area;
Area =
∫31 (X2 – 5) dx
=
[X3/3 – 5X] 31
=
[33 / 3 – 5(3)] – [13 / 3 - 5 (1)]
=
[9 – 15] – [1/3 – 5]
=
[-6] – [-14/3]
=
[-6] + [14/3]
=
-4/3
Simple integration examples for the beginners
Example 01:
Find ∫ dt
∫ dt = ∫ t0
dt
= t(0+1)
/ (0+1)
= t1/1
= t
Example 02:
Find ∫ 8 dt
∫ 8 dt = ∫ 8.1 dt (8*1=8)
= 8 ∫ 1 dt
= 8 ∫ t0
dt (zero power of any function
is equal to 1)
= 8.t + C ;C is a constant
Example 03:
Find ∫ (X-4)(X+4) dx
∫ (X-2)(X+2) dx = ∫ X.(X+2) -2.(X+2) dx
= ∫ X2 + 2X -2X -4 dx
= ∫ X2 - 4 dx
= ∫ X2 – 4.X0 dx
= [X(2+1) / (2+1)] – 4[X1/1]
= [X3 /3] – 4X + C ; C is a constant
Example 04:
Find ∫03 2X dx
∫03 2X dx =
2 ∫30 X1 dx
=
2 [X2 / 2] 30
=
[X2]30
= [32
- 0]
=
9
Example 05:
Find ∫01 √X (1-X) dx
∫01 √X (1-X) dx = ∫01
√X – X (3/2) dx
= ∫01 X (1/2)
– X (3/2) dx
= [X(1/2 + 1) / (1/2 + 1)] 01
– [X(3/2 + 1) / (3/2 + 1)] 01
= [X(3/2) / (3/2) ] 01
– [X(5/2) / (5/2)] 01
Replace X with 1 and 0;
= [1(3/2) / (3/2) - 0(3/2)
/ (3/2)] - [1(5/2) / (5/2) -
0(5/2) / (5/2)]
= [2/3 – 0] – [2/5 - 0]
= [2/3] – [2/5]
= 4/15
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