1. The Fundamental Theorem of Calculus
Allan Zamierowski
6 June 2013
What is the Fundamental Theorem of Calculus?
Integration is one of the two main operations seen in calculus (along with differentiation,
its inverse). In single variable calculus, integration is important for finding the area under
the curve of functions as well as finding the volume of objects rotated around some axis or
line, such as shells and discs. In multivariable calculus, certain types of integration can be
used to find areas or volumes that are more complicated, such as line integration or multi-
ple integration. In these settings, we learn the rules of integration and how to apply them
solve problems, but learning where they come from and how they are proven often occurs in
more advanced classes. The two biggest results are the first two fundamental theorems of
calculus. Very broadly, the first fundamental theorem of calculus tells us that differentiation
and integration undo each other, that they are essentially inverse operations. The second
fundamental theorem allows us to use the antiderivative of a function to evaluate its defi-
nite integral. Both concepts link differentiation and integration and are extremely useful for
solving integral-related problems in calculus.
Formal Description and Proof
Fundamental Theorem of Calculus I
Suppose the function f is defined and continuous on the closed interval [a, b]. Then
d
dx
x
a
f(t) dt = f(x) for all x ∈ (a, b).
Proof. Let x ∈ (a, b) be given. Since (a, b) is open, there exists an h > 0 sufficiently small
such that [x, x+h] ⊂ (a, b) by the definition of an open set. Since f is continuous on (a, b), it
is also continuous of any subset of (a, b). Then f is continuous on [x, x + h]. By the extreme
value theorem, there exist two numbers c, d ∈ [x, x + h] where f(c) is the minimum value
and f(d) is the maximum value of f on this closed interval. Since f(c) ≤ f(y) ≤ f(d) for all
y ∈ [x, x + h],
f(c) · h ≤
x+h
x
f(t) dt ≤ f(d) · h.
Now let g be the function
g(x) =
x
a
f(t) dt
and notice that
g(x + h) − g(x) =
x+h
x
f(t) dt
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2. by manipulation of functions of integrals. If we substitute this result into the original set of
inequalities, we have
f(c) · h ≤ g(x + h) − g(x) ≤ f(d) · h,
and dividing by h yields
f(c) ≤
g(x + h) − g(x)
h
≤ f(d).
As h → 0, c → x and d → x, and we have
f(x) ≤ lim
h→0
g(x + h) − g(x)
h
≤ f(x).
Then by the definition of derivative and squeeze theorem, we finally have
f(x) =
d
dx
g(x) =
d
dx
x
a
f(t) dt.
Fundamental Theorem of Calculus II
Suppose the function f is defined and continuous on the closed interval [a, b], and its an-
tiderivative F is defined on [a, b] such that f = F . Then
b
a
f(x) dx = F(b) − F(a).
Proof. Take [a, b] and divide it into n uniform subintervals with length x = (b − a)/n
where a = x0, x1, ..., xn−1, xn = b are the endpoints. Take F(b) − F(a) and construct a
telescoping sum where
F(b) − F(a) = F(xn) − F(x0)
= F(xn) − F(xn−1) + F(xn−1) − · · · − F(x1) + F(x1) − F(x0)
=
n
i=1
[F(xi) − F(xi−1)]
Since F is differentiable on [a, b], the mean value theorem tells us that each interval [xi−1, xi]
contains an x∗
i such that
F(xi) − F(xi−1) = F (x∗
i ) x = f(x∗
i ) x.
Then we have
F(b) − f(a) =
n
i=1
f(x∗
i ) x.
Finally, taking the limit as n goes to ∞ and using the definition of integral, we have
F(b) − F(a) = lim
n→∞
n
i=1
f(x∗
i ) x =
b
a
f(x) dx.
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3. Uses of the Fundamental Theorem of Calculus
Example 1
Evaluate
d
dx
x
1.4
t7
dt.
This integral looks scary because it deals with two different variables, a decimal lower bound,
and a seventh degree polynomial function. However, since f is defined and continuous
everywhere, we can invoke the first fundamental theorem of calculus and easily see
d
dx
x
1.4
t7
dt = x7
.
Example 2
Evaluate
3
1
3x2
+ 5 dx.
Since f is defined and continuous on [1, 3], and since we know the antiderivative is
F(x) = x3
+ 5x + C (by simple calculus-based arithmetic), which is defined on [1, 3], we can
use the second fundamental theorem of calculus to show
3
1
3x2
+ 5 dx = F(3) − F(1)
= [(3)3
+ 5(3) + C] − [(1)3
+ 5(1) + C]
= 36.
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