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Basic Concepts
of Integration
✏
✒
✑
14
...
The reverse
dx
process is to obtain the function f (x) from knowledge of its derivative
...
Applications of integration are numerous and some of these will be explored in
subsequent Blocks
...
✛
✘
Prerequisites
Before starting this Block you should
...
✙
functions
Learning Style
✓ find some simple integrals by reversing
To achieve what is expected of you
...
Integration as Differentiation in Reverse
dy
Suppose we differentiate the function y = x2
...
Integration reverses this
2
process and we say that the integral of 2x is x
...
The situation is just a little more complicated because there are lots of functions we can differentiate to give 2x
...
5
Now do this exercise
Write down some more functions which have derivative 2x
...
Consequently, when we reverse the process, we have no idea what the
original constant term might have been
...
We state that the integral of 2x is x2 + c
...
In a similar way, when we want to integrate a function we use
a special notation: y(x) dx
...
To integrate 2x we write
integral
sign
2x dx = x2 + c
this term is
called the
integrand
constant of integration
there must always be a
term of the form dx
Note that along with the integral sign there is a term of the form dx, which must always be
written, and which indicates the variable involved, in this case x
...
The function being integrated is called the integrand
...
When you find an indefinite integral your answer should
always contain a constant of integration
...
1: Integration
2
More exercises for you to try
1 a) Write down the derivatives of each of:
x3 ,
x3 + 17,
x3 − 21
b) Deduce that 3x2 dx = x3 + c
...
What is meant by the term ‘integrand’ ?
3
...
Answer
2
...
You could check the entries in this
table using your knowledge of differentiation
...
Table of integrals
function
f (x)
indefinite integral
f (x)dx
constant, k
x
x2
kx + c
1 2
x +c
2
1 3
x +c
3
xn+1
+ c,
n+1
ln |x| + c
xn
x−1 (or
cos x
sin x
cos kx
sin kx
tan kx
ex
e−x
ekx
1
)
x
n = −1
sin x + c
− cos x + c
1
sin kx + c
k
1
− cos kx + c
k
1
ln | sec kx|+c
k
ex + c
−e−x + c
1 kx
e +c
k
When dealing