Integration - Definite Integrals

INTEGRATION: DEFINITE INTEGRALS THE RIEMANN SUMS AND DEFINITE INTEGRAL Let a function f be continuous on the closed interval [a,b] that is from x=a to x=b. We divide this interval in “n” subintervals by inserting (n-1) points x1 ,x 2 , …,x n @ 1 between a and b and relabel a as x 0 and b as x n . We denote the length of the first subinterval @ A @ A x 0 ,x1 by Δ1 x = x1 @ x 0 , of second subinterval x1 ,x 2 by Δ2 x = x 2 @ x1 , …… of last @ A subinterval x n @ 1 ,x n by Δn x = x n @ x n @ 1 . Now we pick a point x1B from x 0 ,x1 and form the product f x1B Δ1 x ; pick a point x B2 @

A

b

c

from x1 ,x 2 and form the product f x B2 Δ2 x and go on tin this manner until we form all @

A

the products

b

c

f x1B Δ1 x, f x B2 Δ2 x, ' , f x Bn Δn x b

c

b

The sum of these products n

c

b

c

S n = X f x Bk Δk x = f x1B Δ1 x + f x B2 Δ2 x + ' + f x Bn Δn x k=1

b

is called a Riemann Sum.

c

b

c

b

c

b

c

` a

B

C

A Riemann Sum may therefore be thought as the sum of n products. If f x >0 on a,b , B

` a

C

then the Riemann Sum is a positive real number, and if f x <0 on a,b , then the Riemann Sum is a negative real number.

FUNDAMENTAL THEOREM & DEFINITE INTEGRAL Evaluating the definite integrals using the limits of Riemann sums is generally very time consuming. Here the fundamental theorem of calculus helps us to evaluate definite integrals without using Riemann sums. If f x is continuous on the interval a,b and if F. x = f x over a,b , then b

B

` a

` ab

C

` a

Z f x dx = F x | = F b @ F a a

` a

` a

a

` a

3

Example : Evaluate Z x 2 dx 2

Solution : 3 xf f f f f f As f is the antiderivative of x 2 : 3 3

3

3

3

3

3

8f xf f f f f f3 3f f f f f 2f f f f f f 19 f f f f f f Z x 2 dx = f |= f @ f =9@ f = f 3

2

2

3

3

2π

Example : Evaluate Z cos x dx πf f f f f 3

Solution : As sin x is the antiderivative of cos x : 2π

2π πf f f f Z cos x dx = sin x | = sin 2π @ sin f

3

πf f f f f 3

πf f f f f 3

=0@ 3

w w w w w w p 3f f f f f f f f f f f

2

=@

w w w w w w p 3f f f f f f f f f f f

2

w w w w w w w

Example : Evaluate Z 3 p x dx 1

Solution : 3f f f f f

w w w w w w w

As 2 x 2 is the antiderivative of 3 p x : 3

w w w w w w w

3f f f f f3

d

3f f f f f

e 3f f f f f

Z 3 p x dx = 2 x 2 | = 2 3 2 @1 2 1 1

w w w w w w

= 6 p3 @ 2

` a

B

C

PROPERTIES OF DEFINITE INTEGRALS Certain properties of definite integrals are needed to solve related problems. The common properties are : a

1 A Z f x dx = 0 a b

` a

a

2 A Z f x dx = @ Z f x dx ` a

a

` a

b

b

b

3 A Z c f x dx = c Z f x dx ` a

a

` a

If f x

` a

a

are continuous on the interval of integration a ≤ x ≤ b, then

` a

and g x b

b

B ` a

` aC

b

4 A Sum Rule : Z f x + g x dx = Z f x dx + Z g x dx a

a

b

` a

a

` a

b

B ` a

` aC

b

5 A Difference Rule : Z f x @ g x dx = Z f x dx @ Z g x dx a

` a

a

If a, b, c are any three points in a closed interval, then b

c

a

` a

b

6 A Z f x dx = Z f x dx + Z f x dx a

` a

` a

a

c

` a

7 A The Mean Value Theorem of Definite Integrals : ` a

B

C

If f x is continuous on the closed interval a,b , then atleast one number c exists b

c

in the open interval a,b such that b

Z f x dx = f c b @ a a

` a

` a`

` a

a ` a

The value of f c is called the average or mean value of the function f x on the B

C

interval a,b and f c = ` a

b

1f f f f f f f f f f f f f f f f fZ ` a

b@a

f x dx

a