Calculus II: The Abridged Version (part 1)
(by Evan Dummit, 2011, v. 1.01)
How to pass the exam:
Step 1: Study the material on this handout.
Step 2: Don't fail the exam.
1
Integration
1.1
•
•
Remember Your Trigonometry
Don't forget the basic identities:
sin(2x)
=
2 sin(x) cos(x)
cos(2x)
=
cos2 (x) − sin2 (x)
sin2 (x) + cos2 (x)
=
1
You really only need to remember integrals of tangent, and secant; the rest are easy from those:
ˆ
sin(x) dx
= − cos(x) + C
cos(x) dx
=
tan(x) dx
= − ln(cos(x)) + C
sec(x) dx
=
csc(x) dx
= − ln(csc(x) + cot(x)) + C
cot(x) dx
=
ˆ
sin(x) + C
ˆ
ˆ
ln(sec(x) + tan(x)) + C
ˆ
ˆ
•
ln(sin(x)) + C
The important inverse-trig functions are arcsine and arctangent:
ˆ
1
√
dx
1 − x2
ˆ
1
dx
1 + x2
ˆ
1
√
dx
x x2 − 1
1.2
=
sin−1 (x) + C
=
tan−1 (x) + C
=
sec−1 (x) + C
Substitution
ˆ
•
The general substitution formula states that
f 0 (g(x)) · g 0 (x) dx = f (g(x)) + C
. It is just the Chain Rule,
written in terms of integration via the Fundamental Theorem of Calculus. We generally don't use the formula
written this way. To do a substitution, follow this procedure:
◦
Step 1: Choose a substitution
◦
Step 2: Compute the dierential
◦
Step 3: Rewrite the original integral in terms of
∗
u = g(x).
du = g 0 (x) dx.
u:
3a: Rewrite the integral to peel o what will become the new dierential
du.
∗
∗
3b: Write the remaining portion of the integrand in terms of
3c: Find the new limits of integration in terms of
limits are
x=a
and
x = b,
the new ones will be
u.
u, if the integral is a denite integral. [If the old
u = g(a) and u = g(b). In a very concrete sense,
these are the same points.]
∗
3d: Write down the new integral.
If the integral is indenite, substitute back in for the original
variable.
•
Substitution is best learned by doing examples; see below.
•
Evaluate
•
´3
0
2
2x ex dx.
◦
Step 1: The exponential has a 'complicated' argument
◦
Step 2: The dierential is
◦
Step 3a: We can rearrange the integral as
◦
Step 3b: The remaining portion of the integrand is
◦
Step 3c: We see that
◦
Step 3d: Putting it all together gives
Evaluate
◦
du = 2x dx.
x=0
´3
0
corresponds to
´9
0
x2 ,
u = x2 .
so we try setting
2
ex · (2x dx).
u = 02
ex
2
, which is just
x=3
and
eu .
corresponds to
eu du = eu |9u=0 = e9 − 1
u = 32 .
.
´ e (ln(x))2
dx.
1
x
Step 1: It might not look like any function has a 'complicated' argument, but if we think carefully we
can see that the numerator is what we get if we plug in
ln(x)
to the squaring function. So we try setting
u = ln(x).
◦
•
Step 2: The dierential is
du =
1
dx.
x
´3
◦
Step 3a: We can rearrange the integral as
◦
Step 3b: The remaining portion of the integrand is
◦
Step 3c: We see that
◦
Step 3d: Putting it all together gives
Evaluate
◦
x=1
0
corresponds to
´1
0
2
[ln(x)] ·
1
dx .
x
2
[ln(x)]
u = ln(1) = 0
u2 du =
, which is just
and
x=e
1
1 31
u |u=0 =
3
3
u2 .
corresponds to
u = ln(e) = 1.
.
´1 √
x 3x2 + 1 dx.
0
Step 1: Here we see that the square root function has the 'complicated' argument
3x2 + 1
so we try
2
u = 3x + 1.
◦
Step 2: The dierential is
◦
Step 3a: We can rearrange the integral as
6·
•
1
6
du = 6x dx.
´1√
0
3x2 + 1 ·
Note that we introduced a factor of
, which is okay since it's just multiplication by 1.
√
◦
Step 3b: The remaining portion of the integrand is
◦
Step 3c: We see that
◦
Step 3d: Putting it all together gives
Evaluate
1
· (6x dx).
6
´3
2
x=0
corresponds to
Step 1: Try the denominator:
◦
Step 2: The dierential is
and
x=1
1
,
6
which is just
corresponds to
´ 4 1 1/2
1 2
7
u du = · u3/2 |4u=1 =
1 6
6 3
9
2x
dx.
x2 − 1
◦
u=1
3x2 + 1 ·
u = x2 − 1.
du = 2x dx.
.
1 1/2
u .
6
u = 4.
1
· (2x dx).
−1
1
−1
The remaining portion of the integrand is
, which is just u
.
x2 − 1
We see that x = 2 corresponds to u = 3 and x = 3 corresponds to u = 8.
´ 8 −1
Putting it all together gives
u du = ln(u)|8u=3 = ln 8 − ln 3 .
3
Step 3a: We can rearrange the integral as
◦
Step 3b:
◦
Step 3c:
◦
Step 3d:
◦
Remark: It's possible to do this one without substitution, by using partial fractions to see that
1
1
+
.
x+1 x−1
1.3
´3
◦
This gives
2
x2
I = [ln(x + 1) + ln(x − 1)] |3x=2 = ln(4) + ln(2) − ln(3) = ln(8) − ln(3) as before.
Integration by Parts
ˆ
•
ˆ
f 0 · g dx = f · g −
The integration by parts formula states
f · g 0 dx
. It is just the Product Rule, re-
arranged and rewritten in terms of integrals using the Fundamental Theorem of Calculus.
integration by parts, all that is required is to pick a function
f 0 (x) · g(x) is equal to
´1
• Evaluate 0 x ex dx.
◦
•
and a function
g(x)
To perform an
such that the product
the original integrand. Examples will make everything clear.
´e
1
x · ex
is an obvious product, and so we need to decide which of
2
[ln(x)] dx.
2
[ln(x)] = ln(x)·ln(x), but this doesn't help unless we remember the antiderivative of ln(x).
2
2
0
Instead we write the integrand as 1 · [ln(x)] , so as to take f = 1 and g = [ln(x)] . Then we get f = x
1
0
by the Chain Rule. So plugging in will yields
and g = 2 · ln(x) ·
x
ˆ e
ˆ e
1
2
2 e
[ln(x)] dx = x · ln(x) |1 −
x · 2 ln(x) · dx
x
1
1
ˆ e
= e−
2 ln(x)
1
ˆ e
1
= e − 2x · ln x|e1 −
(IBP again)
2x · dx
x
1
= e − [2e − (2e − 2)]
We can write
=
Evaluate
◦
f (x)
x and ex should be f 0 .
1 2
0
x
Since x gets more complicated if we take its antiderivative (since we'd get
2 x ) we try ´f = e and
´
1
1
g = x, to get f = ex and g 0 = 1. Plugging into the formula gives 0 x ex dx = x ex |1x=0 − 0 1 · ex dx =
(x ex − ex )|1x=0 = 1 .
The integrand
Evaluate
◦
•
2x
=
x2 − 1
´1
0
e−2 .
(x2 − 2x + 2)e3x dx.
We want
g
to be the thing which gets simpler when we dierentiate. The polynomial
much simpler if we dierentiate it, while the exponential
g = x2 − 2x + 2
and
f 0 = e3x ,
so that
g 0 = 2x − 2
and
3x
x2 − 2x + 2
gets
e stays basically the same. So we should take
f = 13 e3x . Then integrating by parts yields an
expression which we can't evaluate directly we have to integrate by parts again:
ˆ
1
2
(x − 2x + 2)e
3x
dx
0
ˆ 1
1 3x 1
1
= (x − 2x + 2) · e |0 −
(2x − 2) · e3x dx (IBP once)
3
3
0
ˆ 1
2 3 2
1
=
e −
−
(2x − 2) · e3x dx
3
3
3
0
ˆ 1
1
1
2 3 2
e −
− (2x − 2) · e3x |10 −
2 · e3x dx
(IBP again)
=
3
3
9
9
0
2 3 2
2
2 3
2
=
e −
−
−
e −
3
3
9
27
27
2
=
•
Evaluate
◦
´
x3 sin(x) dx.
Here we just need to integrate by parts repeatedly. We get
ˆ
ˆ
x3 sin(x) dx
= −x3 cos(x) +
3x2 cos(x) dx (IBP once)
ˆ
3
2
= −x cos(x) + 3x sin(x) − 6x sin(x) dx
(IBP again)
ˆ
3
2
= −x cos(x) + 3x sin(x) − −6x cos(x) + 6 cos(x) dx (IBP
=
•
Find
20 3 26
e −
.
27
27
´
again)
−x3 cos(x) + 3x2 sin(x) + 6x cos(x) + 6 sin(x) + C
x3 sin(x2 ) dx.
u = x2 . The dierential is du = 2x dx,
´
1
1
and we can rearrange the integral as
x2 sin(x2 ) · · (2x dx) = u sin(u) · du.
2
2
´ 1
1
1
cos(u) du = [−u cos(u) + sin(u)] + C .
◦ Now we integrate by parts, to get − u cos(u) +
2
2
2
1
◦ Finally substitute back for x to get
−x2 cos(x2 ) + sin(x2 ) + C .
2
◦
1.4
•
First we look for a substitution. We try the argument of the sine:
´
Trig Sub
Some kinds of integrals require a more clever sort of substitution to evaluate, that's sort of 'backwards' from
the usual way we try to do substitutions: instead of
function)
f.
u = f (x)
x = f (u) for some appropriate (trig
sin2 (x)+cos2 (x) = 1) to simplify something
we try
The idea is to use one of the basic trig relations (e.g.,
more complicated. As always, examples make everything clear.
•
Evaluate
´√
1 − x2 dx.
◦
Traditional substitution along the lines of
◦
Instead we try
´q
x = sin(u);
u = 1 − x2
doesn't work like we'd hope.
dx = cos(u) du.\
´p
´
1 − sin2 (u) · cos(u) du =
cos2 (u) · cos(u) du = cos2 (u) du.
◦
So we get
◦
Remembering
◦
Finally substitute back for
cos2 (u) =
then
1 + cos(2u)
2
we can evaluate the integral to get
u = sin−1 (x)
simplify this to the equivalent form
to obtain
u sin(2u)
+
+ C.
2
4
sin−1 (x) sin(2 sin−1 (x))
+
+C
2
4
√
sin−1 (x) x 1 − x2
+
+C
2
2
.
. If desired, we can
•
Evaluate
´
1
dx.
(1 + x2 )2
This time we think of the arctangent antiderivative and try
◦
We obtain
◦
Remembering the identity
cos2 (u) =
◦
Finally substitute back for
u = tan−1 (x)
´
´
1
1
· sec2 (u) du =
2
2
sec4 (u)
(1 + tan (u))
simplify this to the equivalent form
•
x = tan(u). Then dx = sec2 (u) du.
´
· sec2 (u) du = cos2 (u) du.
◦
Evaluate
´
1 + cos(2u)
2
we can evaluate the integral to get
to obtain
tan−1 (x) sin(2 tan−1 (u))
+
+C
2
4
tan−1 (x)
x
+
+C
2
2(1 + x2 )
u sin(2u)
+
+ C.
2
4
. If desired, we can
.
2−x
√
dx.
4 − x2
√
x = sin(u) substitution again but it doesn't quite work, since then 4 − x2 isn't nice.
q
√
to try x = 2 sin(u), since then
4 − x2 = 4 − 4 sin2 (u) = 2 cos(u), and we're much
◦
We'd like to do the
◦
Instead we have
happier.
◦
Then we have
dx = 2 cos(u) du, so we get
´ 2 − 2 sin(u)
´
·2 cos(u) du = [2 − 2 sin(u)] du = 2u+2 cos(u)+
2 cos(u)
C.
◦
1.5
•
Substituting back yields
2 sin−1 (x) + 2 cos(sin−1 (x)) + C
, or equivalently,
2 sin−1 (x) + 2
p
1 − x2 + C
.
Partial Fractions
Generally, integrating rational functions is dicult, unless we simplify them. We do this via the partial fraction
decomposition, which breaks down rational functions into simpler parts which we know how to integrate. To
nd the PFD, follow these steps:
◦
Step 1: Factor the denominator.
◦
Step 2: Find the form of the PFD:
For each term
∗
For each non-factorable term
··· +
◦
(x + a)n
∗
Cn x + Dn
.
(x2 + ax + b)n
Step 3: Solve for the coecients
∗
C1
C2
Cn
+
+ ··· +
.
(x + a)n
x + a (x + a)2
C1 x + D1
C2 x + D2
(x2 + ax + b)n , the PFD has terms 2
+
+ 2
(x + ax + b)2
x + ax + b
the PFD has terms
C1 , C2 , . . . , Cn .
The best way is to clear all denominators and then substitute 'intelligent' values for
x (e.g., the roots
of the linear factors).
∗
Sometimes plugging in other values of
x
is necessary, in order to nd the coecients of the higher
terms.
◦
Step 4: Evaluate the integral.
•
As ever, examples make everything easier.
•
Evaluate
◦
´
1
dx.
x2 + 3x
We factor and get
1
x(x + 3)
so we want
1
A
B
= +
.
x(x + 3)
x
x+3
◦
Clear denominators so that
◦
Set
1 = A(x + 3) + B(x).
x = 0 and x = −3 to see that 1 = 3A and 1 = −3B .
´ 1/3
´
1/3
1
1
1
dx =
−
dx =
ln(x) − ln(x + 3) + C
◦ Then
x2 + 3x
x
x+3
3
3
•
•
Evaluate
´
x3
1
dx.
+ x2
1
+ 1)
◦
We factor and get
◦
Clear denominators so that
◦
Set
x=0
◦
Set
x = −1
◦
Set
x=1
◦
Then
Evaluate
´
to get
x2 (x
so we want
D
1
A
B
= + 2+
.
+ 1)
x
x
x+1
x2 (x
1 = Ax(x + 1) + B(x + 1) + Dx2 .
1 = B.
to get
1 = D.
1 = 2A + 2B + D so that A = −1.
´ −1
´
1
1
1
1
dx =
+ 2+
dx = − ln(x) − + ln(x + 1) + C
3
2
x +x
x
x
x+1
x
to get
.
2x
dx.
(x + 1)(x2 + 1)
2x
A
Bx + D
=
+ 2
.
(x + 1)(x2 + 1)
x+1
x +1
◦
We want
◦
Clear denominators so that
◦
Set
◦
.
x = −1
to get
−2 = 2A
2x = A(x2 + 1) + (Bx + D)(x + 1).
so that
A = −1.
2
x + 2x + 1 = (Bx + D)(x + 1) so we see that we need B = D = 1.
´
´
1
−1
2x
x
1
◦ Then
dx =
+
+
dx = − ln(x + 1) + ln(x2 + 1) + tan−1 (x) + C
(x + 1)(x2 + 1)
x + 1 x2 + 1 x2 + 1
2
1.6
•
So we want
Improper Integrals
Sometimes we like to integrate to innity, by which we mean, take the limit as a limit of integration becomes
arbitrarily large. In other words, we write
•
´∞
a
as shorthand for
´q
lim
q→∞ a
f (x) dx.
