integration

Integration

Introduction• This is a very big chapter which covers many

ways of Integrating more difficult expressions

• You are sometimes told which method to use but in addition must be able to identify the most appropriate way

• We also cover using Integration to work out volumes as well as areas, as well as setting up and using equations (similar to the chapter on differentiation)

Teachings for Exercise 6A

IntegrationYou need to be able to integrate standard

functions

You met the following in C3, in the differentiation chapter:

6A

𝑦=𝑥𝑛

𝑑𝑦𝑑𝑥=𝑛𝑥𝑛−1

𝑦=𝑒 𝑓 (𝑥 )

𝑑𝑦𝑑𝑥= 𝑓 ′ (𝑥)𝑒 𝑓 (𝑥)

𝑦= ln ( 𝑓 (𝑥 ))𝑑𝑦𝑑𝑥=

𝑓 ′ (𝑥)𝑓 (𝑥)

𝑦=𝑠𝑖𝑛𝑥𝑑𝑦𝑑𝑥=𝑐𝑜𝑠𝑥

𝑦=𝑐𝑜𝑠𝑥𝑑𝑦𝑑𝑥=−𝑠𝑖𝑛𝑥

𝑦=𝑡𝑎𝑛𝑥𝑑𝑦𝑑𝑥 =𝑠𝑒𝑐2𝑥

𝑦=𝑐𝑜𝑠𝑒𝑐𝑥𝑑𝑦𝑑𝑥 =−𝑐𝑜𝑠𝑒𝑐𝑥𝑐𝑜𝑡𝑥

𝑦=𝑐𝑜𝑡𝑥𝑑𝑦𝑑𝑥 =−𝑐𝑜𝑠𝑒𝑐2𝑥

𝑦=𝑠𝑒𝑐𝑥𝑑𝑦𝑑𝑥 =𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥

IntegrationYou need to be able to integrate standard

functions

Therefore, you already can deduce the following

6A

∫𝑥𝑛=¿¿𝑥𝑛+1

𝑛+1+𝐶

∫𝑒𝑥=¿¿𝑒𝑥+𝐶

∫ 1𝑥=¿¿ln ∨𝑥∨+𝐶

The modulus sign is used here to avoid potential problems with negative numbers…

∫𝑐𝑜𝑠𝑥=¿¿𝑠𝑖𝑛𝑥+𝐶

∫ 𝑠𝑖𝑛𝑥=¿¿−𝑐𝑜𝑠𝑥+𝐶

∫ 𝑠𝑒𝑐2 𝑥=¿¿𝑡𝑎𝑛𝑥+𝐶

∫𝑐𝑜𝑠𝑒𝑐𝑥𝑐𝑜𝑡𝑥=¿¿−𝑐𝑜𝑠𝑒𝑐𝑥+𝐶

∫𝑐𝑜𝑠𝑒𝑐2 𝑥=¿¿−𝑐𝑜𝑡𝑥+𝐶

∫ 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥=¿¿𝑠𝑒𝑐𝑥+𝐶

IntegrationYou need to be able to

integrate standard functions

Find the following integral:

∫(2𝑐𝑜𝑠𝑥+ 3𝑥 −√𝑥)𝑑𝑥∫ 2𝑐𝑜𝑠𝑥

∫(2𝑐𝑜𝑠𝑥+ 3𝑥 −√𝑥)𝑑𝑥

¿2 𝑠𝑖𝑛𝑥∫ 3𝑥

¿3 ln ∨𝑥∨¿∫ √𝑥

¿𝑥32

32

¿∫𝑥12

¿ 23𝑥32

As the terms are separate you can integrate them

separately

Rewrite this term as a

power

¿2 𝑠𝑖𝑛𝑥+3 ln|𝑥|− 23𝑥32+𝐶 Remember

the + C!

IntegrationYou need to be able to

integrate standard functions


∫( 𝑐𝑜𝑠𝑥𝑠𝑖𝑛2𝑥−2𝑒𝑥)𝑑𝑥

∫ 𝑐𝑜𝑠𝑥𝑠𝑖𝑛2𝑥

¿−𝑐𝑜𝑠𝑒𝑐𝑥

∫ 2𝑒𝑥

¿2𝑒𝑥

As the terms are separate you can integrate them

separately

Try to rewrite as an integral you

‘know’

¿−𝑐𝑜𝑠𝑒𝑐𝑥−2𝑒𝑥+𝐶 Remember the + C!

∫( 𝑐𝑜𝑠𝑥𝑠𝑖𝑛2𝑥−2𝑒𝑥)𝑑𝑥

¿∫ 𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝑥

1𝑠𝑖𝑛𝑥

¿∫𝑐 𝑜𝑡𝑥𝑐𝑜𝑠𝑒𝑐𝑥

Teachings for Exercise 6B

IntegrationYou can integrate using the

reverse of the Chain rule

This technique will only work for linear transformations of functions such as f(ax + b)


6B

∫ cos (2𝑥+3 )𝑑𝑥

𝑦=sin (2 𝑥+3)𝑑𝑦𝑑𝑥=2cos (2𝑥+3)

Consider starting with sin(2x + 3) and the

answer that would give

This is double what we are wanting to integrate

Therefore, we must ‘start’ with half the

amount…∫ cos (2𝑥+3 )𝑑𝑥¿12 sin

(2 𝑥+3 )+𝐶Divide the original

‘guess’ by 2

This is a VERY common method of integration – considering what we might start with that would differentiate to our

answer…





6B

∫𝑒4 𝑥+1𝑑𝑥

𝑦=𝑒4 𝑥+1

𝑑𝑦𝑑𝑥 =4 𝑒4𝑥+1

Consider starting with e4x + 1

and the answer that would give

This is four times what we are wanting to integrate

Therefore, we must ‘start’ with a quarter of the

amount…∫𝑒4 𝑥+1𝑑𝑥

Divide the original ‘guess’ by 4





6B

∫ 𝑠𝑒𝑐23 𝑥 𝑑𝑥

𝑦=𝑡𝑎𝑛3 𝑥𝑑𝑦𝑑𝑥 =3𝑠𝑒𝑐23 𝑥

Consider starting with tan3x and the answer that would

give

This is three times what we are wanting to integrate

Therefore, we must ‘start’ with a third of the amount…

∫ 𝑠𝑒𝑐23 𝑥𝑑𝑥

Divide the original ‘guess’ by 3




These three answers illustrate a rule:

