Fundamental Theorem of
Calculus. Part I:
Connection between integration
and differentiation
Motivation:
Problem of finding antiderivatives
Definition: An antiderivative of a function f (x)
is a function F (x) such that F 0(x) = f (x).
In other words, given the function f (x), you want
to tell whose derivative it is.
Example 1. Find an antiderivative of 1.
An answer: x.
1
Example 2. Find an antiderivative of
.
2
1+x
An answer: arctan x.
!!..??!..!?
HOW DO YOU KNOW?
Some antiderivatives can be found by reading
differentiation formulas backwards:
α 0
α−1
[x ]
= αx
[cos x]0
= − sin x
[tan x]0
=
[arctan x]0 =
0
[arcsin x]
=
1
cos2 x
1
1+x2
√ 1
1−x2
Limitation: there aren’t formulas for
2
(x2)
sin(x ),
e ,
q
2
x + sin (1 − x3)
?!?
Do these functions have antiderivatives?
An observation
No matter what object’s velocity v(t) is, its
position d(t) is always an antiderivative of v(t):
d0(t) = v(t). This suggests that all functions
have antiderivatives.
Antiderivative of sin(t2) exists!
Suppose the speed of my car obeys sin(t2)
(do not try it on the road!). The car will move
accordingly and the position of the car F (t)
2
will give the antiderivative of sin(t ).
A hypothesis
Calculating antiderivatives must be similar to
calculating position from velocity
Fundamental Theorem of Calculus
Naive derivation
Let, at initial time t0, position of the car on
the road is d(t0) and velocity is v(t0). After a
short period of time ∆t, the new position of
the car is approximately
d(t1) ≈ d(t0) + v(t0)∆t,
t3
t0
4
hc
v(t0 )
h4
∆t
v(t0 ) ∆t
t1
t
4 h2
4
(t1 = t0 + ∆t)
4
h
After two moments of time
d(t2) ≈ d(t0) + v(t0)∆t + v(t1)∆t,
(t2 = t1 + ∆t)
t3
t0
h
4
h4
∆t
v(t0 ) ∆t
t1
t2
h
4
4
∆t
hc
v(t1 )
4
v(t1 ) ∆t
h
After n moments of time
d(tn) ≈ d(t0) + v(t0)∆t + . . . + v(tn−1)∆t
= d(t0) +
n−1
X
v(ti)∆t
i=0
(tn = t0 + n∆t)
t3
t0
h
4
v(tn-1 )∆t
h4
∆t
v(t0 ) ∆t
t1
4
4
h
∆t
h t2
4
v(t1 ) ∆t
hc
v(tn )
As ∆t → 0, (∆t = (t − t0)/n).
d(t) − d(t0) = lim
n−1
X
∆t→0
v(ti)∆t
i=0
Compare this to
Z t
n−1
X
v(ti)∆t
v(τ )dτ = lim
t0
∆t→0
i=0
Combining the two formulas
Z
t
d(t) − d(t0) =
v(τ )dτ
t0
while
d0(t) = v(t)
Assuming t0 = 0, d(t0) = 0, distance can be
calculated from velocity by
Z t
d(t) =
v(τ )dτ
0
Is this a function in our regular sense? Yes, for
each value of t it defines a unique number.
If d0(t) = v(t)? Yes. This constitutes the
assertion of Fundamental Theorem of Calculus.
THEOREM. Let f (x) be a continuous
function (so, the definite integral of f (x)
exists). Then the function
Z x
F (x) =
f (τ )dτ.
a
is an antiderivative of f (x), F 0(x) = f (x).
Z x
d
f (τ )dτ = f (x)
dx a
EXAMPLES
d
x
Z
f (τ )dτ = f (x)
dx
a
Example 3. Find an antiderivative of
2
f (x) = sin(x ).
Z
An answer: F (x) =
x
2
sin(τ )dτ .
0
d
x
Z
f (τ )dτ = f (x)
dx
a
Example 4. Find an antiderivative of
(x2)
f (x) = e
Z
An answer: G(x) =
x
e
0
(τ 2)
.
dτ .
d
dx
Z
x
f (τ )dτ = f (x)
a
Example 5. Find an antiderivative of
q
2
f (x) = x + sin (1 − x3).
An answer:
Z xq
2
H(x) =
τ + sin (1 − τ 3)dτ .
0
d
dx
Z
x
f (τ )dτ = f (x)
a
Example 6. Calculate F 0(x) if
Z x
F (x) =
sin(τ 2)dτ
0
Answer: sin(x2).
d
dx
Z
x
f (τ )dτ = f (x)
a
Example 7. Calculate G0(x) if
Z x
(τ 2)
G(x) =
e dτ
0
Answer: e
(x2)
.
