The Fundamental Theorem of Calculus
RELATED CALCULATOR: Definite and Improper Integral Calculator
When we introduced definite integrals, we computed them according to the definition as the limit
of Riemann sums and we saw that this procedure is not very easy. In fact, there is a much simpler
method for evaluating integrals.
We already discovered it when we talked about the area problem for the first time.
There, we introduced a function P (x) whose value is the area under the function f on the interval
[a, x] (x can vary from a to b).
x
Now, when we know about definite integrals, we can write that P (x) = ∫a f (t)dt (note that we
changed x to t under the integral in order not to mix it with the upper limit).
Also, we discovered the Newton-Leibniz formula, which states that P ′ (x) = f (x) and P (x) =
′
F (x) − F (a), where F = f .
Here, we are going to formalize this result and give another proof, because these facts are very
important in calculus: they connect differential calculus with integral calculus.
Fundamental Theorem of Calculus. Suppose that f is continuous on [a, b].
x
1. If P (x) = ∫a f (t)dt, then P ′ (x) = f (x).
b
2. ∫a f (x)dx = F (b) − F (a), where F is any antiderivative of f , that is F ′ = f .
d x
Part 1 can be rewritten as dx ∫a f (t)dt = f (x), which says that if f is integrated and then the result
is differentiated, we arrive back at the original function.
This study source was downloaded by 100000898062787 from CourseHero.com on 09-29-2025 04:46:52 GMT -05:00
https://www.coursehero.com/file/250856133/fundamental-theorem-calculuspdf/
, b
Part 2 can be rewritten as ∫a F ′ (x)dx = F (b) − F (a), and it says that if we take a function F , first
differentiate it, and then integrate the result, we arrive back at the original function F but in the
form F (b) − F (a).
The fundamental theorem of calculus says that differentiation and integration are the inverse
processes.
x
Proof of Part 1. Let P (x)= ∫a f (t)dt. If x and x + h are in the open interval (a, b), then
x+h x
P (x + h) − P (x) = ∫a f (t)dt − ∫a f (t)dt.
x+h x
Now, use the adjacency property of the integral: ∫a f (t)dt − ∫a f (t)dt =
x x+h x x+h
(∫a f (t)dt
+ ∫x f (t)dt)
− ∫a f (t)dt
= ∫x f (t)dt.
Now, apply the mean value theorem for integrals:
x+h
∫x f (t)dt = n(x + h − x) = nh, where m′ ≤ n ≤ M ′ (M ′ is the maximum value, and m′ is the
minimum value of f on [x, x + h]).
So, we obtained that P (x + h) − P (x) = nh. If we let h → 0, then P (x + h) − P (x) → 0, or
P ( x + h ) → P ( x) .
This proves that P (x) is a continuous function.
Without the loss of generality, assume that h > 0.
Since f is continuous on [x, x + h], the extreme value theorem says that there are numbers c
and d in [x, x + h] such that f (c)
= m and f (d) = M , where m and M are the minimum and
maximum values of f on [x, x + h].
x+h
According to the comparison property, we have m(x + h − x) ≤ ∫x f (t)dt ≤ M (x + h − h), or
x+h
mh ≤ ∫x f (t)dt
≤ M h.
x+h P (x+h)−P (x)
This can be divided by h > 0: m ≤ 1h ∫x f (t)dt ≤ M , or m ≤
h
≤ M.
P (x+h)−P (x)
Finally, f (c) ≤ h
≤ f (d).
This study source was downloaded by 100000898062787 from CourseHero.com on 09-29-2025 04:46:52 GMT -05:00
https://www.coursehero.com/file/250856133/fundamental-theorem-calculuspdf/
RELATED CALCULATOR: Definite and Improper Integral Calculator
When we introduced definite integrals, we computed them according to the definition as the limit
of Riemann sums and we saw that this procedure is not very easy. In fact, there is a much simpler
method for evaluating integrals.
We already discovered it when we talked about the area problem for the first time.
There, we introduced a function P (x) whose value is the area under the function f on the interval
[a, x] (x can vary from a to b).
x
Now, when we know about definite integrals, we can write that P (x) = ∫a f (t)dt (note that we
changed x to t under the integral in order not to mix it with the upper limit).
Also, we discovered the Newton-Leibniz formula, which states that P ′ (x) = f (x) and P (x) =
′
F (x) − F (a), where F = f .
Here, we are going to formalize this result and give another proof, because these facts are very
important in calculus: they connect differential calculus with integral calculus.
Fundamental Theorem of Calculus. Suppose that f is continuous on [a, b].
x
1. If P (x) = ∫a f (t)dt, then P ′ (x) = f (x).
b
2. ∫a f (x)dx = F (b) − F (a), where F is any antiderivative of f , that is F ′ = f .
d x
Part 1 can be rewritten as dx ∫a f (t)dt = f (x), which says that if f is integrated and then the result
is differentiated, we arrive back at the original function.
This study source was downloaded by 100000898062787 from CourseHero.com on 09-29-2025 04:46:52 GMT -05:00
https://www.coursehero.com/file/250856133/fundamental-theorem-calculuspdf/
, b
Part 2 can be rewritten as ∫a F ′ (x)dx = F (b) − F (a), and it says that if we take a function F , first
differentiate it, and then integrate the result, we arrive back at the original function F but in the
form F (b) − F (a).
The fundamental theorem of calculus says that differentiation and integration are the inverse
processes.
x
Proof of Part 1. Let P (x)= ∫a f (t)dt. If x and x + h are in the open interval (a, b), then
x+h x
P (x + h) − P (x) = ∫a f (t)dt − ∫a f (t)dt.
x+h x
Now, use the adjacency property of the integral: ∫a f (t)dt − ∫a f (t)dt =
x x+h x x+h
(∫a f (t)dt
+ ∫x f (t)dt)
− ∫a f (t)dt
= ∫x f (t)dt.
Now, apply the mean value theorem for integrals:
x+h
∫x f (t)dt = n(x + h − x) = nh, where m′ ≤ n ≤ M ′ (M ′ is the maximum value, and m′ is the
minimum value of f on [x, x + h]).
So, we obtained that P (x + h) − P (x) = nh. If we let h → 0, then P (x + h) − P (x) → 0, or
P ( x + h ) → P ( x) .
This proves that P (x) is a continuous function.
Without the loss of generality, assume that h > 0.
Since f is continuous on [x, x + h], the extreme value theorem says that there are numbers c
and d in [x, x + h] such that f (c)
= m and f (d) = M , where m and M are the minimum and
maximum values of f on [x, x + h].
x+h
According to the comparison property, we have m(x + h − x) ≤ ∫x f (t)dt ≤ M (x + h − h), or
x+h
mh ≤ ∫x f (t)dt
≤ M h.
x+h P (x+h)−P (x)
This can be divided by h > 0: m ≤ 1h ∫x f (t)dt ≤ M , or m ≤
h
≤ M.
P (x+h)−P (x)
Finally, f (c) ≤ h
≤ f (d).
This study source was downloaded by 100000898062787 from CourseHero.com on 09-29-2025 04:46:52 GMT -05:00
https://www.coursehero.com/file/250856133/fundamental-theorem-calculuspdf/
