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Chapter: 12th Maths : UNIT 9 : Applications of Integration

Fundamental Theorems of Integral Calculus and their Applications

These theorems establish the connection between a function and its anti-derivative (if it exists).

Fundamental Theorems of Integral Calculus and their Applications

We observe in the above examples that evaluation of baf ( 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.

Theorem 9.1 (First Fundamental Theorem of Integral Calculus)

If f ( x) be a continuous function defined on a closed interval [a , b] and F (x= xa f (u)dua < x < then, d/dx F (x= f ( x). In other words, F ( x) is an anti-derivative of f ( x).

Theorem 9.2 (Second Fundamental Theorem of Integral Calculus)

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) ba 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 : 1.50 [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


ba f ( x) dx = ba f (a +b − x) dx

Proof

Let a + b − x . Then, we get dx = −du .

When a , u = a + b − a = b . When x = b , we get u = a + b − b = a .

 ba f ( x)dx ab f ( a + b −u)(du) = ba 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

∫ a f ( x) dx = 0a f ( a  x ) dx .


 

Property 7


 

Property 8

If f ( x) is an even function, then aa ( x) dx = 20a f ( x) dx.

(Recall that a function ( x) is an even function if and only if ( x ) = f ( x). )

Proof

By property 3, we have

aa f ( x) dx = a0 f ( x) dx + 0a f ( x) dx .

In the integral 0( 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

a0 f ( x) dx = 0a f (  u)( du) = 0a f ( u) du = 0a f (  x) dx = 0a f ( x) dx . ... (2)

Substituting equation (2) in equation (1), we get

aa f ( x) dx = 0a f (x) dx + 0a f (x) dx = 20a f (x) dx .

 

Property 9

If f ( x) is an odd function, then aa f ( xdx = 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 ∫ 0( x) dx + 0a f ( x) dx .

Consider a0 f ( xdx . 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

a0 f ( x) dx a ∫0 f (  u)( du) = a0 f ( u) du = a ∫0 f (  x) dx =  a ∫0 f ( x) dx . ... (2)

Substituting equation (2) in equation (1), we get

aa f ( x) dx = a0 f ( x) dx  a0 f ( x) dx = 0

 

Property 10

If f (2a  x ) = f ( x), then 2a0 f ( xdx = 2 a0 f ( xdx.

Proof

By property 7, we have

2a0 f ( x) dx = a0 [ f ( x) + f ( 2 x ) dx.       ...(1)

Setting the condition f (2a  x ) = f (x) in equation (1), we get

02a f (x) dx = a0 [ f () + f (x)]dx = 2 a0 f (x) dx.

 

Property 11

If f ( 2a  x ) = − f ( x), then 2a0 xdx = 0.

Proof

By property 7, we have

2a 0 f ( x) dx = a0 [ f ( x) + f ( 2 x ) ]dx.         ... (1)

Setting the condition f (2a  x ) = − f (x) in equation (1), we get

2a0 f ( x) dx = 0a [ f ( x )  f ( x)]dx = 0.

 

Property 12


Note

This property help us to remove the factor x present in the integrand of the LHS.

 

Example 9.20

Show that  , where g (sin x) is a function of sin x .

Solution

We know that

2a0 f ( x) dx = 2 a0 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 = 20a f (x) dx .


Result


Note

The above result is useful in evaluating definite integrals of the type π0 g (sin xdx .

 

Example 9.21


 

Example 9.22

Show that 2π0 g (cos xdx =20π g (cos xdx , where g (cos x) is a function of cos x .

Solution

Take 2a = 2π and (x) = g (cos x) .

Then, (2 x) = (2π  x) = g (cos(2π  x)) = g (cos x) = f (x)

 2a0 f (x) dx = 2 a0 f (x) dx .

 2a0 (cos x) dx 2 π0 g (cos xdx .

Result

0 g (cos x) dx =2 π0 g (cos x) dx.

Note

The above result is useful in evaluating definite integrals of the type 2π0 g (cos xdx.

 

Example 9.23


Substituting x = a + u , we have dx = du ; when x = a, u = 0 and when x = 2a, u = a .

2aa f (x) dx = a0 f (a + u du a0 f (u ) du , since f (x) = f (a + x)

a0 f (x) dx . ... (2)

Substituting (2) in (1), we get

2a0 f (x) dx = 2a0 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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12th Maths : UNIT 9 : Applications of Integration : Fundamental Theorems of Integral Calculus and their Applications |

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