Mathematics Learning Centre
Integration: The anti-derivative
Mary Barnes
c
1999
University of Sydney
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Mathematics Learning Centre, University of Sydney
1
If
Definition of the integral as an anti-derivative
d
(F (x)) = f (x) then f (x)dx = F (x).
dx
In words,
If the derivative of F (x) is f (x), then we say that an indefinite integral of f (x)
with respect to x is F (x).
For example, since the derivative (with respect to x) of x2 is 2x, we can say that an
indefinite integral of 2x is x2 .
In symbols:
d 2
(x ) = 2x,
dx
so
2xdx = x2 .
Note that we say an indefinite integral, not the indefinite integral. This is because the
indefinite integral is not unique. In our example, notice that the derivative of x2 + 3 is
also 2x, so x2 + 3 is another indefinite integral of 2x. In fact, if c is any constant, the
derivative of x2 + c is 2x and so x2 + c is an indefinite integral of 2x.
We express this in symbols by writing
2xdx = x2 + c
where c is what we call an “arbitrary constant”. This means that c has no specified value,
but can be given any value we like in a particular problem. In this way we encapsulate
all possible solutions to the problem of finding an indefinite integral of 2x in a single
expression.
In most cases, if you are asked to find an indefinite integral of a function, it is not necessary
to add the +c. However, there are cases in which it is essential. For example, if additional
information is given and a specific function has to be found, or if the general solution of
a differential equation is sought. (You will learn about these later.) So it is a good idea
to get into the habit of adding the arbitrary constant every time, so that when it is really
needed you don’t forget it.
The inverse relationship between differentiation and integration means that, for every
statement about differentiation, we can write down a corresponding statement about
integration.
For example,
d 4
(x ) = 4x3 ,
dx
so
4x3 dx = x4 + c.
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Exercises 1.1
Complete the following statements:
(i)
d
(sin x)
dx
(ii)
d
(x5 )
dx
=
,
so
(iii)
d
(ex )
dx
=
,
so
(iv)
d
dx
=
,
so
(v)
d
(x)
dx
(vi)
d
(ln x)
dx
= cos x,
so
cos x
dx
= sin x + c.
dx
=
dx
=
dx
=
dx
=
dx
=
1
x2
=
,
so
=
,
so
The next step is, when we are given a function to integrate, to run quickly through all
the standard differentiation formulae in our minds, until we come to one which fits our
problem.
In other words, we have to learn to recognise a given function as the derivative of another
function (where possible).
Try to do the following exercises by recognising the function which has the given function
as its derivative.
Exercises 1.2
i.
(− sin x)dx
ii.
3x2 dx
2dx
iii.
iv.
v.
sec2 xdx
3 1
x 2 dx
2
vi.
−
1
dx
x2
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2
Some rules for calculating integrals
Rules for operating with integrals are derived from the rules for operating with derivatives.
So, because
d
d
(cf (x)) = c (f (x)) , for any constant c,
dx
dx
we have
Rule 1
(cf (x))dx = c
For example
f (x)dx, for any constant c.
10 cos xdx = 10
cos xdx = 10 sin x + c.
It sometimes helps people to understand and remember rules like this if they say them
in words. The rule given above says: The integral of a constant multiple of a function is
a constant multiple of the integral of the function. Another way of putting it is You can
move a constant past the integral sign without changing the value of the expression.
Similarly, from
d
d
d
(f (x) + g(x)) =
(f (x)) + (g(x)),
dx
dx
dx
we can derive the rule
Rule 2
(f (x) + g(x))dx =
For example,
(ex + 2x)dx =
ex dx +
f (x)dx +
g(x)dx.
2xdx
= ex + x2 + c.
In words, the integral of the sum of two functions is the sum of their integrals.
We can easily extend this rule to include differences as well as sums, and to the case where
there are more than two terms in the sum.
Examples
Find the following indefinite integrals:
i.
ii.
(1 + 2x − 3x2 + sin x)dx
1
(3 cos x − ex )dx
2
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Solutions
i.
(1 + 2x − 3x + sin x)dx =
2
1dx +
2xdx −
3x dx −
2
(− sin x)dx
= x + x2 − x3 − cos x + c.
Note: We have written + sin xdx as − (− sin x)dx because (− sin x) is the derivative of cos x.
ii.
1
1 x
(3 cos x − ex )dx = 3 cos xdx −
e dx
2
2
1
= 3 sin x − ex + c.
2
You will find you can usually omit the first step and write the answer immediately.
Exercises 2
Find the following indefinite integrals:
(cos x + sin x)dx
i.
ii.
iii.
iv.
(ex − 1)dx
(1 − 10x + 9x2 )dx
4
(3 sec2 x + )dx
x
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3
Integrating powers of x and other elementary
functions
We can now work out how to integrate any power of x by looking at the corresponding
rule for differentiation:
d n
(x ) = nxn−1 ,
dx
so
d n+1 x
= (n + 1)xn ,
dx
so
nxn−1 dx = xn + c.