6B

∫ cos (2𝑥+3 )𝑑𝑥¿12 sin

(2 𝑥+3 )+𝐶∫𝑒4 𝑥+1𝑑𝑥

∫ 𝑠𝑒𝑐23 𝑥𝑑𝑥

∫ 𝑓 ′ (𝑎𝑥+𝑏 )=¿¿1𝑎 𝑓 (𝑎𝑥+𝑏)+𝐶1) Integrate the function using what you know from C3

2) Divide by the coefficient of x

3) Simplify if possible and add C





6B

∫ 𝑓 ′ (𝑎𝑥+𝑏 )=¿¿1𝑎 𝑓 (𝑎𝑥+𝑏)+𝐶

∫ 13 𝑥+2 𝑑𝑥

∫ 13 𝑥+2 𝑑𝑥

¿13 ln

|3 𝑥+2|+𝐶

1) Integrate the function using what you know from C3

2) Divide by the coefficient of x

3) Simplify if possible and add C





6B

∫ 𝑓 ′ (𝑎𝑥+𝑏 )=¿¿1𝑎 𝑓 (𝑎𝑥+𝑏)+𝐶

∫ ¿¿

𝑦=¿ Consider a function that would leave you with a power 4 and

the same bracket

Simplify after using the Chain rule

As this is 10 times what we want, we need to divide our ‘guess’ by 10

𝑑𝑦𝑑𝑥 =5¿

𝑑𝑦𝑑𝑥 =10¿

∫ ¿¿¿110 ¿

Teachings for Exercise 6C

IntegrationYou can use Trigonometric Identities in Integration

There are some trigonometric expression you cannot integrate yet, but a way to do so is to write

them in terms of ones you can integrate…

Trigonometric Identities are invaluable in this!

Find:

6C

∫𝑡𝑎𝑛2 𝑥𝑑𝑥

∫𝑡𝑎𝑛2 𝑥𝑑𝑥

𝑠𝑖𝑛2𝑥+𝑐𝑜𝑠2𝑥≡1𝑡𝑎𝑛2𝑥+1≡𝑠𝑒𝑐2𝑥𝑡𝑎𝑛2𝑥≡𝑠𝑒𝑐2𝑥−1

Divide by cos

Subtract 1

¿∫ (𝑠𝑒𝑐2𝑥−1)𝑑𝑥

∫ 𝑠𝑒𝑐2 𝑥 ∫1¿ 𝑡𝑎𝑛𝑥 ¿ 𝑥

¿ 𝑡 𝑎𝑛𝑥 −𝑥+𝐶

Using the identity above, replace

tan2x

Integrate each part separately





Find:

6C

∫ 𝑠𝑖𝑛2𝑥 𝑑𝑥

∫ 𝑠𝑖𝑛2𝑥 𝑑𝑥

𝑐𝑜𝑠2𝑥=1−2𝑠𝑖𝑛2𝑥2𝑠𝑖𝑛2𝑥=1−𝑐𝑜𝑠2𝑥

𝑠𝑖𝑛2𝑥≡ 12 −12 𝑐𝑜𝑠2𝑥

Rearrange

Divide by 2

¿∫( 12− 12 𝑐𝑜𝑠2 𝑥)𝑑𝑥∫ 12 ∫ 12 𝑐𝑜𝑠 2 𝑥¿12 𝑥

¿12 𝑥−

14 𝑠𝑖𝑛2 𝑥+𝐶

Using the identity above, replace

sin2x

Integrate each part separately

𝑦=𝑠𝑖𝑛2𝑥𝑑𝑦𝑑𝑥=2𝑐𝑜𝑠2 𝑥

¿14 𝑠𝑖𝑛2 𝑥 Use the

‘guessing’ method

This is 4 times what we want so divide the

‘guess’ by 4





Find:

6C

∫ 𝑠𝑖𝑛3 𝑥𝑐𝑜𝑠3𝑥 𝑑𝑥

s

s

s

12 𝑠𝑖𝑛6 𝑥=𝑠𝑖𝑛3 𝑥𝑐𝑜𝑠3 𝑥

Double angle formula

Follow the pattern…

Divide by 2

∫ 𝑠𝑖𝑛3 𝑥𝑐𝑜𝑠3𝑥 𝑑𝑥∫ 12 𝑠𝑖𝑛6 𝑥 𝑑𝑥

𝑦=𝑐𝑜𝑠6 𝑥𝑑𝑦𝑑𝑥 =−6 𝑠𝑖𝑛6 𝑥

∫ 12 𝑠𝑖𝑛6 𝑥 𝑑𝑥

¿− 112 𝑐𝑜𝑠6 𝑥+𝐶

Replace with the above…

This will give us the sin 6x when differentiating, but is

negative and 12 times too big!

Make the guess negative and divide by 12!





Find:

6C

∫ ¿¿

∫ ¿¿

∫ (𝑠𝑒𝑐2𝑥+𝑡𝑎𝑛2𝑥+2𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 )𝑑𝑥

𝑠𝑖𝑛2𝑥+𝑐𝑜𝑠2𝑥≡1𝑡𝑎𝑛2𝑥+1≡𝑠𝑒𝑐2𝑥𝑡𝑎𝑛2𝑥≡𝑠𝑒𝑐2𝑥−1

Divide by cos

Subtract 1

∫ (𝑠𝑒𝑐2𝑥+(𝑠𝑒𝑐2 𝑥−1)+2𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 )𝑑𝑥∫ (2𝑠𝑒𝑐2 𝑥−1+2 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 )𝑑𝑥∫ 2𝑠𝑒𝑐2𝑥 ∫1 ∫ 2𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥¿2 𝑡𝑎𝑛𝑥 ¿ 𝑥 ¿2 𝑠𝑒𝑐𝑥

Expand the bracket

Integrate separately

Replace tan2x

Simplify

¿2 𝑡𝑎𝑛𝑥− 𝑥+2 𝑠𝑒𝑐𝑥+𝐶

Teachings for Exercise 6D

IntegrationYou can use partial

fractions to integrate expressions

This allows you to split a fraction up – it can

sometimes be recombined after integration…

Find:

6D

∫ 𝑥−5(𝑥+1)(𝑥−2)

𝑥−5(𝑥+1) (𝑥−2)¿

𝐴(𝑥+1)

+𝐵

(𝑥−2)

¿𝐴(𝑥−2)

(𝑥+1)(𝑥−2)+

𝐵(𝑥+1)(𝑥+1)(𝑥−2)

