LECTURE 14: FUNDAMENTAL THEOREM OF CALCULUS
MINGFENG ZHAO
February 04, 2015
Z
0
Definition 1. Let F (x) be an antiderivative of f (x), that is, F (x) = f (x). The indefinite integral,
f (x) dx :=
Z
F (x) + C (where C is arbitrary constant), means find the antiderivatives of f . The function f insider the symbol
is
called the integrand.
Example 1. For any differentiable function f , we have
Z
f 0 (x) dx = f (x) + C.
Theorem 1.
Z
p
x dx
=
Z
cos(ax) dx
=
sin(ax) dx
=
sec2 (ax) dx
=
csc2 (ax) dx
Z
eax dx
=
Z
Z
Z
Z
1
dx
− x2
Z
1
dx
2
x + a2
√
a2
Z
4x39 − 5x−8
dx.
x2
Z
4x39 − 5x−8
dx
x2
Example 2. Compute
=
=
=


xp+1
p+1
+ C,
if p 6= −1
 ln |x| + C,
if p = −1.
1
sin(ax) + C
a
1
− cos(ax) + C
a
1
tan(ax) + C
a
1
− cot(ax) + C
a
1 ax
e +C
a
x
+C
sin−1
a
x
1
tan−1
+C
a
a
In fact, we have
Z
=
(4x37 − 5x−10 ) dx
1
2
MINGFENG ZHAO
Z
=
4
=
4·
=
Example 3. Compute
Z
dx − 5
x−10 dx
x37+1
x−10+1
−5·
+C
37 + 1
−10 + 1
x38
x−9
4·
−5·
+C
38
−9
2 38 5 −9
x + x +C
19
9
=
Z
x
37
By Theorem 1
sec2 (3x) dx.
By Theorem 1, we have
Z
Z
Example 4. Compute
sec2 (3x) dx =
1
tan(3x) + C.
3
e−10x dx.
By Theorem 1, we have
Z
Z
Example 5. Compute
√
e−10x dx = −
1 −10x
e
+ C.
10
4
dx.
9 − x2
In fact, we have
Z
4
√
dx
9 − x2
Z
=
=
=
Z
Example 6. Compute
1
dx
9 − x2
Z
1
4 √
dx
2
3 − x2
x
+ C By Theorem 1
4 sin−1
3
4
√
1
dx.
16x2 + 1
In fact, we have
Z
1
16x2 + 1
Z
dx
=
=
=
=
=
1
dx
1
16 x2 + 16
Z
1
1
1 dx
2
16
x + 16
Z
1
1
2 dx
2
16
x + 41
x
1 1
−1
· 1 tan
+C
1
16 4
4
1
· 4 tan−1 (4x) + C
16
By Theorem 1
LECTURE 14: FUNDAMENTAL THEOREM OF CALCULUS
=
3
1
tan−1 (4x) + C.
4
Fundamental theorem of calculus
Definition 2. Let f be a continuous function for t ≥ a, the area function for f with left endpoint a is:
Z x
A(x) :=
f (t) dt.
a
Example 7. Let f (t) = 2t + 3, then the area function for f with left endpoint 2 gives the area of the trapezoid for
x ≥ 2, and
Z
A(x)
x
f (t) dt
=
2
=
=
=
=
1
· (x − 2) · [f (2) + f (x)]
2
1
· (x − 2)[7 + 2x + 3]
2
1
· (x − 2)(2x + 10)
2
(x − 2)(x + 5)
= x2 + 3x − 10.
Theorem 2 (Fundamental theorem of calculus). Let f be continuous, and F be an antiderivative of f , then
Z b
b
f (x) dx = F (b) − F (a) := F (x)|a .
a
Z
Moreover, let A(x) :=
x
f (t) dt, then A is an antiderivative of f , that is,
a
A0 (x) = f (x).
Z
Example 8. Since
3 sin(x) dx = −3 cos(x) + C, by Theorem 2, then
Z
2π
3 sin(x) dx
2π
=
−3 cos(x)|0
=
−3 cos(2π) + 3 cos(0)
=
−3 + 3
=
0.
0
4
MINGFENG ZHAO
Theorem 3. Let f be continuous, a(x) and b(x) be differentiable, then
Z b(x)
d
f (t) dt = f (b(x))b0 (x) − f (a(x))a0 (x).
dx a(x)
Proof. Let F be an antiderivative of f , by the Fundamental Theorem of Calculus, Theorem 2, we have
Z b(x)
f (t) dt, and F 0 (x) = f (x).
F (b(x)) − F (a(x)) =
a(x)
So we have
b(x)
Z
d
dx
f (t) dt
=
a(x)
d
[F (b(x)) − F (a(x))]
dx
= F 0 (b(x))b0 (x) − F 0 (a(x))a0 (x)
= f (b(x))b0 (x) − f (a(x))a0 (x)
By the Chain rule
Since F 0 = f
d
dx
By Theorem 3, then
x
Z
sin2 (t) dt.
Example 9. Compute
1
d
dx
d
Example 10. Compute
dx
By Theorem 3, then
e2x
Z
d
dx
x−1
Z
e2x
x−1
Z
x
sin2 (t) dt
=
sin2 (x) ·
=
sin2 (x).
1
d
d
(x) − sin2 (1) ·
(1)
dx
dx
sin(t)
dt.
t
sin(t)
dt =
t
=
=
sin(e2x ) d 2x
sin(x − 1) d
·
(e ) −
·
(x − 1)
e2x
dx
x−1
dx
sin(e2x )
sin(x − 1)
· 2e2x −
2x
e
x−1
sin(x
−
1)
2 sin(e2x ) −
.
x−1
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca