7
Integration
7.1
Integration as areas
For well behaved functions we can define its integral using areas:
Z
b
f (x) dx := signed area between the curve y = f (x) and the x-axis for a ≤ x ≤ b.
a
A a C B b Properties. Let’s accept we know what area is and how it behaves. Then
Z b
1 dx = b − a.
•
a
Z
b
•
Z
f (x) dx for c ∈ R.
cf (x) dx = c
a
Z
•
b
a
b
Z
(f (x) + g(x)) dx =
a
b
Z
f (x) dx +
a
b
g(x) dx.
a
Exercise 7.1. Draw a picture to illustrate
a < b if convenient.
1
Rb
a
x dx = 21 b2 − 12 a2 . Asssume 0 <
Z
Example 7.2. The integral
X
x2 dx. Let’s suppose X > 0. Partition the
0
interval [0, X] into N equal parts by using the equally spaced points
0<
1
2
3
N −1
N
X < X < X < ··· <
X < X = X.
N
N
N
N
N
On each block we can approximate the area by using rectangles. Let An be
the area of the rectangle with base the interval [ n−1
X, Nn X] and height ( Nn X)2 .
N
Then the sum A1 + A2 + . . . + AN should approximate the area pretty close and
if we let N → ∞ we should get the required area.
2
We can now deduce a slightly more general formula:
b
Z
b
Z
2
0
b
b 3 a3
1 3
x dx =
−
= x
.
3
3
3
a
2
x dx −
x dx =
a
a
Z
2
0
X
Z
xm dx where m is a positive integer. This
Example 7.3. The integral
0
was answered by Fermat in the 1630s, before Newton and Leibnitz. Fermat’s
idea was to use a partition with division points chosen in a geometric progression,
instead of sticking to equal divisions.
Choose a real number a with 0 < a < 1. Partition the interval [0, X] using the
points X = a0 X > aX > a2 X > · · · > an−1 X > an X > · · · .
Let An denote the area of the rectangle with base [an+1 X, an X] and height
(an X)m .
Then the (infinite) sum A0 + A1 + A2 + . . . should be a good
RX
(over)estimation of the required area and should in fact be 0 xm dx when we
let a → 1.
Now An =
.
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3
.
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.
.
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.
.
.
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.
.
∞
X
X
Z
m
x dx ≈
Hence
0
An
n=0
We can now deduce that
Z
b
m
Z
b
x dx −
x dx =
a
Z
m
0
0
a
m+1 b
am+1
x
bm+1
−
=
x dx =
.
m+1 m+1
m+1 a
m
Clearly we need another way calculating integrals of functions. The answer—
familiar to you already!—is to use antiderivatives. For example
d xm+1 = xm
dx m + 1
Z
and
a
b
xm+1
x dx =
m+1
m
b
.
a
This is a remarkable relationship between tangents and areas! The precise statements are as follows.
Theorem 7.4. The Fundamental Theorem of Calculus. Suppose f is a
continuous function.
Z
• Fix an a in the domain of f and define G(x) :=
f (t) dt. Then
a
d
G(x) = f (x) i.e. G is an antiderivative for f .
dx
• If F is an antiderivative of f i.e. F 0 = f then
Z
b
f (t) dt = F (b) − F (a) = [F (x)]ba .
a
4
x
Note 7.5. The second part of the FTC is the de facto method for calculating
R2
definite integrals. For instance, 0 cos x dx = sin(2).
Z x
d
Example 7.6. What is
(1 + t2016 )ecos t sin t dt? By the first part of the
dx 0
FTC, the integral is
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.
Z tan x
d
(1 + t2016 )ecos t sin t dt? Use the FTC along with the chain
What about
dx 0
rule! If we put u = tan x then
d
dx
Z
.
.
tan x
2016
(1 + t
cos t
)e
0
.
.
.
.
.
.
d Z u
du
sin t dt =
(1 + t2016 )ecos t sin t dt
du 0
dx
.
.
.
.
.
.
.
.
.
.
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.