𝑥−5(𝑥+1) (𝑥−2)

¿𝐴 (𝑥−2 )+𝐵(𝑥+1)

(𝑥+1)(𝑥−2)𝑥−5

(𝑥+1) (𝑥−2)

𝑥−5¿ 𝐴 (𝑥−2 )+𝐵(𝑥+1)Let x = 2 −3¿3𝐵

−1¿𝐵Let x = -1 −6¿−3 𝐴

2¿ 𝐴

𝑥−5(𝑥+1) (𝑥−2)¿

𝐴(𝑥+1)

+𝐵

(𝑥−2)

𝑥−5(𝑥+1) (𝑥−2)¿

2(𝑥+1)

− 1(𝑥−2)

Write as two fractions and make the denominators

equal

Combine

The numerators must be equal

Calculate A and B by choosing

appropriate x values

Replace A and B from the start

∫ 2(𝑥+1)

− 1(𝑥−2)





Find:

6D

∫ 𝑥−5(𝑥+1)(𝑥−2)

∫ 2(𝑥+1)

− 1(𝑥−2)

∫ 2(𝑥+1)

− 1(𝑥−2)

∫ 2(𝑥+1) ∫ 1

(𝑥−2)¿2 ln ∨𝑥+1∨¿ ¿ ln ∨𝑥−2∨¿¿ ln ∨¿

¿ ln ∨¿

¿ ln |(𝑥+1 )2

𝑥−2 |+𝐶


You can combine the natural logarithms as

a division





Find:

6D

∫ 9 𝑥2−3𝑥+29𝑥2−4

9 𝑥2−3𝑥+29 𝑥2−41

9 𝑥2 −4−−3 𝑥+6

1) Divide the first term by the highest power

2) Multiply the answer by the whole expression

you’re dividing by3) Subtract to find the

remainder

4) Remember to write the remainder as a fraction of

the original expression

9𝑥2−3 𝑥+29 𝑥2−4 ¿1

+−3 𝑥+69 𝑥2−4

¿1+6−3𝑥9 𝑥2−4

Looks tidier!





Find:

6D

∫ 9 𝑥2−3𝑥+29𝑥2−4

9𝑥2−3 𝑥+29 𝑥2−4 ¿1

+6−3𝑥9 𝑥2−4

We now need to write the remainder as partial

fractions

6−3 𝑥9𝑥2−4¿

6−3 𝑥(3𝑥+2)(3 𝑥−2)

¿𝐴

(3𝑥+2)+

𝐵(3 𝑥−2)

¿𝐴 (3 𝑥−2 )+𝐵(3 𝑥+2)

(3𝑥−2)(3 𝑥+2)

𝐴 (3𝑥−2 )+𝐵 (3 𝑥+2 )=6−3 𝑥4𝐵=4Let x = 2/3

𝐵=1Let x = -2/3 −4 𝐴=8

𝐴=−21+ 6−3 𝑥9 𝑥2−4 ¿1− 2

3 𝑥+2 +1

3 𝑥−2

Set the numerators equal and solve for A

and B

Write the final answer with the remainder

broken apart!





Find:

6D

∫ 9 𝑥2−3𝑥+29𝑥2−4

9𝑥2−3 𝑥+29 𝑥2−4

¿1− 23 𝑥+2 +

13 𝑥−2

∫1− 23 𝑥+2+

13 𝑥−2

∫1 ∫ 23 𝑥+2 ∫ 1

3 𝑥−2¿ 𝑥 ¿ (2 )( 13 ) ln ∨3 𝑥+2∨¿ ¿ ( 13 ) ln ∨3𝑥−2∨¿

¿ ( 13 ) ln ∨ (3𝑥+2 )2∨¿

¿ 𝑥− 13 ln|(3𝑥+2 )2|+ 13 ln|3𝑥−2|+𝐶

¿ 𝑥+ 13 | (3 𝑥−2)(3𝑥+2)2|+𝐶


You can combine the natural logarithms

(be careful, the negative goes on the

bottom…)

Teachings for Exercise 6E

IntegrationYou can Integrate by using

standard patterns

You have seen how to integrate fractions of the form:

Including using partial fractions where an expression can be

factorised

However, this method will not work for integrals of the form:

Some expressions like this can by integrated by using the ‘standard

patterns’ technique

6E

12𝑥+3

1𝑥2+1

IntegrationYou can Integrate by

using standard patterns


Some expressions can by integrated by using the

‘standard patterns’ technique

Find:

6E

∫ 2 𝑥𝑥2+1

𝑑𝑥

∫ 2 𝑥𝑥2+1

𝑑𝑥Notice that the denominator would

differentiate to become the numerator This is a pattern we can use to figure

out what the integral is…

Remember…𝑖𝑓 𝑦=ln ∨ 𝑓 (𝑥 )∨¿

𝑑𝑦𝑑𝑥=

𝑓 ′ (𝑥)𝑓 (𝑥)

So imagine starting with ln|denominator|

𝑦=ln ∨𝑥2+1∨¿𝑑𝑦𝑑𝑥 =

2𝑥𝑥2+1

In this case, we get straight to the answer!

∫ 2 𝑥𝑥2+1

𝑑𝑥¿ ln|𝑥2+1|+𝐶






Find:

6E

∫ 𝑐𝑜𝑠𝑥3+2𝑠𝑖𝑛𝑥 𝑑𝑥


𝑦=ln ∨3+2 𝑠𝑖𝑛𝑥∨¿

𝑑𝑦𝑑𝑥 =

2𝑐𝑜𝑠𝑥3+2𝑠𝑖𝑛𝑥


¿12 ln

|3+2 𝑠𝑖𝑛𝑥|+𝐶

Start by trying y = ln|

denominator|

Differentiate

This is double what we want so multiply the ‘guess’ by 1/2

𝑖𝑓 𝑦=ln ∨ 𝑓 (𝑥 )∨¿

𝑑𝑦𝑑𝑥 =

𝑓 ′ (𝑥)𝑓 (𝑥)






Find:

6E

∫ 3𝑐𝑜𝑠𝑥𝑠𝑖𝑛2𝑥 𝑑𝑥

∫ 3𝑐𝑜𝑠𝑥𝑠𝑖𝑛2𝑥 𝑑𝑥

𝑦=𝑠𝑖𝑛3𝑥

In this case consider the power of sine.

