Cambridge Additional Mathematics

Integration (Chapter 15) 413

THE DEFINITE INTEGRAL

Consider the lower and upper rectangle sums for a function which is positive and increasing on the interval

a 6 x 6 b.

We divide the interval intonsubintervals, each of width w=

b¡a

n

.

Since the function is increasing:

AL=wf(x 0 )+wf(x 1 )+::::+wf(xn¡ 2 )+wf(xn¡ 1 )=w

nX¡ 1

i=0

f(xi)

AU=wf(x 1 )+wf(x 2 )+::::+wf(xn¡ 1 )+wf(xn)=w

Xn

i=1

f(xi)

Notice that AU¡AL=w(f(xn)¡f(x 0 ))

=

1

n

(b¡a)(f(b)¡f(a))

) lim

n!1

(AU¡AL)=0 fsince lim

n!1

1

n

=0g

) lim

n!1

AL= lim

n!1

AU fwhen both limits existg

) since AL<A<AU for all values ofn, it follows that

lim

n!1

AL=A= lim

n!1

AU

This fact is true for all positive continuous functions on an interval a 6 x 6 b.

The valueAis known as the “definite integralof f(x) fromatob”, written A=

Zb

a

f(x)dx.

If f(x)> 0 for all a 6 x 6 b, then

Zb

a

f(x)dx is equal to the shaded area.

lim

n!1

means we

have infinitely

many subintervals.

x

y

ab

y = f(x)

O

A

abx 1 x 2 x 3

x 0

xn-2

xn-1xn

y

x

y = (x)f

....

abx 1 x 2 x 3 O

x 0

xn-2

xn-1xn

y

x

y = f(x)

....

O

Pn

i=1

ai=a 1 +a 2 +::::+an

4037 Cambridge

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