—
1
—
Problem.
Averages
Consider the plot below in which we have drawn a continuous function of time and evaluated it at
the center of 10 intervals.
a) Write an expression for the average of these 10 points.
Solution.
< f
(
t
)
>
=
f
(
t
1
) +
f
(
t
2
) +
· · ·
+
f
(
t
10
)
10
b) Generalize this expression to
N
points and write it in “sigma” notation.
Solution.
< f
(
t
)
>
=
f
(
t
1
) +
· · ·
+
f
(
t
N
)
N
=
1
N
N
k
=1
f
(
t
k
)
c) Multiply and divide the expression above by the interval length
Δ
t
. What is
N
Δ
t
equal to?
Rewrite the expression using this fact.
Solution.
We use the fact that
N
Δ
t
=
T
to obtain
< f
(
t
)
>
=
1
N
Δ
t
N
k
=1
f
(
t
k
)Δ
t
=
1
T
N
k
=1
f
(
t
k
)Δ
t
d) Take the limit as
N
goes to infinity.
Note that
Δ
t
goes to
dt
, and the sum becomes an
integral. This limit is defined as the average value of a function on the interval
[0
, T
]
.
Solution.
Note that the expression in part (c) is a Riemann sum. Thus, when we take the
limit as
N
→ ∞
we obtain a definite integral:
< f
(
t
)
>
= lim
N
→∞
1
T
N
k
=1
f
(
t
k
)Δ
t
=
1
T
T
0
f
(
t
)
dt
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 Fall '07
 BERMAN
 Calculus, Energy, Average Power, energy company, trigonometric identity

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