Other times, we like to integrate through a 'singularity' of a function that is, through a point where a
function is undened because it blows up.
´1 1
dx
0 x
•
f (x) dx
as shorthand for
´1 1
lim+ q dx.
x
x→0
[E.g.,
x = 0
on
y = 1/x.]
Again, we will write something like
Integrals with either of these two kinds of 'bad behaviors' are called improper integrals. (I can only assume
that the terminology stems from the fact that trying to evaluate such integrals is not something done in polite
company.)
•
Typically we will be interested in asking (i) if the integral actually converges to a nite value, and (ii) if it
does, what the value is.
•
There are approximately two ways to solve problems like this:
◦
Method 1: Find the indenite integral, and then evaluate the limit. If the limit is hard, try rewriting
the function to make the limit doable.
◦
Method 2: Compare the integrand to another function whose integral is easier to evaluate. We are aided
in the following theorem:
∗
0 < f (x) < g(x) for
´b
if
f (x) dx diverges,
a
Theorem (Comparison Test for Integrals): If
one or both of
a
and
b
converges, then so does
can be innite, then
´b
a
f (x) dx.
all
x
[a, b] where
´b
g(x) dx. If a g(x) dx
in the interval
so does
´b
a
.
∗
The idea behind the theorem is to say that if
f (x)-integral
so if the
∞ then so must
f (x)-integral.
goes to
stays nite then so must the
0 < f (x) < g(x)
the g(x)-integral,
´b
f (x) dx < a g(x) dx, and
a
and conversely if the g(x)-integral
then
´b
Example time!
•
Evaluate
◦
•
´∞
0
e−x dx.
We just evaluate to see
Evaluate
´∞
1
x2
´∞
0
e−x dx = lim (−e−x )|qx=0 = lim −e−q + 1 = 1
q→∞
We do partial fractions to see
◦
Now to take the limit as
as
ln
.
1
dx.
+x
◦
q→∞
x
x+1
´
x2
1
dx = ln(x) − ln(x + 1) + C .
+x
x → ∞ is not so easy, unless we notice that we can rewrite the indenite integral
.
´∞
x
x
1
1
q
dx
=
lim
ln
|
=
lim
ln
−
ln(
)
= ln(2)
x=1
q→∞
q→∞
x2 + x
x+1
x+1
2
◦
Then we have
◦
If the problem had only asked about convergence, we could have observed that
x,
so
1
1
1
< 2.
2
x +x
x
Then the Comparison Test would have said that, because
converges, then so does
´∞
1
1
dx.
x2 + x
.
x2 + x > x2 for positive
´∞ 1
1
dx = − |∞
=1
1 x2
x x=1
However, the Comparison Test doesn't help us in computing the
actual value.
•
Determine whether the integral
◦
´∞
0
1
dx
(x + 1)(x + 2)(x + 3)(x + 4)
converges.
Doing partial fractions is too much work (although it does give the correct answer, if the trick in the
previous example is used repeatedly).
1
1
1
1
<
<
<
.
x+4
x+3
x+2
x+1
´q
´q
1
1
1
◦ Then 0
dx < 0
dx = − (x + 1)−3 |qx=0 .
4
(x + 1)(x + 2)(x + 3)(x + 4)
(x + 1)
4
1
nite (in fact, it is
).
4
◦ Therefore this integral converges .
◦
•
2
But it's easier to notice that
Determine whether the integral
q→∞
this remains
converges.
´1 1
dx = − ln(q).
q x
◦
We integrate to get
◦
So as
◦
Therefore this integral
q → 0+
´1 1
dx
0 x
As
−∞.
we see that this diverges to
diverges to
−∞
.
Sequences and Series
2.1
•
Sequences, Series, Convergence
We say a sequence
small.
a1 , a2 , . . . , ak , . . .
converges to 0 if the terms of the sequence eventually become arbitrarily