Section 5.4 The Fundamental Theorem of Calculus
2010 Kiryl Tsishchanka
The Fundamental Theorem of Calculus
EXAMPLE: If f is a function whose graph is shown below and g(x) =
Zx
f (t)dt, find the values
0
of g(0), g(1), g(2), g(3), g(4), and g(5). Then sketch a rough graph of g.
Solution: First we notice that g(0) =
Z0
f (t)dt = 0. From the figure above we see that g(1) is
0
the area of a triangle:
g(1) =
Z1
1
f (t)dt = (1 · 2) = 1
2
0
To find g(2) we add to g(1) the area of a rectangle:
g(2) =
Z2
0
f (t)dt =
Z1
f (t)dt +
Z2
f (t)dt = 1 + (1 · 2) = 3
1
0
We estimate that the area under f from 2 to 3 in about 1.3, so
g(3) = g(2) +
Z3
f (t)dt ≈ 3 + 1.3 = 4.3
2
For t > 3, f (t) is negative and so we start subtracting areas:
g(4) = g(3) +
Z4
f (t)dt ≈ 4.3 + (−1.3) = 3
g(5) = g(4) +
3
Z5
4
1
f (t)dt ≈ 3 + (−1.3) = 1.7
Section 5.4 The Fundamental Theorem of Calculus
2010 Kiryl Tsishchanka
We use these values to sketch the graph of g:
g(0) = 0
g(1) = 1
g(2) = 3
g(3) ≈ 4.3
g(4) ≈ 3
g(5) ≈ 1.7
EXAMPLE: If g(x) =
Zx
f (t)dt, where a = 0 and f (t) = sin t, find a formula for g(x) and
a
calculate g ′ (x).
Solution: By the Evaluation Theorem we have:
g(x) =
Zx
sin tdt = − cos t]x0 = − cos x − (− cos 0) = − cos x + 1
0
d
(− cos x + 1) = sin x.
dx
REMARK: To see why this might be generally true we consider a continuous function f with
f (x) ≥ 0. Then
Zx
g(x) = f (t)dt
Then g ′ (x) =
a
To compute g ′ (x) from the definition of derivative we first observe that, for h > 0, g(x+h)−g(x)
is obtained by subtracting areas, so it is the area under the graph of f from x to x + h (the
gold area). For small h you can see that this area is approximately equal to the area of the
rectangle with height f (x) and width h:
g(x + h) − g(x) ≈ hf (x)
=⇒
g(x + h) − g(x)
≈ f (x)
h
Intuitively, we therefore expect that
g(x + h) − g(x)
= f (x)
h→0
h
g ′ (x) = lim
The fact that this is true, even when f is not necessarily positive, is the first part of the
Fundamental Theorem of Calculus.
2
Section 5.4 The Fundamental Theorem of Calculus
2010 Kiryl Tsishchanka
THEOREM (The Fundamental Theorem Of Calculus, Part I): If f is continuous on [a, b], then
the function g defined by
Zx
a≤x≤b
g(x) = f (t)dt
a
is an antiderivative of f, that is g ′ (x) = f (x) for a < x < b.
Proof: If x and x + h are in the open interval (a, b), then

 x
x+h
x+h
x+h
Z
Zx
Z
Z
Zx
Z
f (t)dt
f (t)dt − f (t)dt =
f (t)dt − f (t)dt =  f (t)dt +
g(x + h) − g(x) =
a
x
a
a
a
and so, for h 6= 0,
g(x + h) − g(x)
1
=
h
h
x
x+h
Z
f (t)dt
(1)
x
For now let’s assume that h > 0. Since f is continuous on [x, x+h], the Extreme Value Theorem
says that there are numbers u and v in [x, x + h] such that f (u) = m and f (v) = M, where
m and Mare the absolute minimum and maximum
values of f on [x, x + h]. By Property 8 of

Zb
integrals m(b − a) ≤ f (x)dx ≤ M (b − a) we have
a
x+h
Z
mh ≤
f (t)dt ≤ M h
=⇒
x
x+h
Z
f (u)h ≤
f (t)dt ≤ f (v)h
x
Since h > 0, we can divide this inequality by h:
1
f (u) ≤
h
x+h
Z
f (t)dt ≤ f (v)
x
Now we use (1) to replace the middle part of this inequality:
g(x + h) − g(x)
≤ f (v)
(2)
h
Inequality (2) can be proved in a similar manner for the case h < 0. Now let h → 0. Then
u → x and v → x, since u and v lie between x and x + h. Thus
f (u) ≤
lim f (u) = lim f (u) = f (x) and
h→0
u→x
lim f (v) = lim f (v) = f (x)
h→0
v→x
because f is continuous at x. We conclude, from (2) and the Squeeze Theorem, that
g(x + h) − g(x)
= f (x)
(3)
h→0
h
If x = a or b, then (3) can be interpreted as a one-sided limit. We know that if f is differentiable
at a, then f is continuous at a. If we adopt this theorem for one-sided limits, we obtain that
g is continuous on [a, b]. g ′ (x) = lim
3
Section 5.4 The Fundamental Theorem of Calculus
2010 Kiryl Tsishchanka
EXAMPLE: Find the derivative of the function g(x) =
Zx
t4 dt.
3
Solution: Since f (t) = t4 is continuous, Part 1 of the Fundamental Theorem of Calculus gives
g ′ (x) = x4
EXAMPLE: Find the derivative of g(x) =
Zx
2
et dt.
−1
2
Solution: Since f (t) = et is continuous, Part 1 of the Fundamental Theorem of Calculus gives
g ′ (x) = ex
EXAMPLE: Find the derivative of g(x) =
Zx2
2
2
et dt.
−1
2
Solution: Since f (t) = et is continuous, Part 1 of the Fundamental Theorem of Calculus gives
d
dx
Zx2
−1
2
et dt =
d
dx
Zu
2
et dt =