.
Here are “sketch proofs” of the FTC. For the first part
a x For the second part, observe that F 0 = f = G0 . Therefore F (x) = G(x) + c for
Rb
some constant c, and so F (b)−F (a) = G(b)+c−G(a)−c = G(b) = a f (x) dx.
Remarks.
(i) The “indefinite integral”
antiderivative for f .
5
R
f (x) dx = F (x) + c just means an
(ii) It is often useful to think of dG = f (x)dx as expressing an equality between
the infinitesimal area dG and the area under the graph over an infenitesimal
width dx around x. Integration is then the process of summing up infinitely
many of these small quantities to produce an agreeable answer. You will
exploit this idea in MAS111 and use it to calculate volumes, surface areas
and more.
(iii) We can write
[F (x)]ba
Z
=
a
b
dF
dx =
dx
Z
x=b
dF .
x=a
In the examples we looked at we overestimated areas by using rectangles from
above. What happens if we change our way of approximation? What if we
underestimated areas by using rectangles from below? Will we get the same answer? What about if we chose to approximate by using rectangles using function
values from somewhere inside each block? What happens if we change the way
we partition the interval? What do we even mean by ‘area’ ?
Answer: Being able to integrate should mean that we get the same answer if
we use thinner and thinner parts with the maximum width tending to 0. Once
we know the function can be integrated then we can choose how we partition
the underlying interval and where to choose sample points.
(N.b. The class of integrable functions cover most of the functions you will meet
in university. Do you know of any non-integrable function?)
7.2
Techniques of integration
Recall the Fundamental Theorem of Calculus:
dF
If
= f then
dx
Z
b
f (x) dx =
.
.
a
6
.
.
.
.
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.
.
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.
.
.
So we need to consider how we’d go about finding a suitable anti-derivative F (x)
given the integrand f (x). This can get really tricky but most of the integrals
we meet can be computed by manipulating it into bits that look like standard
elementary integrals. As for how to manipulate, well, there are essentially just two
general techniques: making a substitution (or change of variables) and integration
by parts.1 If you think of integration as the reverse process to differentiation then
these two methods are the counterparts of the chain rule and the product rule
respectively.
Let’s look at some simple examples to start with.
Rb
x3 dx. We need to figure out a function which differend 4
tiates to x3 . Let’s try x4 :
x = 4x3 and we have to adjust the facdx
Rb
d
tor of 4. This leads us to
= x3 . Therefore, a x3 dx =
.
.
.
dx
Example 7.7.
.
.
.
.
.
a
.
.
.
.
.
.
.
R
d 4
x = 4x3 gives x4 = 4x3 dx up to a constant. Therefore
Alternatively,
dx
R 3
x dx = . . . . . . . . . .
Example 7.8.
R
(1 + 4x)3 dx. One method would be to expand out the bracket.
Another method would be to proceed as in the previous example: they look
d
somewhat similar. So we could try
(1 + 4x)4 = . . . . . . . which
dx
R
gives (1 + 4x)3 dx = . . . . . . . . . .
Here’s a list of known indefinite integrals which you should memorise.
Z
1
•
xa dx =
xa+1 + C, provided a 6= −1.
a+1
Z
1
•
dx = ln x + C for x > 0.
x
1
There are of course specialized methods for dealing with particular classes of integrals e.g.
partial fractions for integrating rational functions.
7
Z
•
sin x dx = − cos x + C.
Z
•
cos x dx = sin x + C.
Z
•
Z
•
sec2 x dx = tan x + C.
csc2 x dx = − cot x + C.
Z
•
sec x tan x dx = sec x + C.
Z
•
csc x cot x dx = − csc x + C.
Z
1
x
dx = sin−1 + C, a > 0.
a
a2 − x 2
Z
1
1
x
•
dx = tan−1 + C.
2
2
a +x
a
a
Z
1
•
eax dx = eax + C.
a
Z
1 ax
•
bax dx =
b + C.
a ln b
•
√
These are verified by differentiating the right hand sides.