If it has been differentiated, it must have been sin3x originally…

𝑦=¿𝑑𝑦𝑑𝑥 =3¿(𝑐𝑜𝑠𝑥)

𝑑𝑦𝑑𝑥 =3𝑐𝑜𝑠𝑥𝑠𝑖𝑛2𝑥

∫ 3𝑐𝑜𝑠𝑥𝑠𝑖𝑛2𝑥 𝑑𝑥¿ 𝑠𝑖𝑛3𝑥+𝐶

Write as a cubed bracket

Differentiate using the chain rule

Rewrite – this has given us exactly what we

wanted!

Don’t forget the + C!






Find:

6E

∫𝑥 ¿¿

∫𝑥 ¿¿

𝑦=¿

Consider the power on the bracket

As it is a power 3, it must have been a power 4 before

differentiation

𝑑𝑦𝑑𝑥 =4 ¿(2 𝑥)

𝑑𝑦𝑑𝑥=8 𝑥¿

∫𝑥 ¿¿¿18 ¿

Differentiate the bracket to the power 4 using the chain rule

Simplify

This is 8 times too big so multiply the ‘guess’ by 1/8

Don’t forget to add C!






Find:

6E

∫ 𝑐𝑜𝑠𝑒𝑐2𝑥¿¿ ¿


𝑦=¿

∫(𝑐𝑜𝑠𝑒𝑐2𝑥 )¿¿

𝑑𝑦𝑑𝑥=−2¿(−𝑐𝑜𝑠𝑒𝑐2 𝑥)

𝑑𝑦𝑑𝑥=2𝑐𝑜𝑠𝑒𝑐2𝑥¿

𝑑𝑦𝑑𝑥=

2𝑐𝑜𝑠𝑒𝑐2𝑥 ¿¿


¿12 ¿

Write using powers

Imagine how we could end up with a -3 as a power...

Use the chain rule

Rewrite

This is double what we want so multiply

the ‘guess’ by 1/2






Find:

6E

∫5 𝑡𝑎𝑛𝑥𝑠𝑒𝑐4 𝑥 𝑑𝑥

∫5 𝑡𝑎𝑛𝑥𝑠𝑒𝑐4 𝑥 𝑑𝑥𝑦=𝑠𝑒𝑐5𝑥𝑦=¿𝑑𝑦𝑑𝑥 =5¿(𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥)

𝑑𝑦𝑑𝑥 =5 𝑡𝑎𝑛𝑥𝑠𝑒𝑐5𝑥

Consider using a power 5

Write as a bracket to the power 5

Differentiate using the chain

ruleWe have an extra

secx

HOWEVER:

We cannot just add this to our ‘guess’ as before, as the differentiation will need to be performed using

the product rule from C3, rather than the Chain rule!

We need to find another way!






Find:

6E

∫5 𝑡𝑎𝑛𝑥𝑠𝑒𝑐4 𝑥 𝑑𝑥

∫5 𝑡𝑎𝑛𝑥𝑠𝑒𝑐4 𝑥 𝑑𝑥𝑦=𝑠𝑒𝑐4 𝑥𝑦=¿𝑑𝑦𝑑𝑥=4 ¿(𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥)

𝑑𝑦𝑑𝑥 =4 𝑡𝑎𝑛𝑥𝑠𝑒𝑐4 𝑥

Consider using a power 4

Write as a bracket to the power 4

Differentiate using the chain

rule

This is what we want, but 4/5 of the amount

Multiply by the guess by 5/4

𝑦=54 𝑠𝑒𝑐

4 𝑥+𝐶∫5 𝑡𝑎𝑛𝑥𝑠𝑒𝑐4 𝑥 𝑑𝑥

Teachings for Exercise 6F

IntegrationIt is sometimes possible to

simplify an integral by changing the variable. This is known as integration by substitution.

Use the substitution:

To find:

6F

𝑢=2𝑥+5

∫𝑥 √2 𝑥+5𝑑𝑥

∫𝑥 √2 𝑥+5𝑑𝑥𝑢=2𝑥+5

To integrate this, you need to replace the x terms with equivalent u terms, and replace the dx with an

equivalent du

𝑑𝑢𝑑𝑥=2

𝑑𝑢2 =𝑑𝑥

𝑢−5=2 𝑥12 (𝑢−5)=𝑥

Differentiate Rearrange to find x

Rearrange to get dx

∫𝑥 √2 𝑥+5𝑑𝑥∫ 12 (𝑢−5)√𝑢

𝑑𝑢2

∫ 14

(𝑢−5 ) √𝑢𝑑𝑢

Replace each ‘x’ term with an equivalent ‘u’

term

Rearrange – you should leave ‘du’ at the end

∫ 14𝑢32− 54𝑢12 𝑑𝑢

Combine terms including the square root, changed to a

power ‘1/2’




To find:

6F

𝑢=2𝑥+5

∫𝑥 √2 𝑥+5𝑑𝑥

∫ 14𝑢32− 54𝑢12 𝑑𝑢

¿ ( 14 )𝑢52

52

−( 54 ) 𝑢32

32

¿ ( 14 ) 25 𝑢52−( 54 ) 23𝑢

52

¿ 15𝑢52− 56𝑢52+𝐶

+𝐶

+𝐶Differentiate

terms separately

Flip the dividing fractions

Calculate the fraction parts

¿ 15(2𝑥+5)

52− 56(2𝑥+5)

52+𝐶

Finally, replace with

u with its equivalent from the

start!




To find:

6F

𝑢=𝑠𝑖𝑛𝑥+1

∫𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝑥¿¿

∫𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝑥¿¿𝑢=𝑠𝑖𝑛𝑥+1

𝑑𝑢𝑑𝑥=𝑐𝑜𝑠𝑥 𝑢−1=𝑠𝑖𝑛𝑥𝑑𝑢𝑐𝑜𝑠𝑥=𝑑𝑥

Differentiate Rearrange to find Sinx

Rearrange to get dx

∫𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝑥¿¿∫𝑐𝑜𝑠𝑥(𝑢−1)𝑢3𝑑𝑢𝑐𝑜𝑠𝑥

∫ (𝑢−1 )𝑢3𝑑𝑢∫𝑢4−𝑢3𝑑𝑢¿15𝑢

5− 14𝑢4+𝐶

¿15 (𝑠𝑖𝑛𝑥+1)5− 14 (𝑠𝑖𝑛𝑥+1)4+𝐶

Replace each ‘x’ term with an equivalent ‘u’

termCancel the Cosx terms

Multiply out

Integrate

Replace u with x terms again!


simplify an integral by changing the variable. This is

known as integration by substitution.

Use integration by substitution to find:

Sometimes you will have to decide on a substitution yourself.

In this case, the bracket would be hardest to integrate so it makes sense

to use the substitution:

6F

∫0

2

𝑥 ¿¿

𝑢=𝑥+1

𝑢=𝑥+1

∫0

2

𝑥 ¿¿

𝑑𝑢𝑑𝑥=1

𝑑𝑢=𝑑𝑥

𝑢−1=𝑥Differentiate Rearrange to

find x

Rearrange to get dx

𝑢=𝑥+1You also need to recalculate

limits in terms of u

𝑥=2 ,𝑢=3𝑥=0 ,𝑢=1

∫1

3

(𝑢−1)𝑢3𝑑𝑢

∫1

3

𝑢4−𝑢3𝑑𝑢

¿ [𝑢55 −𝑢4

4 ]1

3

¿ ( 35

5− 3

4

4 )−( 15

5− 1

4

4 )¿28.4

Replace x limits with u limits and the x terms with u terms

∫0

2

𝑥 ¿¿

Multiply out bracket

Integrate

Sub in limits and

calculate

An alternative method is to replace the ‘u’ terms with x terms at the end

and then just use the original ‘x’ limits – either way is fine!

Teachings for Exercise 6G

IntegrationYou can use integration by

parts to integrate some expressions

6G

𝑑𝑑𝑥 (𝑢𝑣 )=𝑢 𝑑𝑣𝑑𝑥 +𝑣 𝑑𝑢𝑑𝑥

In C3 you met the following: (the product rule)

This is the differential of two functions multiplied

together You could think of it as:

𝑑 (𝑢𝑣 )𝑑𝑥

∫𝑢 𝑑𝑣𝑑𝑥=∫ 𝑑𝑑𝑥 (𝑢𝑣 )−∫𝑣 𝑑𝑢𝑑𝑥

𝑢 𝑑𝑣𝑑𝑥=𝑑𝑑𝑥 (𝑢𝑣 )−𝑣 𝑑𝑢𝑑𝑥

∫𝑢 𝑑𝑣𝑑𝑥=𝑢𝑣−∫𝑣 𝑑𝑢𝑑𝑥

Rearrange by subtracting vdu/dx

Integrate each term with respect to x

The middle term is just a differential.

Integrating a differential cancels

them both out!

The other terms do not cancel as only part of them are differentiated…

∫𝑢 𝑑𝑣𝑑𝑥=𝑢𝑣−∫𝑣 𝑑𝑢𝑑𝑥This is the formula used for

Integration by parts! You get given this in the

booklet



Find:

You can recognise that Integration by parts is needed as we have two

functions multiplied together…

6G


∫𝑥𝑐𝑜𝑠𝑥 𝑑𝑥

∫𝑢 𝑑𝑣𝑑𝑥=𝑢𝑣−∫𝑣 𝑑𝑢𝑑𝑥 𝑢=𝑥𝑑𝑢𝑑𝑥=1

𝑣=𝑠𝑖𝑛𝑥𝑑𝑣𝑑𝑥=𝑐𝑜𝑠𝑥

Unlike when using the product rule, we now have

one function to differentiate, and one to integrate…

Differentiate Integrate

¿ (𝑥)(𝑠𝑖𝑛𝑥)−∫(𝑠𝑖𝑛𝑥) (1)¿ 𝑥𝑠𝑖𝑛𝑥−(−𝑐𝑜𝑠𝑥)

¿ 𝑥𝑠𝑖𝑛𝑥+𝑐𝑜𝑠𝑥+𝐶

The integral here is simpler!

Be careful with negatives

here!

Now replace the relevant parts to find the integral…

As a general rule, it is easiest to let u = anything of the form xn. The exception is when there is a lnx term, in which case this should be used as

u



Find:



6G


∫𝑥2 𝑙𝑛𝑥𝑑𝑥

Let u be lnx!

∫𝑢 𝑑𝑣𝑑𝑥=𝑢𝑣−∫𝑣 𝑑𝑢𝑑𝑥 𝑢=𝑙𝑛𝑥𝑑𝑢𝑑𝑥=

1𝑥

𝑣=𝑥33

𝑑𝑣𝑑𝑥=𝑥2



¿ (𝑙𝑛𝑥 )(𝑥3

3 )−∫( 𝑥3

3 )( 1𝑥 )¿𝑥33𝑙𝑛𝑥−∫ 13 𝑥

2

¿𝑥33𝑙𝑛𝑥− 𝑥

2

9+𝐶

Simplify terms

Integrate the second part



Find:



Sometimes you will have to use the process twice!

This happens if the new integral still has two functions multiplied

together…

6G


∫𝑥2𝑒𝑥𝑑𝑥

∫𝑢 𝑑𝑣𝑑𝑥=𝑢𝑣−∫𝑣 𝑑𝑢𝑑𝑥𝑢=𝑥2𝑑𝑢𝑑𝑥=2𝑥

𝑣=𝑒𝑥

𝑑𝑣𝑑𝑥=𝑒𝑥

Differentiate IntegrateNow replace the relevant parts to find the integral…

¿ (𝑥2)(𝑒𝑥 )−∫(𝑒𝑥 )(2 𝑥)¿ 𝑥2𝑒𝑥−∫ 2𝑥𝑒𝑥 𝑢=2𝑥

𝑑𝑢𝑑𝑥=2

𝑣=𝑒𝑥

𝑑𝑣𝑑𝑥=𝑒𝑥


¿ 𝑥2𝑒𝑥− [2𝑥 𝑒𝑥−∫ 2𝑒𝑥 ]¿ 𝑥2𝑒𝑥− [2 𝑥𝑒𝑥−2𝑒𝑥 ]

¿ 𝑥2𝑒𝑥−2 𝑥𝑒𝑥+2 𝑥𝑒𝑥+𝐶

Work out the square bracket which is the

second integration by parts

Careful with negatives!!



Evaluate:

Leave your answer in terms of natural logarithms…

You will be asked to leave exact answers a lot so make sure you

know your log laws!!

