Cambridge Additional Mathematics

Integration (Chapter 15) 411

The next diagram showslower rectangles, which are rectangles

with top edges at the minimum value of the curve on that

subinterval.

The area of the lower rectangles,

AL=1£f(1) + 1£f(2) + 1£f(3)

=2+5+10

=17units^2

Now clearly AL<A<AU, so the areaAlies between 17 units^2 and 32 units^2.

If the interval 16 x 64 was divided into 6 subintervals, each of width^12 , then

AU=^12 f(1^12 )+^12 f(2) +^12 f(2^12 )+^12 f(3) +^12 f(3^12 )+^12 f(4)

=^12 (^134 +5+^294 +10+^534 + 17)

=27: 875 units^2

and AL=^12 f(1) +^12 f(1^12 )+^12 f(2) +^12 f(2^12 )+^12 f(3) +^12 f(3^12 )

=^12 (2 +^134 +5+^294 +10+^534 )

=20: 375 units^2

From this refinement we conclude that the areaAlies between 20 : 375 and 27 : 875 units^2.

As we create more subintervals, the estimatesALandAUwill become more and more accurate. In fact, as

the subinterval width is reduced further and further, bothALandAUwillconvergetoA.

We illustrate this process by estimating the areaAbetween the graph of y=x^2 and thex-axis for

06 x 61.

This example is of historical interest.Archimedes( 287 - 212 BC) found the exact area. In an article that

contains 24 propositions, he developed the essential theory for what is now known as integral calculus.

Consider f(x)=x^2 and divide the interval 06 x 61 into 4 subintervals of equal width.

AL=^14 (0)^2 +^14 (^14 )^2 +^14 (^12 )^2 +^14 (^34 )^2 and AU=^14 (^14 )^2 +^14 (^12 )^2 +^14 (^34 )^2 +^14 (1)^2

¼ 0 : 219 ¼ 0 : 469

Now suppose there arensubintervals between x=0and x=1, each of width

1

n

.

We can use thearea findersoftware to help calculateALandAUfor large values ofn.

AREA

FINDER

y

x

(1 1),

1

1

¦¡¡(x) = x 2

O

y

x

(1 1),

1

1

¦¡¡(x) = x 2

O

1 2 3 4

20

15

10

5

y

x

2 5

10

17

11

f(x) = x + 1 2

O

4037 Cambridge

cyan magenta yellow black Additional Mathematics

(^05255075950525507595)
100 100
(^05255075950525507595)
100 100
Y:\HAESE\CAM4037\CamAdd_15\411CamAdd_15.cdr Monday, 7 April 2014 3:56:27 PM BRIAN