LECTURE 13: DEFINITE INTEGRALS AND ANTIDERIVATIVES
MINGFENG ZHAO
February 02, 2015
Evaluating definite integrals
If f is integrable on [a, b], then
b
Z
f (x) = lim
a
n→∞
n
X
f (x∗k )∆x,
k=1
where x0 = a < x1 < · · · < xn = b is a regular partition of [a, b]. For the choice of x∗k , it can be arbitrary. In particular,
n
X
the Riemann sum
f (x∗k )∆x can be left, right or midpoint Riemann sum.
k=1
Example 1. Assume that lim
∆→0
n
X
(3x∗k 2 + 2x∗k + 1)∆xk is the limit of a Riemann sum for a function f on [1, 3]. Then
k=1
f (x) = 3x2 + 2x + 1 and
lim
∆→0
n
X
(3x∗k 2 + 2x∗k + 1)∆xk =
Z
3
(3x2 + 2 + 1) dx.
1
k=1
Z 4p
Example 2. Compute
1 − (x − 3)2 dx.
3
p
p
Let y = 1 − (x − 3)2 , then y 2 + (x − 3)2 = 1, that is, the graph of y = 1 − (x − 3)2 is the upper semi-circle
centered at (3, 0) with radius 1. So we get
4
Z
p
1 − (x − 3)2 dx =
3
Z
1
π
· π · 12 = .
2
2
4
Example 3. Compute
(2x + 3) dx.
Z 4
Approach I: The integral
(2x + 3) dx is just the area of the trapezoid bounded by the x-axis and the line y = 2x + 3
2
2
from x = 2 to x = 4. The width of its base is 4 − 2 = 2, and the lengths of its two parallel sides are f (2) = 7 and
f (4) = 11. So we get
Z
4
(2x + 3) dx =
2
1
· 2 · [7 + 11] = 18.
2
1
2
MINGFENG ZHAO
Approach II: Since f (x) := 2x + 3 is continuous on [2, 4], then f is integrable on [2, 4]. So we can use right Riemann
Z 4
sum to compute
(2x + 3) dx. For any n, the right Riemann sum of the regular partition of [2, 4] with n, we have
2
4−2
2
∆x =
= and
n
n
xk = 2 + k∆x = 2 +
2k
,
n
∀k = 0, 1, 2, ·, n
Then the right Riemann sum is: x∗k = xk for all k = 1, · · · , n and
n
X
f (x∗k )∆x
=
k=1
n
X
f (xk )∆x
k=1
=
∆x
n
X
f (xk−1 )
k=1
=
=
=
=
=
By taking n → ∞, we get
Z 4
(2x + 3) dx = lim
2
n→∞
n
X
n 2k
2 X
·
+3
2· 2+
n
n
k=1
n
2 X
4k
·
7+
n
n
k=1
#
" n
n
2 X
4 X
k
·
7+ ·
n
n
k=1
k=1
2
4 n(n + 1)
· 7n + ·
n
n
2
n+1
n
14 + 4 ·
f (x∗k )∆x = lim
n→∞
k=1
n+1
14 + 4 ·
= 18.
n
Theorem 1. If f is continuous on [a, b] or bounded on [a, b] with a finite number of discontinuities, then f is integrable
on [a, b].