6G



∫1

2

𝑙𝑛𝑥𝑑𝑥

𝑢=𝑙𝑛𝑥𝑑𝑢𝑑𝑥=

1𝑥

𝑣=𝑥𝑑𝑣𝑑𝑥=1


When integrating lnx, you MUST think of it as ‘lnx times 1, and use lnx as ‘u’ and 1 as

‘dv/dx’


¿ (𝑙𝑛𝑥 )(𝑥)−∫ (𝑥)( 1𝑥 )¿ 𝑥𝑙𝑛𝑥−∫1¿ [ 𝑥𝑙𝑛𝑥− 𝑥 ]1

2

¿ (2 𝑙𝑛2−2 )−(1 𝑙𝑛1−1)

¿2 𝑙𝑛2−1

Simplify terms

Integrate and use a square bracket with

limitsSub in the limits

Calculate and leave in terms

of ln2

Teachings for Exercise 6H

IntegrationYou can use numerical

integration

In C2 you saw how to estimate the area under a curve by using

the trapezium rule

This method can take time and is only an approximation, but it

allows you to find areas of functions that otherwise can be

extremely hard to Integrate

This is because using the trapezium rule actually avoids

Integration altogether!

6H

∫𝑎

𝑏

𝑦 𝑑𝑥≈ 12 h [ 𝑦𝑜+2 ( 𝑦1+𝑦 2+…+𝑦𝑛−1 )+ 𝑦𝑛 ]

h=𝑏−𝑎𝑛

𝑦 0=¿𝑦 𝑛=¿

The first value for yThe last value for y

The central bracket will contain all the values of y in-between. The more vales

there are, the more accurate the approximation is!y

x

y0 y1 y2

h

y4y3 y5

h h h hba

n is the number of strips the area is split into!


integration

Complete the table of values and use it to find an estimate

for:

6H

∫0

𝜋3

𝑠𝑒𝑐𝑥𝑑𝑥

x

y

π 12

π 6

π 4

π 30

1 1.035

1.155

1.414

2

Calculate the values in the table by substituting the x-values into the

equation above…

≈ 12 h [ 𝑦𝑜+2 ( 𝑦1+𝑦 2+…+ 𝑦𝑛−1 )+ 𝑦𝑛 ]∫0

𝜋3

𝑠𝑒𝑐𝑥𝑑𝑥

h is the height of each strip In the table it is given by the gaps between the x values

usedThe y values correspond to y0, y1 etc…

y0 y1 y2 y3 y4

≈ 12 ( 𝜋12 ) [1+2 (1.035+1.155+1.414 )+2 ]

≈ ( 𝜋24 ) [10.208 ]

≈1.34

Be very careful here – it is easy to make an error

on your calculator!

Calculate and round the answer!

The trapezium rule isn’t any different to in C2 really, just

slightly more tricky functions to use!

h h h h

≈ 12 h [ 𝑦𝑜+2 ( 𝑦1+𝑦 2+…+ 𝑦𝑛−1 )+ 𝑦𝑛 ]


integration

Use the trapezium rule with 4 strips to find an approximation

for:

6H

∫0

2

𝑥𝑠𝑖𝑛𝑥 𝑑𝑥

h=𝑏−𝑎𝑛

h=2−04

h=0.5

Sub in b, a and n (number of strips)

We can then find the height of each

strip!

x

y

0 0.5 1.51 2

0 0.24 0.84 1.50 1.82y0 y1 y2 y3 y4

≈ 12(0.5 ) [ 0+2 (0.24+0.84+1.50 )+1.82 ]

≈ (0.25 ) [6.98 ]

≈1.745

Be very careful here!

Calculate and round

the answer!

≈ 12 h [ 𝑦𝑜+2 ( 𝑦1+𝑦 2+…+ 𝑦𝑛−1 )+𝑦𝑛 ]

Teachings for Exercise 6I

IntegrationYou can use Integration to find

areas and volumes

6I

𝑦=√𝑥

x

y

a b

You already know how to find the area under a curve by

IntegrationImagine we rotated the area shaded around

the x-axis What would be the shape of the solid

formed?

y

x

This would be the solid formed

In this section you will learn how to find the volume of any solid created in this way. It also

involves Integration!


areas and volumes

6J

x

y

a b

In the trapezium rule we thought of the area under a curve being split into trapezia.

To simplify this explanation, we will use rectangles now instead

The height of each rectangle is y at its x-coordinate

The width of each is dx, the change in x valuesSo the area beneath the curve is the sum of ydx

(base x height)

The EXACT value is calculated by integrating y with respect to x (y dx)

y

dx

y

x

For the volume of revolution, each rectangle in the area would become a ‘disc’, a cylinder

The radius of each cylinder would be equal to yThe height of each cylinder is dx, the change in xSo the volume of each cylinder would be given

by πy2dx

The EXACT value is calculated by integrating y2 with respect to x, then multiplying by π.

a b

y

dx


areas and volumes

The volume of revolution of a solid rotated 2π radians around the x-axis between x = a and x = b is

given by:

1) The region R is bounded by the curve y = sin2x, the x-axis and the

vertical lines x = 0 and x = π/2.

Find the volume of the solid formed when the region is rotated

2π radians about the x-axis.

6J

𝑉𝑜𝑙𝑢𝑚𝑒=𝜋∫𝑎

𝑏

𝑦2𝑑𝑥


𝑏

𝑦2𝑑𝑥

𝑉𝑜𝑙𝑢𝑚𝑒=𝜋∫0

𝜋2

(𝑠𝑖𝑛 2𝑥)2𝑑𝑥


𝜋2

𝑠𝑖𝑛22𝑥 𝑑𝑥

𝑐𝑜𝑠2 𝐴≡1−2 𝑠𝑖𝑛2𝐴𝑐𝑜𝑠4 𝐴≡1−2 𝑠𝑖𝑛22𝐴

2𝑠𝑖𝑛22 𝐴≡1−𝑐𝑜𝑠4 𝐴𝑠𝑖𝑛22 𝐴≡ 12 (1−𝑐𝑜𝑠 4 𝐴)


𝜋212(1−𝑐𝑜𝑠 4 𝑥)𝑑𝑥

𝑉𝑜𝑙𝑢𝑚𝑒=12𝜋∫

0

𝜋2

(1−𝑐𝑜𝑠4 𝑥)𝑑𝑥

𝑉𝑜𝑙𝑢𝑚𝑒=12𝜋 [𝑥− 14 𝑠𝑖𝑛4 𝑥 ]0

𝜋2

𝑉𝑜𝑙𝑢𝑚𝑒=12𝜋 [( 𝜋2 − 14 𝑠𝑖𝑛2𝜋)−(0− 14 𝑠𝑖𝑛0) ]

𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 2

4

Sub in a, b and y

Square the bracket

Using the identity Cos2A = 1 – 2sin2A, replace sin22x with

something equivalent

The 1/2 can be put outside the integral

Integrate and use a square bracket with the

limits

Sub in the two limits

And finally we have the volume!


areas and volumes

The volume of revolution can also be calculated when x and y are

given parametrically. In this case you must also include dx/dt in the

integral.

