2.2 INTEGRATION
m
f(x)
a b x
Figure 2.10 The mean valuemof a function.
Find the mean valuemof the functionf(x)=x^2 between the limitsx=2andx=4.
Using (2.40),
m=
1
4 − 2
∫ 4
2
x^2 dx=
1
2
[
x^3
3
] 4
2
=
1
2
(
43
3
−
23
3
)
=
28
3
.
Finding the length of a curve
Finding the area between a curve and certain straight lines provides one example
of the use of integration. Another is in finding the length of a curve. If a curve
is defined byy=f(x) then the distance along the curve, ∆s, that corresponds to
small changes ∆xand ∆yinxandyis given by
∆s≈
√
(∆x)^2 +(∆y)^2 ; (2.41)
this follows directly from Pythagoras’ theorem (see figure 2.11). Dividing (2.41)
through by ∆xand letting ∆x→0weobtain§
ds
dx
=
√
1+
(
dy
dx
) 2
.
Clearly the total lengthsof the curve between the pointsx=aandx=bis then
given by integrating both sides of the equation:
s=
∫b
a
√
1+
(
dy
dx
) 2
dx. (2.42)
§Instead of considering small changes ∆xand ∆yand letting these tend to zero, we could have
derived (2.41) by considering infinitesimal changesdxanddyfrom the start. After writing (ds)^2 =
(dx)^2 +(dy)^2 , (2.41) may be deduced by using the formal device of dividing through bydx. Although
not mathematically rigorous, this method is often used and generally leads to the correct result.