d 
du
Zu
−1
−1

2
et dt
du
2 du
2 2
4
= eu
= e(x ) · 2x = 2xex
dx
dx
In short,
d
dx
Zx2
2
et dt = e(x
2 )2
· (x2 )′ = 2xex
−1
EXAMPLE: Find the derivative of g(x) =
sin x
Z
√
3
tdt.
2
EXAMPLE: Find the derivative of g(x) =
Z1
sec tdt.
x4
4
4
Section 5.4 The Fundamental Theorem of Calculus
2010 Kiryl Tsishchanka
sin x
Z
√
3
EXAMPLE: Find the derivative of g(x) =
tdt.
2
Solution: Part 1 of the Fundamental Theorem of Calculus gives

 u
sin x
Z √
Z
Zu √
√
d
d  3  du √
du √
d
3
3
3
= 3u
= sin x cos x
tdt =
tdt =
tdt
dx
dx
du
dx
dx
2
2
2
In short,
d
dx
sin x
Z
√
3
tdt =
√
3
sin x(sin x)′ =
√
3
sin x cos x
2
EXAMPLE: Find the derivative of g(x) =
Z1
sec tdt.
x4
Solution: By Property 1 of the Definite Integral and Part 1 of the Fundamental Theorem of
Calculus we have

 u
Z
Z1
Zx4
Zu
d
du
du
d
d
d
= − sec u
sec tdt = −
sec tdt = −  sec tdt
sec tdt = −
dx
dx
dx
du
dx
dx
1
x4
1
1
= − sec(x4 ) · 4x3
= −4x3 sec(x4 )
In short,
d
dx
Z1
x4
sec tdt = −
d
dx
Zx4
sec tdt = − sec(x4 )(x4 )′ = −4x3 sec(x4 )
1
EXAMPLE: Find the derivative of g(x) =
cos
Z x
t2 dt.
sin x
5
Section 5.4 The Fundamental Theorem of Calculus
2010 Kiryl Tsishchanka
EXAMPLE: Find the derivative of g(x) =
cos
Z x
t2 dt.
sin x
Solution: We have
cos
Z x
Z0
2
t dt =
2
t dt +
0
sin x
sin x
cos
sin x
Z x
Z
2
t2 dt
t dt +
t dt = −
cos
Z x
2
0
0
therefore
d
dx
cos
Z x
d
t dt = −
dx
2
sin x
cos
Z
Z x
d
2
t dt +
t2 dt
dx
0
sin x
0
= − sin2 x(sin x)′ + cos2 x(cos x)′
= − sin2 x cos x + cos2 x(− sin x)
= − sin2 x cos x − cos2 x sin x
THEOREM (The Fundamental Theorem Of Calculus, Part II): If f is continuous on [a, b], then
Zb
f (x)dx = F (b) − F (a)
a
where F is any antiderivative of f, that is F ′ = f.
Proof: Put
g(x) =
Zx
f (t)dt
a
By the Fundamental Theorem Of Calculus, Part I, g(x) is an antiderivative of f (x). Therefore
any other antiderivative F (x) of f (x) can be written as
F (x) = g(x) + C =
Zx
f (t)dt + C
a
It follows that
F (a) =
Za
f (t)dt + C = 0 + C = C
=⇒
F (b) =
Zb
f (t)dt + C =
thus
F (b) =
Zb
f (t)dt + F (a)
=⇒
a
F (b) − F (a) =
Zb
a
6
f (t)dt + F (a)
a
a
a
Zb
f (t)dt Section 5.4 The Fundamental Theorem of Calculus
2010 Kiryl Tsishchanka
THEOREM (The Mean Value Theorem for Integrals): If f is continuous on [a, b], then there
exists a number c in [a, b] such that
f (c) = fave
1
=
b−a
Zb
f (x)dx
Zb
or
a
f (x)dx = f (c)(b − a)
a
EXAMPLE: Find the average value of the function f (x) =
√
x on the interval [1, 4].
Solution: We have
fave
1
=
4−1
Z4
√
1
xdx =
3
1
Z4
x
1/2
1 x1/2+1
dx = ·
3 1/2 + 1
1
4
1
1 2
= · x3/2
3 3
4
1
1
=
3
2 3/2 2 3/2
4 − 1
3
3
=
We now find c:
f (c) = fave
14
=
9
=⇒
√
14
c=
9
=⇒
c=
14
9
2
=
196
142
=
2
9
81
EXAMPLE: Find the average value of the function f (x) = 1 + x2 on the interval [−1, 2].
EXAMPLE: Find the average value of the function f (x) =
7
√
4 − x2 on the interval [−2, 2].
14
9
Section 5.4 The Fundamental Theorem of Calculus
2010 Kiryl Tsishchanka
EXAMPLE: Find the average value of the function f (x) = 1 + x2 on the interval [−1, 2].
Solution: We have
fave
1
=
2 − (−1)
Z2
2
1
x3
(1 + x )dx =
x+
=2
3
3 −1
2
−1
We now find c:
f (c) = fave = 2
=⇒
1 + c2 = 2
=⇒
c2 = 1
EXAMPLE: Find the average value of the function f (x) =
Solution: We have
fave
1
=
2 − (−2)
Z2 √
4 − x2 dx =
√
=⇒
c = ±1
4 − x2 on the interval [−2, 2].
π
1 π · 22
·
=
4
2
2
−2
We now find c:
f (c) = fave
π
=
2
=⇒
√
π
4 − c2 =
2
π2
4−c =
4
2
=⇒
8
=⇒
c=±
r
π2
4−
≈ ±1.23798
4