Cambridge International AS and A Level Mathematics Pure Mathematics 1

P1^

6

Area as the limit of a sum

Notice  that    in  the limit:

●●●    is  replaced    by  =

●●●δx   is  replaced    by  dx

●●●Σ●is replaced    by  ∫,  the integral    sign    (the    symbol  is  the Old English letter  S)

●●●instead  of  summing for i = 1   to  n the   process is  now carried out over    a   range   

of  values  of  x (in   this    case    1   to  5), and these   are called  the limits of   the integral.   

(Note   that    this    is  a   different   meaning of  the word    limit.)

This    method  must    give    the same    results as  the previous    one which   used    d

d

A

x

= y,    

and at  this    stage   the notation    F x a

b

[]()

is  used    again.

In  this    case

∫

5

1 (x

(^2) + 1) dx = x^3 x
1
5
3 +









.

Recall  that    this    notation    means:  find    the value   of  x

3

(^3)

• x when x = 5 (the upper
  
  limit) and subtract the value of x
  3
  (^3)


• x when x = 1 (the lower limit).
  x^3 x
  1
  (^533)
  3

5

3 5

1

3 145

1

+ 3









=+ – + =.

So  the area    A is    45^13

square  units.

EXAMPLE 6.6  Find the area under the curve y = 4 x^3 + 4 between x = −1 and x = 2.

SOLUTION

The graph   is  shown   in  figure  6.11.

The shaded  part,   A

= (^) ∫−^21 (4x^3 + 4) dx
The limits have now
moved to the right of the
square brackets.
y
2
4
–1 x
A
O
Figure 6.11 

=+

=+ +

=

[]

(())–((–)(–))

xx–

4

1

2

4 4

4

2421 41

27 squareunitts.