MAT 266 Calculus for Engineers II
Notes on Chapter 5
Professor: John Quigg
Semester: spring 2017
Review of Sections 5.1–5.4
Let f be a bounded function on [a, b].
P = {x0 , x1 , . . . , xn } such that
A partition of [a, b] is a finite set
a = x0 < x1 < x2 < · · · < xn = b.
We write ∆xi = xi − xi−1 for the distance between consecutive points, equivalently the
length of the ith interval [xi−1 , xi ]. Note: the book also calls the set of intervals [xi−1 , xi ] the
partition. For each i = 1, . . . , n let x∗i be a sample point in the ith interval [xi−1 , xi ]. The
associated Riemann sum is
n
X
f (x∗i )∆xi .
i=1
Definition: If the limit
lim
max ∆xi →0
n
X
f (x∗i )∆xi
i=1
exists, we call it the definite integral of f from a to b, we write it as
say that f is integrable on [a, b].
Rb
a
f (x) dx, and we
The integrand is f (x). The limits of integration are a and b, with a the lower
limit and b the upper limit. The process of computing the value of the integral is called
integration. The variable of integration is x, and can be replaced by any other letter.
Theorem: If f is continuous on [a, b], or more generally if f has only a finite number of
jump discontinuities, then f is integrable.
The partition P is called regular if all the intervals [xi−1 , xi ] have the same length
∆x =
b−a
,
n
in which case
xi = a + i ∆x.
Using regular partitions simplifies the limit for the integral:
Theorem: If f is integrable on [a, b] then
Z b
n
X
f (x) dx = lim
f (xi )∆x
a
n→∞
1
i=1
2
Rb
If f is nonnegative, then a f (x) dx is the area under the curve y = f (x) for a ≤ x ≤ b. If
Rb
f can take negative values, then a f (x) dx is interpreted as the net area between the curve
and the x-axis:
Z b
f (x) dx = A1 − A2 ,
a
where A1 is the area of the region above the x-axis and below the graph of f , and A2 is the
area of the region below the x-axis and above the graph of f .
We define
Z
a
Z
b
f (x) dx = −
b
f (x) dx
a
and
a
Z
f (x) dx = 0.
a
Theorem: If f and g are integrable on [a, b] and c is a constant, then
(1)
Rb
a
c dx = c(b − a)
Rb
Rb
[f (x) + g(x)] dx = a f (x) dx + a g(x) dx
Rb
Rb
(3) a cf (x) dx = c a f (x) dx
Rb
Rb
Rb
(4) a [f (x) − g(x)] dx = a f (x) dx − a g(x) dx
(2)
Rb
a
Theorem: If f is integrable on [a, b] and a ≤ c ≤ b then
Z b
Z b
Z c
f (x) dx =
f (x) dx +
f (x) dx
a
a
c
Theorem: Let f and g be integrable on [a, b], and let m and M be constants.
(1) If f (x) ≥ 0 for a ≤ x ≤ b then
Rb
f (x) dx ≥ 0
Rb
Rb
(2) If f (x) ≥ g(x) for a ≤ x ≤ b then a f (x) dx ≥ a g(x) dx
a
(3) If m ≤ f (x) ≤ M for a ≤ x ≤ b then
Z b
m(b − a) ≤
f (x) dx ≤ M (b − a)
a
Fundamental Theorem of Calculus: Let f be continuous on [a, b].
3
(1) The function g defined by
Z
g(x) =
x
f (t) dt
a
is an antiderivative of f . In other words,
d
dx
Z
x
f (t) dt = f (x).
a
(2) If F is any antiderivative of f then
Z
b
b
f (x) dx = F (b) − F (a) = F (x) .
a
a
If v(t) is the velocity of an object and s(t) is its position at time t, then s0 (t) = v(t), so by
the Fundamental Theorem of Calculus the displacement, or net change in position, is
Z
b
v(t) dt = s(b) − s(a).
a
If the object always moves in the positive direction, so that v(t) ≥ 0 for a ≤ t ≤ b, then
Rb
s(b)−s(a) is the distance that the object travels. In any case, a |v(t)| dt is the total distance
traveled.
In general, we regard F 0 as the rate of change of F , and
the net change of F from a to b.
Rb
a
F 0 (x) dx = F (b) − F (a) as
Definition:
The indefinite integral of f is the collection of all antiderivatives of f , and is
R
written f (x) dx. If F is any particular antiderivative of f , then every antiderivative of f
is of the form F + C for some constant C, so we write
Z
f (x) dx = F (x) + C
We call C the constant of integration.