Similarly
(n + 1)xn dx = xn+1 + c.
Therefore
n
1
· (n + 1)xn dx
n+1
1 (n + 1)xn dx
=
n+1
1
xn+1 + c.
=
n+1
1
← notice that n+1
· (n + 1) is just 1
when we cancel
x dx =
← take
1
n+1
outside the
sign
We should now look carefully at the formula we have just worked out and ask: for which
values of n does it hold?
d
(xn ) = nxn−1 holds whether n is positive or
Remember that the differentiation rule dx
negative, a whole number or a fraction or even irrational; in other words, for all real
numbers n.
We might expect the integration rule to hold for all real numbers also. But there is one
snag: in working it out, we divided by n + 1. Since division by zero does not make sense,
the rule will not hold when n + 1 = 0, that is, when n = −1. So we conclude that
Rule 3
1
xn dx =
xn+1 + c
n+1
for all real numbers n, except n = −1.
When n = −1, xn dx becomes x−1 dx = x1 dx. We don’t need to worry that the rule
above doesn’t apply in this case, because we already know the integral of x1 .
Since
1
1
d
(ln x) = ,
we have
dx = ln x + c.
dx
x
x
Examples
Find
i.
x3 dx
dx
x2
√
iii.
xdx
ii.
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Solutions
x3 dx =
i
ii
iii
1
1
x4 + c = x4 + c.
(3 + 1)
4
1
1
dx −2
= x dx =
x−2+1 + c = − + c.
2
x
−2 + 1
x
√
xdx =
1
x 2 dx =
1
2
1
1
2 3
x 2 +1 + c = x 2 + c.
3
+1
← replacing n by 3 in the
formula
← replacing n by −2 in the
formula
← replacing n by
1
2
Exercises 3.1
1. Find anti-derivatives of the following functions:
i
x5
ii
x9
iii
x−4
iv
1
x2
v
√1
x
√
vi
vii x
√
3
x
√
viii x x
2
1
xπ
ix
2. Find the following integrals:
i.
ii.
−3xdx
iii.
iv.
v.
vi.
vii.
x3 + 3x2 + x + 4 dx
1
dx
x
1 2
x−
dx
x √ 2
x
√ +
dx
x
2
2x4 + x2
dx
x
3 + 5x − 6x2 − 7x3
dx
2x2
x−
Hint: multiply out the expression
Hint: divide through by the denominator
Hint: divide through by the denominator
At this stage it is very tempting to give a list of standard integrals. However, you are NOT
encouraged to memorise integration formulae, but rather to become VERY familiar with
the list of derivatives and to practise recognising a function as the derivative of another
function.
If you try memorising both differentiation and integration formulae, you will one day
mix them up and use the wrong one. And there is absolutely no need to memorise the
integration formulae if you know the differentiation ones.
It is much better to recall the way in which an integral is defined as an anti-derivative.
Every time you perform an integration you should pause for a moment and check it by
differentiating the answer to see if you get back the function you began with. This is a
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very important habit to develop. There is no need to write down the checking process
every time, often you will do it in your head, but if you get into this habit you will avoid
a lot of mistakes.
Examples
Find the following indefinite integrals:
i.
ii.
5
ex + 3x 2 dx
(5 csc2 x + 3 sec2 x)dx
Solutions
i.
x
e + 3x
5
2
dx =
x
e dx + 3
= ex + 3 ·
5
2
5
x 2 dx
5
1
x 2 +1 + c
+1
2 7
= ex + 3 · x 2 + c
7
7
6
= ex + x 2 + c.
7
ii.
(5 csc x + 3 sec x)dx = −5 (− csc x)dx + 3
2
2
2
sec2 xdx
= −5 cot x + 3 tan x + c.
Exercises 3.2
Integrate the following functions with respect to x:
10ex − 5 sin x
√ 2
ii.
x(x + x + 1)
i.
iii. 1
+√
x
(1 − x2 )
5
Hint: Multiply through by
fractional exponents.
√
x, and write with
iv.
x3 + x + 1
1 + x2
Hint: Divide through by 1 + x2 , and consult table
of derivatives.
v.
tan x
sin x cos x
Hint: Write tan x as
sin x
cos x
and simplify.
Hint: Remember the formula 1 + tan2 x = sec2 x.
vi. tan2 x
(If you are not familiar with inverse trig functions, omit parts iii and iv.)
Hint: In order to get some of the functions above into a form in which we can recognise
what they are derivatives of, we may have to express them differently. Try to think of
ways in which they could be changed that would be helpful.
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4
Solutions to exercises
Exercises 1.1
i
d
(sin x)
dx
= cos x, so
ii
d
(x5 )
dx
= 5x4 , so
iii
d
(ex )
dx
= ex ,
iv
d
dx
v
1
x2
1
x2
vi
d
(x)
dx
= 1,
d
(ln x)
dx
d
(x−2 )
dx
cos xdx= sin x + c.
