420 Integration (Chapter 15)

²

Zb

a

f(x)dx+

Zc

b

f(x)dx=

Zc

a

f(x)dx

²

Zb

a

[f(x)§g(x)]dx=

Zb

a

f(x)dx§

Zb

a

g(x)dx

Example proof:

Example 4 Self Tutor

Use the fundamental theorem of calculus to find the area between:

a thex-axis and y=x^2 from x=0to x=1

b thex-axis and y=

p

x from x=1to x=9.

f(x)=x^2 has antiderivative F(x)=

x^3

3

) the area=

Z 1

0

x^2 dx

=F(1)¡F(0)

=^13 ¡ 0

=^13 units^2

b f(x)=

p

x=x

1

(^2) has antiderivative
F(x)=
x
3
2
3
2
=^23 x
p
x
) the area=
Z 9
1
x
1
(^2) dx
=F(9)¡F(1)
=^23 £ 27 ¡^23 £ 1
=17^13 units^2
a
1
y
x
y=x 2
O
y
19 x
y = ~`x
O
Zb
a
f(x)dx+
Zc
b
f(x)dx
=F(b)¡F(a)+F(c)¡F(b)
=F(c)¡F(a)

Zc
a
f(x)dx
y
x
y = f(x)
A 1 A 2
O abc
In particular, for the case where a 6 b 6 c and f(x)> 0 for a 6 x 6 c, we observe that
Zb
a
f(x)dx+
Zc
b
f(x)dx=A 1 +A 2 =
Zc
a
f(x)dx
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(^05255075950525507595)
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Additional Mathematics
Y:\HAESE\CAM4037\CamAdd_15\420CamAdd_15.cdr Monday, 7 April 2014 3:58:28 PM BRIAN