4
Some indefinite integrals:
Z
Z
Z
[f (x) + g(x)] dx = f (x) dx + g(x) dx
Z
xn+1
xn dx =
+ C (n 6= −1)
n+1
Z
ex dx = ex + C
Z
sin x dx = − cos x + C
Z
sec2 x dx = tan x + C
Z
sec x tan x dx = sec x + C
Z
1
dx = tan−1 x + C
2
x +1
Z
Z
cf (x) dx = c
f (x) dx
Z
1
dx = ln |x| + C
x
Z
ax
+C
ax dx =
ln a
Z
cos x dx = sin x + C
Z
csc2 x dx = − cot x + C
Z
csc x cot x dx = − csc x + C
Z
1
√
dx = sin−1 x + C
1 − x2
If f is integrable on [a, b], we define the average value of f on [a, b] by
Z b
1
f (x) dx.
fave =
b−a a
Mean Value Theorem for Integrals: If f is continuous on [a, b] then there exists c in
[a, b] such that
Z b
1
f (c) =
f (x) dx
b−a a
Section 5.5: Substitution Rule
Substitution Rule: If g 0 is continuous and f is continuous on the range of g then
Z
Z
0
f (g(x))g (x) dx = f (u) du
Proof. Let F be an antiderivative of f . Then
Z
f (u) du = F (u) + C.
On the other hand, by the Chain Rule, F (g(x)) is an antiderivative of f (g(x))g 0 (x), so
Z
f (g(x))g 0 (x) dx = F (g(x)) + C.
Making the change of variable u = g(x), we get
Z
Z
0
f (g(x))g (x) dx = F (u) + C = f (u) du.
5
Example: We will apply the Substitution Rule to evaluate
Then du = −6x dx, so x dx = − 16 du, and hence
Z
Z
2
−3x2
xe
dx e−3x (x dx)
Z
1
u
= e − du
6
Z
1
=−
eu du
6
1
= − eu + C
6
2
e−3x
=−
+C
6
Example: We evaluate
R
R
2
xe−3x dx. Let u = −3x2 .
tan x dx by substituting u = cos x. We have du = − sin x dx, so
Z
Z
sin x
tan x dx =
dx
cos x
Z
1
=−
du
u
= − ln |u| + C
= − ln | cos x| + C
1 +C
= ln cos x = ln | sec x| + C
Example: Similar techniques show that
Z
cot x dx = ln | sin x| + C
Substitution Rule for Definite Integrals: If g 0 is continuous on [a, b] and f is continuous
on the range of g then
Z b
Z g(b)
0
f (g(x))g (x) dx =
f (u) du.
a
Example: We evaluate
R e3
e
g(a)
1
x ln x
dx. Let u = ln x. Then du =
x = e implies u = ln e = 1
x = e3
implies u = ln e3 = 3
1
x
dx. Also,
6
Thus
e3
Z
e
3
Z
1
du
1 u
3
= ln |u|
1
dx =
x ln x
1
= ln 3 − ln 1
= ln 3
R e3
R e3
Warning: do not write something like e x ln1 x dx = e u1 du; this is wrong! Once you make
the substitution in a definite integral, the limits of integration must change accordingly!
Theorem: Let f be continuous on [−a, a].
Ra
Ra
(1) If f is even, that is, f (−x) = f (x), then −a f (x) dx = 2 0 f (x) dx.
Ra
(2) If f is odd, that is, f (−x) = −f (x), then −a f (x) dx = 0.
Proof. We have
Z
Z a
f (x) dx =
−a
0
Z
Z
−a
f (x) dx = −
f (x) dx +
−a
a
Z
0
0
a
f (x) dx.
f (x) dx +
0
Let u = −x. Then du = −dx, and
x = 0 implies u = 0
x = −a implies u = a
so
−a
Z
−
a
Z
f (x) dx =
f (−u) du
0
(1) When f is even,
Z
0
a
Z
a
f (−u) du =
(2) When f is odd,
Z
f (u) du
0
0
a
Z
f (−u) du = −
0
In either case the conclusion follows.
a
f (u) du
0
Remember: when substituting in an indefinite integral, you have to substitute back after
integrating, but in a definite integral you just change the limits of integration and forget
about the old variable.