Begin2.DVI

We know that the functions f(x)and f−^1 (x)are symmetric about the line y=x

as illustrated in the figure 12-1. Examine the figure 12-1 and note that one can

express the area A=

∫b

af(x)dx as

A=

∫b

a

f(x)dx = (b−a)d

︸ ︷︷ ︸

orange+green area

−

∫d

c

[f−^1 (x)−a]dx

︸ ︷︷ ︸

green area

= (b−a)d−

∫d

c

f−^1 (x)dx + (d−c)a

(12.1)

which shows that the area A is given by the rectangular area (orange plus green

area) minus the area under the inverse curve (green plus grey area) corrected by the

rectangular grey area. Hence, if you know

∫b

a

f(x)dx , then you can find

∫d

c

f−^1 (x)dx

and vice-versa.

The use of integration to sum infinite series

There is a definite relation between certain infinite series and definite integrals.

For example, consider the relationship between the problem of finding the sum of

the alternating infinite series

1

a

−^1

a+b

+^1

a+ 2 b

−^1

a+ 3b

+^1

a+ 4 b

−···+ (−1)n^1

a+nb

+··· (12.2)

where a > 0 ,b > 0 and the associated problem of evaluating the definite integral

∫ 1

0

ta−^1

1 + tbdt (12.3)

Use the well known series expansion

1

1 + x= 1 −x+x

(^2) +x (^3) −x (^4) +··· (12.4)

with xreplaced by tbto write the equation (12.3) in the form

∫ 1

0

ta−^1

1 + tb

dt =

∫ 1

0

ta−^1

[

1 −tb+t^2 b−t^3 b+t^4 b−···

]

dt (12.5)

and then integrate each term to produce the result

∫ 1

0

ta−^1

1 + tbdt =

[ta

a−

ta+b

a+b+

ta+2b

a+ 2 b+···+ (−1)

nta+nb

a+nb +···

] 1

∫^0

1

0

ta−^1

1 + tbdt =

1

a−

1

a+b+

1

a+ 2 b+···+ (−1)

n^1

a+nb +···

(12.6)

This demonstrates that if the infinite series on the right-hand side converges, then

it can be evaluated by calculating the integral on the left-hand side.