# The MeanValue Theorem for Integrals and the First Fundamental by t0231232

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The Mean Value Theorem for Integrals
and the
First Fundamental Theorem of Calculus

This handout will present a proof of the First FTC that is simpler to understand than the one in the text.
The proof uses the so-called Mean Value Theorem for Integrals, which is stated and proved below. In order
to follow the proofs, you need to be familiar with the properties of the integral (text, pp.373–375 and class
notes from December 3).
Mean Value Theorem for Integrals. Let v be continuous on the interval [a, b]. Then there is some
number t∗ on [a, b] such that
b
1
v(t∗ ) =         v(t) dt.                                    (1)
b−a a
Proof . Since v is continuous on [a, b], v has minimum value m and maximum value M on [a, b], and

b
m(b − a) ≤                   v(t) dt ≤ M (b − a);
a

then, dividing through by (b − a) gives

b
1
m ≤                              v(t) dt ≤ M.
b−a           a

Now apply the Intermediate Value Theorem: since v assumes the values m and M , it must assume all values
b
1
in between m and M , including the particular value          v(t) dt. In other words: there is at least one
b−a a
∗
number t on the interval that satisﬁes (1).
Theorem (First FTC). Let v be continuous on [a, b], and for any a ≤ x ≤ b, put
x
s(x) :=                    v(t) dt.
a

Then for each x in the interval [a, b], s (x) = v(x).
Proof . By deﬁnition,
s(x + h) − s(x)
s (x) = lim                           ,                                   (2)
h→0                 h
[A]

so we need to ﬁnd a way to evaluate this limit. The ﬁrst step is to observe that
x+h                              x                x+h
s(x + h) − s(x) =              v(t) dt −                      v(t) dt =          v(t) dt.
a                                a                   x

Second, divide by h (to get [A]).. This gives

x+h
s(x + h) − s(x)   1
[A] =                   =                                  v(t) dt.
h          h                         x

Next, apply the Mean Value Theorem for Integrals in the interval [x, x + h]1 get yet another expression
for [A]: for some t∗ in this interval, we have
1
Writing the interval this way assumes that h > 0; for h < 0, the proof is identical, but the notation has
to be rearranged slightly.

1
x+h
1
[A] =             v(t) dt = v(t∗ ).
h   x

Now, we can take the limit in (2), at least from the right:

lim [A] = lim+ v(t∗ )
h→0+       h→0
(because t∗ → x as h → 0 =⇒)   +

= lim v(t∗ )
∗t →x
(because v is continuous at x =⇒)
= v(x).

For negative h, as noted in the footnote, there is a completely analogous argument on intervals [x + h, x],
which shows that
lim [A] = v(x).
h→0−

That means that the two-sided limit also equals v(x); that is,

s (x) = lim [A] = v(x).
h→0

2

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