You will also need to change limits so they are in terms of t rather

than x!

6J


𝑏

𝑦2𝑑𝑥


𝑏

𝑦2 𝑑𝑥𝑑𝑡 𝑑𝑡


areas and volumes

The volume of revolution can also be calculated when x and y are given

parametrically. In this case you must also include dx/dt in the integral.

The curve C has parametric equations:

Where t ≥ 0.

The region R is bounded by C, the x-axis and the lines x = 0 and x = 2.

Find the volume of the solid formed when R is rotated 2π radians about the

x-axis. 6J


𝑏



𝑏

𝑦2𝑑𝑥

𝑥=𝑡 (1+𝑡 ) 𝑦=11+𝑡


𝑏



𝑏

( 11+𝑡 )2

𝑥=𝑡+𝑡2𝑑𝑥𝑑𝑡 =1+2𝑡

(1+2 𝑡 ) 𝑑𝑡


𝑏 1+2 𝑡(1+𝑡)2

𝑑𝑡

1+2 𝑡(1+𝑡 )2

≡ 𝐴(1+𝑡 )2

+𝐵1+𝑡

𝐴+𝐵(1+𝑡)(1+𝑡 )2

1+2 𝑡=𝐴+𝐵(1+𝑡 )−1=𝐴t = -1

t = 0 1=𝐴+𝐵2=𝐵 We know A = -1 from before

Replace y, and calculate dx/dt

Square the bracket and combine them

We need to use partial fractions

here

Combine with a common

denominator

Sub in values to find A and B

1+2 𝑡(1+𝑡)2

≡ −1(1+𝑡)2

+21+𝑡

1+2 𝑡(1+𝑡 )2

≡

− 1(1+𝑡)2

21+𝑡


areas and volumes




Where t ≥ 0.



x-axis. 6J


𝑏



𝑏

𝑦2𝑑𝑥

𝑥=𝑡 (1+𝑡 ) 𝑦=11+𝑡


𝑏



𝑏

( 11+𝑡 )2


(1+2 𝑡 ) 𝑑𝑡


𝑏 1+2 𝑡(1+𝑡)2

𝑑𝑡




here𝑉𝑜𝑙𝑢𝑚𝑒=𝜋∫𝑎

𝑏 21+𝑡 −

1(1+𝑡 )2

𝑥=𝑡 (1+𝑡 )We also need to calculate the limits for t rather than

x Sub in the x limits and

solve for tx = 0

0=𝑡 (1+𝑡)𝑡=0𝑜𝑟 −1𝑡=0

x = 2

2=𝑡(1+𝑡)0=𝑡 2+𝑡−20=(𝑡+2)(𝑡−1)𝑡=−2𝑜𝑟 1

𝑡=1


areas and volumes




Where t ≥ 0.



x-axis. 6J


𝑏



𝑏

𝑦2𝑑𝑥

𝑥=𝑡 (1+𝑡 ) 𝑦=11+𝑡


𝑏



𝑏

( 11+𝑡 )2


(1+2 𝑡 ) 𝑑𝑡


𝑏 1+2 𝑡(1+𝑡)2

𝑑𝑡




here𝑉𝑜𝑙𝑢𝑚𝑒=𝜋∫𝑎

𝑏 21+𝑡 −

1(1+𝑡 )2

Use t-limits𝑉𝑜𝑙𝑢𝑚𝑒=𝜋∫

0

1 21+𝑡 −

1(1+𝑡 )2


areas and volumes




Where t ≥ 0.



x-axis. 6J


𝑏



𝑏

𝑦2𝑑𝑥

𝑥=𝑡 (1+𝑡 ) 𝑦=11+𝑡𝑥=𝑡+𝑡2

𝑑𝑥𝑑𝑡 =1+2𝑡


1 21+𝑡 −

1(1+𝑡 )2

¿𝜋 [2 ln|1+𝑡|+ 11+𝑡 ]0

1

¿𝜋 [(2 𝑙𝑛2+ 12 )−(2𝑙𝑛1+1)]¿𝜋 (2 𝑙𝑛2− 12 )

Integrate and write as a square bracket

Sub in limits

separately

Simplify/Calculate

Teachings for Exercise 6J

IntegrationYou can use Integration to solve

differential equations

You can solve differential equations by a process known as the ‘separation of variables.’

When:

you can write

6J

𝑑𝑦𝑑𝑥 = 𝑓 (𝑥 )𝑔(𝑦 )

𝑑𝑦= 𝑓 (𝑥 )𝑔 (𝑦 )𝑑𝑥1

𝑔(𝑦 )𝑑𝑦= 𝑓 (𝑥 )𝑑𝑥

∫ 1𝑔 (𝑦 )

𝑑 𝑦=∫ 𝑓 (𝑥 )𝑑𝑥

Multiply both sides by dx

Divide by g(y)

Take Integrals of both sides


∫ 1𝑔 (𝑦 )

𝑑 𝑦=∫ 𝑓 (𝑥 )𝑑𝑥 Effectively you are just putting the x terms on

one side with dx, and the y terms on the other side, with dy




Find a general solution to the following differential equation:

Finding the general solution means finding a formula for y in terms of x

and C (the unknown part).

Effectively working back to the equation that generated the curve

originally (Integrating!)

6J


∫ 1𝑔 (𝑦 )


(1+𝑥2 ) 𝑑𝑦𝑑𝑥=𝑥𝑡𝑎𝑛𝑦


𝑑𝑦𝑑𝑥=

𝑥𝑡𝑎𝑛𝑦(1+𝑥2)

1𝑡𝑎𝑛𝑦

𝑑𝑦𝑑𝑥=

𝑥(1+𝑥2)

1𝑡𝑎𝑛𝑦 𝑑 𝑦=

𝑥(1+𝑥2)

𝑑𝑥

𝑐 𝑜𝑡𝑦 𝑑 𝑦=𝑥

(1+𝑥2)𝑑𝑥

Divide by (1 + x2)

Divide by tany

Multiply by dx

Replace 1/tany with coty

As the x and y terms are separate, we can now

integrate∫𝑐𝑜𝑡𝑦 𝑑𝑦=∫ 𝑥

(1+𝑥2)𝑑𝑥




Find a general solution to the following differential equation:

Finding the general solution means finding a formula for y in terms of x

and C (the unknown part).