Z
d
dx
x
f (τ )dτ = f (x)
a
Example 8. Calculate H 0(x) if
Z xq
2
H(x) =
τ + sin (1 − τ 3)dτ
0
Answer:
q
2
x + sin (1 − x3).
d
dx
Z
x
f (τ )dτ = f (x)
a
0
Example 9. Calculate F (x) if
Z 0
F (x) =
sin(τ 2)dτ
x
Solution:
d
F (x) =
d
"Z
#
0
sin(τ 2)dτ
dx
dx x
Z x
d
−
sin(τ 2)dτ = − sin(x2)
dx
=
d
Z
x
f (τ )dτ = f (x)
dx
a
0
Example 10. Calculate G (x) if
Z 5x
G(x) =
sin(τ 2)dτ
1
Trick. Introduce F (x) =
G(x) = F (5x).
0
Rx
1
2
sin(τ )dτ , so,
2
Use FTC to calculate F (x) = sin(x ).
Use the Chain Rule to find the derivative
d
dx
0
G(x) =
0
d
dx
F (5x)
2
= F (5x)[5x] = sin((5x) )5.
d
dx
Z
x
f (τ )dτ = f (x)
a
0
Example 11. Calculate H (x) if
Z x3
2
H(x) =
sin(τ )dτ
π
Trick. Introduce F (x) =
H(x) = F (x3).
Rx
π
sin(τ 2)dτ , so,
Use FTC to calculate F 0(x) = sin(x2).
Use the Chain Rule to find the derivative
d
dx
0
H(x) =
3
3 0
d
dx
3
F (x )
3 2
2
= F (x )[x ] = sin((x ) )3x .
Example 12. Calculate G0(x) if
Z x3
2
G(x) = √ sin(τ )dτ
x
Solution: First, notice that
Z 0
Z
G(x) = √ sin(τ 2)dτ +
x
x3
sin(τ 2)dτ
0
Rx
Trick. Introduce F (x) = 0 sin(τ 2)dτ , so,
√
3
G(x) = −F ( x) + F (x ).
Use FTC to calculate F 0(x) = sin(x2).
Use the Chain Rule to find the derivative
d
G(x) = −
d
√
F ( x) +
d
3
F (x )
dx
dx
√ √ 0
0
= −F ( x)[ x] + F 0(x3)[x3]0
√ 2 1
3 2
2
= − sin(( x) ) √ + sin((x ) )3x .
2 x
dx
Control exercises
Example 13. Calculate F 0(x) if
Z 1+x2
F (x) =
sin(τ 2)dτ
1−x2
Example 14. Calculate G0(x) if
Z x
G(x) =
x2 sin(τ 2)dτ
1
Fundamental Theorem of Calculus
Proof of Part I
We will be using two facts
Interval Additive Property:
Z b
Z c
Z
f (τ )dτ =
f (τ )dτ +
a
a
b
f (τ )dτ
c
Comparison Property: If m ≤ f (x) ≤ M on
[a, b]
Z b
m(b − a) ≤
f (τ )dτ ≤ M (b − a)
a
x
Z
Let F (x) =
f (τ )dτ .
a
We will now show that F 0(x) = f (x).
By definition
0
F (x) = lim
F (x + h) − F (x)
h→0
R x+h
= lim
h→0
a
f (τ )dτ −
h
h
Rx
a
f (τ )dτ
Notice, that by the Interval Additive Property
Z x+h
Z x
Z x+h
f (τ )dτ −
f (τ )dτ =
f (τ )dτ
a
a
x
R x+h
Rx
Therefore,
0
F (x) = lim
a
f (τ )dτ −
h
h→0
= lim
h→0
a
1
h
Z
x+h
f (τ )dτ
x
f (τ )dτ
Also, by the Comparison Property, for
m=
min
τ ∈[x,x+h]
f (τ ),
M =
max f (τ )
τ ∈[x,x+h]
one has
Z
x+h
mh ≤
f (τ )dτ ≤ M h
x
or, dividing by h,
Z x+h
1
m≤
f (τ )dτ ≤ M
h x
Finally, by the Squeeze Theorem, expression
Z x+h
1
m≤
f (τ )dτ ≤ M
h x
converges to
f (x) ≤ lim
h→0
1
h
Z
x+h
f (τ )dτ ≤ f (x)
x
since both m = min f (τ ) and M = max f (τ )
on [x, x + h] coincide with f (x) as h → 0
The theorem is proved.