SME
Calculus I · Problem type
Average value of a function
Use this for “Find the average value of f(x) = 2x² on [1, 4], and the c in [1, 4] where f(c) equals it.”
Chart
Shaded: ∫ from 0 to 3 of f(x) dx = 9
Level
favg
=
3
meets
the
curve
at
c
=
√
3
This is an example. Type your own values to start a new problem.
f
avg
=
1
−
∫
p
x²
dx
=
favg
f(
c
)
=
favg
I
Integral of f from a to b
worked out (not typed)
Formulas
I
=
p
×
b
3
3
−
p
×
a
3
3
9
=
1
×
3
3
3
−
1
×
0
3
3
favg
=
I
b
−
a
3
=
9
3 − 0
p
×
c
2
=
favg
1
×
1.7321
2
=
3
Assumptions
•
The average value of f on [a, b] is favg = I ÷ (b − a), where I is the integral of f from a to b: the area spread evenly across the interval.
•
An antiderivative of f(x) = px² is F(x) = px³ ÷ 3, so I = F(b) − F(a), by the Fundamental Theorem.
•
The Mean Value Theorem for integrals: a continuous f equals its average at some c in [a, b]. The rectangle of height favg over [a, b] has area I.
•
f(c) = pc² = favg gives c = √(favg ÷ p) or −√(favg ÷ p). Only a root inside [a, b] counts.
Step-by-step
WE KNOW
p = 1, a = 0, b = 3
FIND
integral of f from a to b (I), average value of f on [a, b] (favg), point where f(c) = favg (c)
Step 1 · Find integral of f from a to b (I)
Use I = p × b³ ÷ 3 − p × a³ ÷ 3
Raise
the
power
by
1
and
divide
by
the
new
power:
F(x)
=
x
3
3
.
Take
F(
b
)
−
F(
a
).
I
=
3
3
3
−
0
3
3
3
∫
0
x
2
dx
=
[
x
3
3
]
from
0
to
3
=
9
I
=
9
Step 2 · Find average value of f on [a, b] (favg)
Use favg = I ÷ (b − a)
Divide
the
integral
by
the
width
of
the
interval:
the
height
of
a
rectangle
over
[
a
,
b
]
with
the
same
area.
favg
=
9
3 − 0
favg = 3
Step 3 · Find point where f(c) = favg (c)
Use p × c² = favg
Set
f
(
c
)
=
pc²
equal
to
favg
and
divide
by
p
,
then
take
the
square
root.
Keep
the
root
that
lies
in
[
a
,
b
].
c
=
√(
favg
p
)
c
=
√(
3
1
)
c
=
1.7321
Check
Put the numbers back in. Both sides should match.
9
=
1
×
3
3
3
−
1
×
0
3
3
✓
3
=
9
3 − 0
✓
1
×
1.7321
2
=
3
✓
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