Effectively working back to the equation that generated the curve

originally (Integrating!)

6J


∫ 1𝑔 (𝑦 )



∫𝑐𝑜𝑡𝑦 𝑑𝑦=∫ 𝑥(1+𝑥2)

𝑑𝑥

ln|𝑠𝑖𝑛𝑦|=¿12 ln

|1+𝑥2|+𝐶

ln|𝑠𝑖𝑛𝑦|=¿12 ln

|1+𝑥2|+ 𝑙𝑛𝑘

ln|𝑠𝑖𝑛𝑦|=¿ln ¿ ¿

ln|𝑠𝑖𝑛𝑦|=¿ln|𝑘√1+𝑥2|

𝑠𝑖𝑛𝑦=¿𝑘√1+𝑥2

Integrate each side using whatever

methods you have (including the formula

booklet!)As C is just a number, it can be written as lnk, the natural logarithm of another number

Use the power law

Use the multiplication

lawAs the whole of both

sides is written as natural logarithms, we

can remove them

You can only remove the natural logarithm one the whole of each side

is combined in this way!




Find the particular solution of the differential equation:

given that x = 1 when y = 4

Finding the particular solution means you are also able to find the

unknown value C (or lnk or whatever it is called!)

You start by finding the general solution as before…

6J


∫ 1𝑔 (𝑦 )


𝑑𝑦𝑑𝑥 =

−3(𝑦−2)(2 𝑥+1)(𝑥+2)

𝑑𝑦𝑑𝑥 =

−3(𝑦−2)(2 𝑥+1)(𝑥+2)

1𝑦−2 𝑑𝑦=

−3(2 𝑥+1)(𝑥+2)

𝑑𝑥

∫ 1𝑦−2𝑑𝑦=∫ −3

(2𝑥+1)(𝑥+2)𝑑𝑥

∫ 1𝑦−2𝑑𝑦=∫ 1

𝑥+2−2

2𝑥+1 𝑑𝑥

𝑙𝑛|𝑦−2|=𝑙𝑛|𝑥+2|− 𝑙𝑛|2 𝑥+1|+𝑙𝑛𝑘

𝑙𝑛|𝑦−2|=𝑙𝑛| 𝑥+22 𝑥+1|+ 𝑙𝑛𝑘

𝑙𝑛|𝑦−2|=𝑙𝑛|𝑘(𝑥+2)2𝑥+1 |

𝑦−2=𝑘(𝑥+2)2𝑥+1

Divide by (y – 2)Multiply by dx

Separate the right hand side into partial

fractions

We need to Integrate each side

Now integrate and include lnk

Combine 2 terms using the division

lawInclude the lnk using the multiplication law

Finally remove the logarithms (you could also

move the -2 across by adding 2)




Find the particular solution of the differential equation:

given that x = 1 when y = 4

Finding the particular solution means you are also able to find the

unknown value C (or lnk or whatever it is called!)

You start by finding the general solution as before…

6J


∫ 1𝑔 (𝑦 )


𝑑𝑦𝑑𝑥 =

−3(𝑦−2)(2 𝑥+1)(𝑥+2)

𝑦−2=𝑘(𝑥+2)2𝑥+1

𝑦=𝑘(𝑥+2)2 𝑥+1

+2

4=𝑘(1+2)2(1)+1

+2

4=3𝑘3 +2

2=3𝑘3

6=3𝑘2=𝑘

𝑦=𝑘(𝑥+2)2 𝑥+1

+2 𝑦=2(𝑥+2)2𝑥+1

+2

Rearrange to get y = f(x) (this isn’t essential but can

help!)Sub in y = 4 and x

= 1 from the question

Simplify the fraction parts

Subtract 2

Multiply by 3

Divide by 3

k = 2

General Solution Particular Solution for y = 4 when x = 1

Teachings for Exercise 6K

IntegrationSometimes the differential

equation will arise out of context and the solution might need

interpretation

The rate of increase of a population of micro-organisms at time t is given

by:

Given that at t = 0 the population was size 8, and at t = 1 the

population was 56, find the size of the population at time t = 2.

You are looking to find the original formula linking P and t, so you need

to integrate.

Once you have this formula you can then use it…

6K

𝑑𝑃𝑑𝑡 =𝑘𝑃


1𝑃 𝑑𝑃=𝑘𝑑𝑡

∫ 1𝑃 𝑑𝑃=∫𝑘𝑑𝑡

𝑙𝑛|𝑃|=𝑘𝑡 +𝐶𝑙𝑛|8|=𝐶

𝑙𝑛|𝑃|=𝑘𝑡 +𝑙𝑛 ∨8∨¿𝑙𝑛|56|=𝑘+𝑙𝑛 ∨8∨¿𝑙𝑛|7|=𝑘

𝑙𝑛|𝑃|=𝑘𝑡 +𝐶 𝑙𝑛|𝑃|=𝑡𝑙𝑛7+𝑙𝑛 8

Separate the variables (remember that k just represents an unknown

number, not a variable like x or y…)

We need to integrate both sides

As k is just a number, the integral is kt

Sub in t = 0 and P = 8, this gives us C

We can rewrite the equation, replacing C with

ln8Sub in the second pair of

values

Rearrange to find k

We now have an equation linking P and t

IntegrationSometimes the differential

equation will arise out of context and the solution might need

interpretation

The rate of increase of a population of micro-organisms at time t is given

by:

Given that at t = 0 the population was size 8, and at t = 1 the

population was 56, find the size of the population at time t = 2.

You are looking to find the original formula linking P and t, so you need

to integrate.

Once you have this formula you can then use it…

6K


𝑙𝑛|𝑃|=𝑡𝑙𝑛7+𝑙𝑛 8

𝑙𝑛|𝑃|=2 𝑙𝑛7+𝑙𝑛8𝑙𝑛|𝑃|=𝑙𝑛 49+ 𝑙𝑛8𝑙𝑛|𝑃|=𝑙𝑛 392𝑃=392

Sub in t = 2

Use the power law

Use the multiplication

lawRemove the logarithms

Summary• We have covered how to Integrate

many more functions

• We have seen how Integration can be used to calculate volumes

• We have also seen how it links in with differential equations

integration

Documents

standard functions

fax b

answer integrationyou

c integrationyou

original guess

differentiation chapter

coefficient of x3

big chapter