7.2.1
Method of substitution
Recall the chain rule:
d
f (g(x)) = f 0 (g(x))g 0 (x). Thus
dx
Z
f 0 (g(x))g 0 (x) dx = f (g(x)) + C.
The formalism of the above is what we usually call making a substitution: To
R
calculate the integral f (g(x))g 0 (x) dx make a substitution u = g(x). Then
8
du
= g 0 (x) or, as differentials, du = g 0 (x)dx. Therefore
dx
Z
Z
0
f (g(x))g (x) dx =
f (u) du.
This is integrated (as a function of u), and finally u = g(x) is substituted back.
Z
cos(3 ln x)
dx. Clearly cos(3 ln x) is not something we’d recogExample 7.9.
x
nise as a basic integral unless we got rid of the ln x inside. So we should try the
substitution u = 3 ln x. Now
du
=
dx
Z
Hence
i.e. du =
cos(3 ln x)
dx =
x
.
.
.
Z
Exercise 7.10. Evaluate: (a)
.
.
.
.
.
.
.
.
x
dx ,
2
x +1
.
.
.
.
.
Z
(b)
.
.
.
.
x2 (1 + x3 )4 dx .
If we make a substitution in a definite integral then we will need to change
the limits of Z
integration accordingly. For example, if we make the substitution
b
u = g(x) in
f (g(x))g 0 (x) dx, then u will vary between the limits g(a) and
a
g(b). To prevent errors creeping in you might
in the limits
Z want to retain a variable
Z
b
x=b
f (g(x))g 0 (x) dx as
f (u) du or
of integration. For example, if we write
x=a
a
Z u=g(b)
f (u) du, then it might be clearer where to evaluate the antiderivative.
u=g(a)
Z
Exercise 7.11. Calculate
0
Exercise 7.12. Calculate
1
x2
x
dx.
+1
d
ln(sec x + tan x). Hence or otherwise evaluate the
dx
integral
Z
0
eπ/3
1
dx.
x cos(ln x)
9
Z
Exercise 7.13. (Sometimes the substitution might be hidden.) Evaluate
R
R
What about cos5 x dx? cos7 x dx?
π/2
cos3 x dx.
0
Trigonometric substitutions. These rely on the two identities sin2 θ +cos2 θ =
1 and 1 + tan2 θ = sec2 θ. There are two things to watch out for:
• if the integrand involves a2 −x2 try a substitution x = a sin θ or x = a cos θ,
• if the integrand involves a2 + x2 try x = a tan θ.
Z
dx
√
Example 7.14. Consider
. Put x = a sin θ. Then dx = a cos θ dθ
2
a − x2
and a2 − x2 = a2 − a2 sin2 θ = a2 cos2 θ. So
Z
dx
√
=
a2 − x2
Z
a cos θ dθ
=
a cos θ
Z
dθ = θ + C = sin−1
x
+ C.
a
Example 7.15. Consider the upper half semi-circular region of the circle x2 +
y 2 = R2 above the x-axis. Its area is the integral
x=R
Z
I :=
Z
x=R
y dx =
x=−R
√
R2 − x2 dx.
x=−R
Put x = R sin θ. Then dx = R cos θ dθ. For the limits, we can take θ =
π
2
when
x = R and θ = − π2 when x = −R. So
Z
θ= π2
I=
Z
p
2
2
2
2
R − R sin θ R cos θ dθ = R
θ=− π2
θ= π2
cos2 θ dθ
θ=− π2
=R
2
Z
θ= π2
θ=− π2
π
1 + cos 2θ
πR2
sin 2θ 2
2 θ
=
dθ = R
+
.
2
2
4
2
−π
2
10
7.2.2
Integration by parts
Start with the product rule:
d
d
d
f (x)g(x) = f (x) g(x) + g(x) f (x).
dx
dx
dx
So up to a constant f (x)g(x) =
R
f 0 (x)g(x) dx +
R
f (x)g 0 (x) dx. Rearrange-
ment then gives
Z
dg
f (x) dx = f (x)g(x) −
dx
Z
df
g(x) dx + C.
dx
For definite integrals the above becomes
Z
a
b
Z
b
df
g(x) dx
a dx
Z b
df
= f (b)g(b) − f (a)g(a) −
g(x) dx.
a dx
dg
f (x) dx = [f (x)g(x)]ba −
dx
dg
Remarks.
