Fundamental Theorems of Integral Calculus and their Applications
We observe in the above examples that evaluation of b∫af ( x)dx as a limit of the sum is quite tedious, even if f ( x) is a very simple function. Both Newton and Leibnitz, more or less at the same time, devised an easy method for evaluating definite integrals. Their method is based upon two celebrated theorems known as First Fundamental Theorem and Second Fundamental Theorem of Integral Calculus. These theorems establish the connection between a function and its anti-derivative (if it exists). In fact, the two theorems provide a link between differential calculus and integral calculus. We state below the above important theorems without proofs.
If f ( x) be a continuous function defined on a closed interval [a , b] and F (x) = x∫a f (u)du, a < x < b then, d/dx F (x) = f ( x). In other words, F ( x) is an anti-derivative of f ( x).
If f ( x) be a continuous function defined on a closed interval [a , b] and F ( x) is an anti derivative of f ( x), then,
bfa ( x)dx = F (b) − F ( a).
Note
Since F (b) − F ( a) is the value of the definite integral (Riemann integral) b∫a f ( x)dx, any arbitrary constant added to the anti-derivative F ( x) cancels out and hence it is not necessary to add an arbitrary constant to the anti-derivative, when we are evaluating definite integrals. As a short-hand form, we write F (b) − F ( a) = [ F ( x)]ba . The value of a definite integral is unique.
By the second fundamental theorem of integral calculus, the following properties of definite integrals hold. They are stated here without proof.
i.e., definite integral is independent of the change of variable.
i.e., the value of the definite integral changes by minus sign if the limits are interchanged.
This property is used for evaluating definite integrals by making substitution.
We illustrate the use of the above properties by the following examples.
Example 9.14
Evaluate : 1.5∫0 [x2] dx, where [ x] is the greatest integer function.
Solution
We know that the greatest integer function [ x] is the largest integer less than or equal to x. In other words, it is defined by [ x ] = n , if n ≤ x < (n +1) , where n is an integer.
We note that the above function is not continuous on [0,1.5] .
But, it is continuous in each of the sub-intervals [0,1) , [1, √2 ) and [ √2,1.5] ; that is, it is piece-wise continuous on [0,1.5] .
See Fig. 9.6. Hence, we get
Next, we give examples to illustrate the application of Property 5.
We derive some more properties of definite integrals.
Property 6
b∫a f ( x) dx = b∫a f (a +b − x) dx
Proof
Let u = a + b − x . Then, we get dx = −du .
When x = a , u = a + b − a = b . When x = b , we get u = a + b − b = a .
∴ b∫a f ( x)dx = a∫b f ( a + b −u)(−du) = b∫a f ( a + b −u)du
= ∫ab f ( a + b − x)dx .
Note
Replace a by 0 and b by a in the above property we get the following property
0 ∫ a f ( x) dx = ∫0a f ( a − x ) dx .
Property 7
Property 8
If f ( x) is an even function, then −a∫a f ( x) dx = 20∫a f ( x) dx.
(Recall that a function f ( x) is an even function if and only if f (− x ) = f ( x). )
Proof
By property 3, we have
−a∫a f ( x) dx = −a∫0 f ( x) dx + 0∫a f ( x) dx .
In the integral 0∫−a f ( x) dx , let us make the substitution, x = −u. Then, dx = −du.
When x = −a , we get u = a , when x = 0 , we get u = 0 , So, we get
−a∫0 f ( x) dx = 0∫a f ( − u)( −du) = 0∫a f ( −u) du = 0∫a f ( − x) dx = 0∫a f ( x) dx . ... (2)
Substituting equation (2) in equation (1), we get
−a∫a f ( x) dx = 0∫a f (x) dx + 0∫a f (x) dx = 20∫a f (x) dx .
Property 9
If f ( x) is an odd function, then −a∫a f ( x) dx = 0.
(Recall that a function f ( x) is an odd function if and only if f (− x ) = − f ( x). )
Proof
By property 3, we have
−a ∫a f ( x) dx = −a ∫ 0f ( x) dx + 0∫a f ( x) dx .
Consider −a∫0 f ( x) dx . In this integral, let us make the substitution, x = −u. Then, dx = −du.
When x = −a , we get u = a ; when x = 0 , we get u = 0 . So, we get
−a∫0 f ( x) dx = a ∫0 f ( − u)( −du) = a∫0 f ( −u) du = a ∫0 f ( − x) dx = − a ∫0 f ( x) dx . ... (2)
Substituting equation (2) in equation (1), we get
a∫−a f ( x) dx = a∫0 f ( x) dx − a∫0 f ( x) dx = 0
Property 10
If f (2a − x ) = f ( x), then 2a∫0 f ( x) dx = 2 a∫0 f ( x) dx.
Proof
By property 7, we have
2a∫0 f ( x) dx = a∫0 [ f ( x) + f ( 2a − x ) ] dx. ...(1)
Setting the condition f (2a − x ) = f (x) in equation (1), we get
0∫2a f (x) dx = a∫0 [ f (x ) + f (x)]dx = 2 a∫0 f (x) dx.
If f ( 2a − x ) = − f ( x), then 2a∫0 ( x) dx = 0.
By property 7, we have
2a ∫0 f ( x) dx = a∫0 [ f ( x) + f ( 2a − x ) ]dx. ... (1)
Setting the condition f (2a − x ) = − f (x) in equation (1), we get
2a∫0 f ( x) dx = ∫0a [ f ( x ) − f ( x)]dx = 0.
Note
This property help us to remove the factor x present in the integrand of the LHS.
Show that , where g (sin x) is a function of sin x .
We know that
2a∫0 f ( x) dx = 2 a∫0 f ( x) dx if f (2a − x ) = f (x) .
Take 2a = π and f (x) = g (sin x) .
Then, f (2a − x) = g (sin(π − x)) = g (sin x) = f (x) .
∴∫02a f (x) dx = 2∫0a f (x) dx .
The above result is useful in evaluating definite integrals of the type π∫0 g (sin x) dx .
Example 9.21
Example 9.22
Show that 2π∫0 g (cos x) dx =20∫π g (cos x) dx , where g (cos x) is a function of cos x .
Solution
Take 2a = 2Ï€ and f (x) = g (cos x) .
Then, f (2a − x) = f (2π − x) = g (cos(2π − x)) = g (cos x) = f (x)
∴ 2a∫0 f (x) dx = 2 a∫0 f (x) dx .
∴ 2a∫0 g (cos x) dx = 2 π∫0 g (cos x) dx .
Result
2π∫0 g (cos x) dx =2 π∫0 g (cos x) dx.
Note
The above result is useful in evaluating definite integrals of the type 2π∫0 g (cos x) dx.
Example 9.23
Substituting x = a + u , we have dx = du ; when x = a, u = 0 and when x = 2a, u = a .
∴2a∫a f (x) dx = a∫0 f (a + u ) du = a∫0 f (u ) du , since f (x) = f (a + x)
= a∫0 f (x) dx . ... (2)
Substituting (2) in (1), we get
2a∫0 f (x) dx = 2a∫0 f (x) dx .
Example 9.24
Solution
Let f(x) = x cos x .Then f(-x) = (− x) cos(− x) = xcos = − f(x).
So f(x) = x cos x is an odd function.
Hence, applying the property, for odd function f(x)
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