The Chemistry Maths Book, Second Edition

146 Chapter 5Integration

The integral as a length

Lety 1 = 1 f(x)be continuous in the rangea 1 ≤ 1 x 1 ≤ 1 b, and let sbe the length of its graph

from point A, atx 1 = 1 a, to point B, atx 1 = 1 b(Figure 5.14). To calculate this length, we

divide the arc ABinto segments of length ∆s. Then, by Pythagoras’ theorem, an

approximate value of the length of the segment betweenxandx 1 + 1 ∆xis

The corresponding element of length is

(5.37)

and the total length of arc ABis

(5.38)

EXAMPLE 5.12The circumference of a circle

The equation of the circle of radius aand centre at the origin isx

2

1 + 1 y

2

1 = 1 a

2

so that

y 1 = 1 ±(a

2

1 − 1 x

2

)

122

anddy 2 dx 1 = 13 x 2 (a

2

1 − 1 x

2

)

122

. Therefore,

and the circumference of the circle is

sdxa

dx

a

aa

dy

dx

=+

































41 = 4

0

12

2

0

2

ZZ

−−x

2

11

2

2

22

2

22

• 
  













=+

−

=

−

dy

dx

x

ax

a

ax

sds

dy

dx

dx

s

a

b

==+

































ZZ

0

12

2

1

ds

dy

dx

=+ dx

































12

2

1

∆∆ ∆

∆

∆

sx y

y

x

≈+ =+



































12

22

12

2

() () 1















∆x

• 
  
  
  
  • 
    
  
  
  
  


• 
  


• 
  

A

B

a x x+∆ x b

∆ x

∆ y

∆ s

..

...

..

.

..

...

..

.

..

...

..

.

..

...

..

.

..

...

..

.

..

...

..

.

..

...

..

.

..

...

..

.

..

...

..

.

..

...

..

.

..

...

...

..

...

...

...

.

................................................

..

...

..

.

..

...

..

.

..

...

..

.

..

...

..

.

..

...

..

.

..

...

..

.

..

...

..

.

..

...

...

..

...

...

...

..

...

...

..

...

...

..

...

...

..

...

...

..

...

...........................

.................................

..................................

............................

..........................

............................

........................

......................

........................

....................

....................

.....................

..................

.................

...................

................

................

.................

..............

...............

...............

..............

.............

..............

.............

............

..............

...........

............

............

...........

...........

............

...........

..........

...........

..........

..........

...........

.........

.........

...........

.........

.........

..........

.........

........

..........

........

.........

.........

........

........

.........

........

........

.........

.......

........

.........

.......

........

........

.......

........

........

.......

.......

........

.......

.......

........

.......

.......

.......

....

Figure 5.14