• dg =
dx as differentials. This gives a neater formula:
R
R dx
f dg = f g − g df .
R
R
R
R
Alternatively uv = u v − (u0 v) is an easy to remember sloppy
dg
version (think f = u and dx
= v in the product rule).
R
R
R 0R
• In uv = u v − (u v), which function should we choose to be the u
and which to be v when we meet a product?
– If the product involves a polynomial taking u to be the polynomial is
worth a try. This is because u0 will be a polynomial of smaller degree
and if we repeat then we migth be able to make the polynomial part
disappear altogether.
– If the product involves a logarithm or an inverse trig or some function
11
which is hard to integrate but easy to differentiate, that should be u.
Obviously the above are guides; they might not work all the time.
• Just because the integral involves a product does not mean you have to
use integration Zby parts!
Z
x
−1
For example:
dx = x tan x − tan−1 x dx doesn’t make
1 + x2
things any easier.
Exercise 7.16. Use integration by parts to evaluate:
Z
(a)
Z
ln x dx ,
(b)
1
2
−1
sin
Z
x dx ,
(c)
−1
x tan
Z
x dx ,
(d)
sec3 x dx.
0
7.2.3
Other techniques
Rational functions. A rational function is a function of the form polynomial
divided by a polynomial. To calculate
Z
f (x)
dx
g(x)
where f (x), g(x) are polynomials, the integrand needs to be prepared so that the
integral resembles forms we are familiar with. The general method is to start out
by dividing out improper fractions, and then use partial fraction decompositions.
We illustrate using examples.
Z 3 2
4x − 7
Example 7.17. Evaluate
dx.
0 2x + 1
Example 7.18. Calculate:
Z
(a)
Z
dx
;
2
x − 2x − 3
(b)
x3
x
dx.
−1
Example 7.19. To handle integrals involving root of a polynomial, you could
12
Z
1
try completing the square. For example:
√
x − x2 dx.
0
Recursive formulas. This is a very useful trick for handling integrals which
come in a family indexed in a natural way.
R 5 x
Example 7.20.
Z To calculate x e dx we can start with the more general integral In = xn e−x dx. Integration by parts then gives In = −xn e−x + nIn−1
and we can use this to to figure out I5 .
7.3
Improper integrals
We will often need to calculate integrals/areas over unbounded intervals and/or
unbounded functions. These are called improper integrals and evaluated as limits
of standard definite integrals. Here are the relevant (slightly informal) definitions.
Z ∞
Z t
•
f (x) dx = lim
f (x) dx, provided the limit exists.
t→∞
a
Z
a
a
•
Z
f (x) dx = lim
t→−∞
−∞
Z ∞
•
Z
a
f (x) dx, provided the limit exists.
t
a
f (x) dx =
∞
Z
f (x) dx +
−∞
f (x) dx, provided both integrals exist.
−∞
a
In each case we say that the integral is convergent if the corresponding limit(s)
exists and divergent if the limit does not exist.
Z t
dx
Example 7.21. If t ≥ 1 then
= . .
1 x
Z ∞
dx
the improper integral
.
.
.
.
.
.
x
1
Z
.
.
.
.
.
.
.
.
.
.
So
.
t
dx
=
2
1 x
Z ∞
dx
the improper integral
x2
1
On the other hand,
.
.
.
.
.
.
.
.
.
.
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.
.
13
.
.
.
.
.
for t ≥ 1. So
Z
More generally, let p > 0. Then the integral
1
diverges when p ≤ 1.
Z
∞
Exercise 7.22. Evaluate
−∞
dx
.
1 + x2
14
∞
dx
converges when p > 1 and
xp