On-line Math 21

6.2 Average value of a function

Example 1 Find the average value of sin(x) over

[0,p] ,
[0,2p]

Solution

For the first interval,

f_av

: =

1
p-0
ó
õ p

0
sin(x) dx

=

1
p
( -cos(x)) | ^p₀

=

1
p
( -cos(p)+cos(0))

=

2
p
.

You should first check to see that this is a reasonable answer. It is, since the graph of sin(x) on that interval looks like:

Note that an answer greater than 1/2 should be right (and of course, it should be bigger than 0 and less than 1!), since most of the time the graph is above the mark for height 0.5. Always verify your results by checking to make sure they are reasonable. Areas and the like have to be positive (averages don't, of course). Averages should certainly be between the maximum and the minimum of the function on the interval.

For that same function on the interval [0,2p] , things are a little different:

f_av

: =

1
2p-0
ó
õ 2p

0
sin(x) dx

=

1
2p
( -cos(x)) | ^2p₀

=

1
2p
( -cos(2p)+cos(0))

=

0.

0? Is that a reasonable answer? Actually, of course it is, since the graph:

spends just about as much time above and below the x -axis. By symmetry, it spends exactly as much time above the axis as below, so an average of 0 is to be expected.

File translated from T_EX by T_TH, version 2.61.
On 11 Jan 2001, 23:13.