5x4 dx = x5 + c.
x
e dx
so
= − x23 , so
= x−2 and
(Note:
− x23 dx =
1
x2
= −2x−3 = − x23
so
1dx
1dx is usually written as
= x1 ,
= ex + c.
1
dx
so
x
+ c.
= x + c.
dx.)
= ln x + c.
Exercises 1.2
Note: All these answers can be checked by differentiating!
i
ii
iii
iv
v
vi
(− sin x)dx = cos x + c.
3x2 dx
= x3 + c.
2dx
= 2x + c.
sec2 xdx
3 1
x 2 dx
2
1
− 2 dx
x
= tan x + c.
3
= x 2 + c.
1
=
+ c.
x
Exercises 2
i
ii
iii
iv
(cos x + sin x)dx
= sin x − cos x + c.
(ex − 1)dx
= ex − x + c.
(1 − 10x + 9x2 )dx
4
(3 sec2 x + )dx
x
= x − 5x2 + 3x3 + c.
= 3 tan x + 4 ln x + c.
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Exercises 3.1
1.
i
ii
iii
iv
v
vi
x5 dx
=
1 6
x + c.
6
x9 dx
=
1 10
x + c.
10
x−4 dx
=
1
− x−4+1
3
1
dx
x2
=
1
√ dx
x
√
3
=
xdx
=
=
1
− x−3 + c
3
=
−
x−2 dx
=
1 −1
x +c
−1
=
1
− + c.
x
x− 2 dx
=
2x 2 + c
=
√
2 x + c.
=
3 1 +1
x3
4
=
3 4
x 3 + c.
4
1
1
x 3 dx
1
1
+ c.
3x3
(Note: In exercises like v and vi above, it is easier to work out what power of x is
required, and then to work out what coefficient is needed to give the correct answer on
1
xn+1 . So v is more
differentiating. This is usually better than substituting for n in
n+1
1
1
easily done by saying (mentally) “ − 12 + 1 = 12 , so the answer will involve x 2 . Now 2x 2
1
1
will give a coefficient of 1 when differentiated so x− 2 dx = 2x 2 + c ”).
vii
viii
ix
2.
i
ii
iii
iv
√
x 2 dx
=
√
x xdx
=
√
1
x 2+1 + c.
2+1
2 2√
x x + c.
5
x 2 dx
=
2 5
x2 + c
5
x−π dx
=
1
x−π+1 + c =
−π + 1
−3xdx
=
1
3
−3 · x2 + c = − x2 + c.
2
2
(x3 + 3x2 + x + 4)dx
=
1 4
1
x + x3 + x2 + 4x + c.
4
2
1
(x − )dx
x
=
1 2
x − ln x + c.
2
1
dx
xπ
=
1
(x − )2 dx
x
v
√
√ 2
x
√ +
dx
x
2
3
=
(x2 − 2 +
=
1 3
1
x − 2x − + c.
3
x
=
2
x− 2 dx +
1
1
+ c.
(π − 1)xπ−1
1
)dx
x2
=
−
1 1
x 2 dx
2
(Recall,
1
1
dx = − )
2
x
x
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vi
vii
1
2 · 2x 2 +
=
√
1 √
4 x + x x + c.
3
2x4 + x2
dx
x
=
(2x3 + x)dx
=
1
1
2 · x4 + x2 + c
4
2
=
1 4 1 2
x + x + c.
2
2
3 + 5x − 6x2 − 7x3
dx
2x2
1 2 3
· x2 + c
2 3
=
=
5
3
7x
+
dx
−3−
2
2x
2x
2
=
3 −2
5 1
7
x dx +
xdx
dx − 3 dx −
2
2 x
2
=
5
7 1
3 1
(− ) + ln x − 3x − · x2 + c
2 x
2
2 2
=
−
3
5
7
+ ln x − 3x − x2 + c.
2x 2
4
Exercises 3.2
i
ii
(10ex − 5 sin x)dx
√
=
x(x2 + x + 1)dx
=
=
iii
iv


5
1

+ √  dx =
x
(1 − x2 )
x3 + x + 1
dx
1 + x2
10ex + 5 cos x + c.
tan x
dx
sin x cos x
3
1
2 7 2 5 2 3
x 2 + x 2 + x 2 + c.
7
5
3
√
5 sin−1 x + 2 x + c.
=
x(x2 + 1) + 1
dx
1 + x2
v
5
x 2 + x 2 + x 2 dx
=
1
x+
dx
1 + x2
=
1 2
x + tan−1 x + c.
2
=
1
sin x
·
dx
cos x sin x cos x
Dividing through by (1 + x2 )
Writing tan x =
sin x
cos x
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=
=
=
vi
tan2 xdx
=
=
sec2 xdx
tan x + c.
=
1
dx
cos2 x
(sec2 x − 1)dx
sec2 xdx −
tan x − x + c.
1dx
Using tan2 x = sec2 x − 1