P1 Chapter P1 Chapter 1616

CIE Centre A-levelCIE Centre A-level Pure Maths Pure Maths

© Adam Gibson

IntegrationIntegration

Differentiation

dy

dx

Integration

ydx

SumDifference

Gradient Area

Calculus

Calculus

Integrals

Definite Integral Indefinite Integral

( )f x dx( )x b

x af x dx

Noun (Action): integration

Verb: integrateNoun: integral

Derivatives

Read it as “integral [of] f [of] x dx [between a and b]”

Integration as the reverse of differentiation

The first, and easiest, way to understand integration:What function g(x) has f(x) as its gradient at all points x in the domain?

The indefinite integral helps us find the answer.

Example:

2What function ( ) has gradient given by 2 1 ?g x y x

We remember the rule that

1 n ndyy ax nax

dx

Think….

Integration as the reverse of differentiation – contd.

We note that we can change to (I won’t give a formal proof here, but you can verify it).

323( )g x x x ! Warning …

What if 323( ) 10g x x x ?

So we construct the rule:

1

1n na

ax dx x kn

where k is an “arbitrary

constant”.This is the indefinite integral.

Implications

•Does the formula apply for all rational numbers n?

•No! At n = -1 the formula “explodes”. We will consider this in detail later.

•What does “arbitrary constant k” actually mean?

•For this point, think about moving the graph of a function up and down by adding a number. Does the gradient change?

•So there are an infinite number of functions with the same gradient formula. We can only calculate k if we are giventhe coordinates of at least one point on the graph. If you don’t have specific information, always write “+k”.

TASKSTASKS

1. Find the integral of

2. Find

3. Find

4. Try Q15 and Q16 p. 240

2 1x x

( )g x xdx given that g(1)=0

32

1x dx

x

Integration as areaarea

1x 2x 3x

Area Area of each trapeziumx

4x

How do we make it more accurate?

…

Integration as area – area – continued.continued.

Just like finding the gradient, this is a limit problem.

The exact area is given by

0lim (area of trapezium with base width x)x

x

A

We take the sum of an infinite number of trapezia, each with infinitely narrow width.

Thus, if the function is f(x), we get:

0

11

1 120

0

( ( ) ( )).( )

lim ( ).

( )n

n

k k k kk

xx

x

x

A f x f x x x

A f x x

A f x dx

Remember – dx is an infinitely small ∆x

Section 16.2 and Section 16.3 givea more detailedderivation

Integration as area – area – continued.continued.

0

2

4

6

8

10

12

0 0.5 1 1.5 2 2.5

3 2y x

Let’s find the area underthe curve between x=0 and x=2

“Limits of integration”

Integration as area – area – continued.continued.

We write this quantity as the definite integral:

2 3

0( 2)x dx

Higher limit at the top

Lower limit at the bottom

24

0

24

xx

You should perform the integration,but not include the arbitraryconstant k (can you see why?).Then you should put the answer insquare brackets with the limitsmarked on the right.

Integration as area – area – continued.continued.

24

0

24

xx

Now for the last step. Evaluatethe expression inside the bracketsat x=upper limit and x=lower limit,and subtract the latter from the former.

4 42 02 2 2 0

4 4

8

What are the unitsfor the answer?

To be clear: the definite integral is the difference between the indefinite integral evaluated at the upper and lower limits … here we could write it as I(2) – I(0). The constant “k” is not needed as it’s the same in both cases.

Your turn …

10 2

1Question: Evaluate the definite integral ( 3 4)x x dx

Answer:

103 212 1

3 212

12

12

4

( 1000 50 40) ( ( 1) ( 1) 4( 1))

( 990) (1 4)

995

x x x

Be careful!

The area is negative?! Is it a mistake …?

-350

-300

-250

-200

-150

-100

-50

0

-1-0.7-0.4-0.1 0.

20.50.81.11.41.7 2

2.32.62.93.23.53.84.14.44.7 5

5.35.65.96.26.56.87.17.47.7 8

8.38.68.99.29.59.8

No mistake! The area is just a sum of function values. Itcan be any real number, positive, negative or zero. So be careful when interpreting the result!

You have just calculated the blue area.

23 4y x x

Infinite and improper integrals

Let’s consider difficult cases like:

1

21

1

0

1

1

1

dxx

dxx

dxx

We don’t have a formula forthis (n+1=0), so we’ll ignore itfor now (actually )1

ln | |dx x kx

Improper integrals are integrals over an interval includinga value of x where the function is undefined.Infinite integrals are those involving at least one limit at±∞

This has a finite value

This also has a finite value

Calculate please

A

B

C

Infinite and improper integrals continued

In order to better understand the difference between Aand B, let’s take a close look at the graph:

0

1

2

3

4

5

6

7

8

9

10

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5

1/x

1/x2

1/rootx

The area under between 1 and ∞ is finite but unbounded.

2

1

x

Infinite and improper integrals continued

The problem of infinite integrals is of course similar to theproblem for infinite sums:

2

21

1

6n n

but1

1

n

… we say it “diverges”

Example

Evaluate:83

1 2 1

xdx

x

523 3

1

2 53 3

33

5

1 3 13 0

51 1

12

5

x x

Notes on Integration

Integration by substitution – ONLY use this for examples where the “inner” function is linear. Other examples are too advanced for P1.

2.5 3.52.5

3 33

3.5dx x dx x

x

Remember the “n+1 over n+1” rule and use it correctly!

1.52x

•Indefinite integrals – you MUST use +k.32217 2

2 3(17 )x x dx x x k !

4(10 2)x dx n = -4. a = 10, so divide by (-3x10)

31(10 2)

30x k

Notes on Integration

Areas found by definite integration are positive if..… the function values are positive (and vice versa)

Areas of regions bounded by two functions can be calculated… by subtracting one function from the other

and then integrating. Result will be positive if … you subtract the “lower” function from the higher one.

( 4)b

ax x dx

Is the integral (area) positive for:a=8 b=10a=-10 b=-8a=1 b=3a=1 b=7 ?