 1,
if 0 ≤ x ≤ 1,
, then f is a bounded with a finite number of discontinuities. So f is
Example 4. Let f (x) =
 −2, if 1 < x ≤ 3.
integrable on [0, 3]. And we have
Z 3
f (x) dx = 1 · 1 − 2 · 2 = −3.
0
Definition 1. For a < b, if f is integrable on [a, b], we define
Z a
Z
f (x) dx = −
b
a
b
f (x) dx.
LECTURE 13: DEFINITE INTEGRALS AND ANTIDERIVATIVES
3
Also we define
Z
a
f (x) dx = 0.
a
Theorem 2. Suppose f (x) and g(x) are integrable, for any constants a, b, c, p, then
Z b
Z b
Z b
[f (x) + g(x)] dx =
f (x) dx +
g(x) dx
a
a
b
Z
a
Z
= c·
cf (x) dx
f (x) dx
a
Z
b
a
b
Z
f (x) dx
p
Z
p
a
a
b
f (x) dx.
f (x) dx +
=
Z
b
|f (x)| dx is the sum of the
Theorem 3. Let f (x) be integrable on [a, b], then |f (x)| is also integrable on [a, b] and
a
areas of the region bounded by the graph of f and the x-axis on [a, b].
Example 5. The following picture shows the graph of a continuous function f (x) with the areas of the regions bounded
by its graph and the x-axis given.
Figure 1. The graph of f (x) on [a, d]
Then
Z
b
f (x) dx
=
12
f (x) dx
=
−10
a
Z
c
b
Z
c
Z
f (x) dx
a
=
b
Z
f (x) dx +
a
c
f (x) dx
b
By Theorem 2
4
MINGFENG ZHAO
Z
=
12 − 10
=
2
Z
d
f (x) dx
b
Z
c
=
Z
f (x) dx
b
|f (x)| dx
−10 + 8
=
−2
Z b
Z
b
Z
12 + 10 + 8
=
30.
5
d
f (x) dx
c
7
Z
f (x) dx = −10, then
f (x) dx = 3 and
0
Z
Z
f (x) dx +
b
=
By Theorem 2
c
c
f (x) dx −
a
d
|f (x)| dx
b
=
Example 6. Assume that
Z
|f (x)| dx +
a
Z
c
Z
|f (x)| dx +
=
a
By Theorem 2
c
=
d
d
f (x) dx +
0
7
7
Z
2f (x) dx
=
2
f (x) dx
0
By Theorem 2
0
Z
=
2 · (−10)
=
−20
Z 0
Z
f (x) dx +
7
f (x) dx
=
5
5
f (x) dx
By Theorem 2
0
Z
=
7
5
−
Z
f (x) dx +
0
=
−3 − 10
=
−13.
7
f (x) dx
0
Antiderivatives
Definition 2. Let f (x) and F (x) be two functions, we say that F (x) is an antiderivative of f (x) if F 0 (x) = f (x).
Example 7.
d 2
(x ) = 2x =⇒
dx
d
(− cos(x)) = sin(x)
dx
=⇒
x2 is an antiderivative of 2x
− cos(x) is an antiderivative of sin(x)
LECTURE 13: DEFINITE INTEGRALS AND ANTIDERIVATIVES
d
1
tan−1 (x) = 2
dx
x +1
=⇒
5
1
is an antiderivative of tan−1 (x)
x2 + 1
Theorem 4. Any two antiderivatives of one function only differ by a constant, that is, if F (x) and G(x) are antiderivatives of f (x), then F (x) = G(x) + C for some constant C.
Indefinite integrals
Definition 3. Let F (x) be an antiderivative of f (x), that is, F 0 (x) = f (x). The indefinite integral,
Z
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 8. For Example 7, we have
Z
2x dx
=
x2 + C
sin(x) dx
=
− cos(x) + C
1
dx
+1
=
tan−1 (x) + C.
Z
Z
x2
Theorem 5 (Linear properties). Let a be a constant, then
Z
Z
af (x) dx = a f (x) dx.
Also we have
Z
Z
[f (x) + g(x)] dx =
Z
f (x) dx +
g(x) dx.
Some useful indefinite integrals
Theorem 6.
Z
xp dx
Z
cos(ax) dx
Z
sin(ax) dx
 p+1

 x
+ C,
p+1
=

 ln |x|,
1
sin(ax) + C
a
1
= − cos(ax) + C
a
=
if p 6= −1
if p = −1.
6
MINGFENG ZHAO
Z
Z
Z
sec2 (ax) dx
=
csc2 (ax) dx
Z
eax dx
=
1
dx
− x2
Z
1
dx
2
x + a2
√
a2
=
=
=
1
tan(ax) + C
a
1
− cot(ax) + C
a
1 ax
e +C
a
x
+C
sin−1
a
x
1
+C
tan−1
a
a
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