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Integration is the sum of the area and the definite integral is used as a tool to identify the area within an interval or limit. The mathematical expression of definite integral is ∫baf(x)dx.
This article covers the measurement of definite integral, the concept of definite integral, a thorough explanation on properties of definite integral, formula of definite integral and so on.
Definition of Definite Integral
A definite integral is given using this notation here, where is a function f (x) is continuous on the closed interval, from a to b, [ a, b], then the definite integral from a to b is defined as below:
∫ab f (x) dx = limn→∞ ∑nr=1 f (ci) ΔXi
Properties of Definite Integral
In this section you will find a brief of the basic properties of definite integral, its functions to understand the definite integral more effectively. So, let’s have a walk through these properties.
1. ∫ba f(x) dx= ∫ca f(x) dx + ∫bc f(x) dx
Proof: Let g (x) be the antiderivative of f(x). Then,
LHS = g(b) – g(a)
RHS = (g (c) – g (a) ) + (g (b) – g (c) )
= g(b) – g(a)
2. ∫ba f(x) dx = -∫abf(x) dx also, ∫aa f(x) dx = 0
Proof: Let g (x) be the antiderivative of f(x). Then,
LHS = g(b) – g(a)
RHS = – (g (a) – g (b) )
= g(b) – g(a)
3. ∫ba f(x) dx = ∫ba f(t) dt
Proof: Let g be the antiderivative of (f). Then,
LHS = g(b) – g(a)
RHS = g(b) – g(a)
Hence proved LHS = RHS
The value of definite integral is always free from the variable of integration, due to this fact we can call the variable of definite integral in definite integral- dummy variable.
4. ∫ba f(x) dx = ∫ba f (a + b -x) dx
Proof: Take RHS and then, put a+b-x = t, -dx = dt
x- a to b
t- b to a
RHS = -∫ab f(t) dt = ∫ba f(t) dt = ∫ba f(x) dx = LHS
5. ∫a0 f(x) dx = ∫a0 f (a-x) dx
Proof : As we can see that property 5 is similar to property 4 we need to put a = 0 and b = a.
This is the most commonly used property in definite integral.
6. ∫2a0 f(x) dx = ∫a0 f (x) dx + ∫a0 f (2a – x) dx
Proof: Using P -I we will get ….
∫2a0 f(x) dx = ∫a0 f (x) dx + ∫2aa f (x) dx
In I, put x = 2a – t
. dx = – dt
X – a to 2a
T – 2a to 0
∫2a0 f(x) dx = ∫a0 f (x) dx + ∫0a f (2a – t) dt
∫2a0 f(x) dx = ∫a0 f (x) dx + ∫a0 f (2a – x) dx
Specific cases of this property……
a. ∫2a0 f(x) dx = 2 ∫a0 f (x) dx [ if f (x) = f (2a – x) ]
b. ∫2a0 f(x) dx = 0 [ if f (x) = -f (2a-x)]
How to calculate a definite integral
Step 1. Calculate the antiderivative F(x)
Step 2. Find values of F(b) and F(a)
Step 3. Calculate F(b) – F(a).
Let’s look at some examples for our better understanding of measurement or calculation of definite integral.
Example 1. F = ∫64 3x2 dx
= 3∫64 x2 dx (as per the rule , of a constant getting multiplied by a variable it can be taken outside)
= 3 [x3 / 3]64 ….(as per the power rule ∫ xn dx = x n+1 / (n +1) + c, n ≠ -1)
= [ x3 / ]64
= [6]3 – [4]3 …….( replace x with upper limit and lower limit)
= 216 – 64
= 152
Example 2. F = ∫42 6x2 – 3x +11 dx
= [ 6x3/3 – 3x2/2 + 11x]4 2 ( 1 dx = x , so 11dx = 11x )
= [ 2x3 – 3/2 x2 + 11x ] 4 2
= [ 2 (4)3 – 3/2 (4)2 + 11 (4)] – [ 2 (2)3 – 3/2 (2)2 + 11 (2)]
= [128 – 24 + 44] – [ 16 – 6 + 22]
=148 – 32
= 116
Conclusion
In this article describing the definite integral, we studied the measurement of the definite integral in much detail. We covered several other topics such as properties of definite integral, important formulas of definite integral and other related topics along with the solved examples. We hope you find this study material helpful for better understanding on measurement of definite integral.
