C1 Chapter 7 Differentiation - Aquinas Maths

C1 Chapter 7 Differentiation - Aquinas Maths

Exercises Given the gradient , find the original function . Q 1 1 2 3 2 4 3 5 4 6 5 7 8 6 9 7 10 8 9 10 ? ? ? ? ? ? ? ? ? ? Further Examples ? ?

Notice that at this point, we havent actually integrated yet, only simplified, hence the integral symbol remains! Working out the c Suppose that for a curve , and the curve passes through the point . Work out . By integrating, Substituting: . So . ? A curve passes through the point and so Thus ? Schoolboy ErrorsTM Whats wrong with these workings? They forgot to put on ? the third line. Theyve integrated on the second line, so they dont want? the integral symbol! Evaluating Definite Integrals 1 2 3 3

2 +2 ? 1 2 4 +3 2 ? Harder Examples Sketch Sketch the the curve curve with with equation equation and and find find the the area area between between the the curve curve and and the the -axis. axis. The Sketch The number crunching Adding: -3 ?

1 ? Integration Integration You need to be able to integrate standard functions You met the following in C3, in the differentiation chapter: = = = 1 = = = = () = = ()

= () = 2 = =ln ( ( ) ) = = ( ) = ( ) 2 = = 6A Integration You need to be able to integrate standard functions Therefore, you already can deduce the following +1 +

+1 + = + = = 1 = ln + The modulus sign is used here to avoid potential problems with negative numbers (More info on the next slide!) + = 2 + = + = 2 + = +

= 6A Integration = You need to be able to integrate standard functions Therefore, you already can deduce the following 1 = ln + This is saying when we Integrate either of the following, we get the same result: ( )= 1 ( )= 1 As we are integrating to find the Area, you can see for any 2 points, the area will be the same for either graph Therefore you can use either x or x in the Integral

= However, you cannot find ln of a negative, just use the positive value instead! 6A Integration You need to be able to integrate standard functions Find the following integral: ( 3 2 + ) As the terms are separate you can integrate them separately (2 + 3 ) 22 3 3 ln

1 2 Rewrite this term as a power 3 2 3 2 3 2 2 3 3 2 2 + 3 ln || 2 + 3 Remember the + C! 6A Integration You need to be able to integrate standard functions Find the following integral: (

2 2 ) Try to rewrite as an integral you know ( As the terms are separate you can integrate them separately 2 2 ) 2 2 1

2 2 + Remember the + C! 6A Ex 6A C4 Integration You can integrate using the reverse of the Chain rule This technique will only work for linear transformations of functions such as f(ax + b) Find the following integral: cos (2+3 ) =sin (2 +3) =2 cos (2 +3) cos (2+3 ) 1 sin ( 2 + 3 ) + 2 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 Divide 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 Integration You can integrate using the reverse of the Chain rule This technique will only work for linear transformations of functions such as f(ax + b) Find the following integral: 4 +1 = 4 +1 4 + 1 =4 4 +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 Divide the original guess by 4 6B Integration You can integrate using the reverse of the Chain rule This technique will only work for linear transformations of functions such as f(ax + b) Find the following integral: 2 3 =3 2 =3 3 2 3 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 Divide the original

guess by 3 6B Integration You can integrate using the reverse of the Chain rule cos (2+3 ) This technique will only work for linear transformations of functions such as f(ax + b) 1 sin ( 2 + 3 ) + 2 4 +1 2 3 These three answers illustrate a rule: (+ )= 1 ( + ) + 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 Integration You can integrate using the reverse of the Chain rule This technique will only work for linear transformations of functions such as f(ax + b) 1 3 +2 1 ln|3 + 2|+ 3 1 ( + ) + (+ )= 1) Integrate the function using what you know from C3 2) Divide by the coefficient of x 3) Simplify if possible and add C Find the following integral: 1 3 +2 6B (+ )=

Integration You can integrate using the reverse of the Chain rule This technique will only work for linear transformations of functions such as f(ax + b) = =5 =10 1 ( + ) + (+ )= Consider a function that would leave you with a power 4 and the same bracket Simplify after using the Chain rule Find the following integral: As this is 10 times what we want, we need to divide our guess by 10 1 10 6B

Ex 6B C4 Integration You can use Trigonometric Identities in Integration There are some trigonometric expressions 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: 2 Divide by cos Subtract 1 2 + 2 1 2 +1 2 2 2 1 2 2 ( 1) 2 1 Using the identity above, replace tan2x Integrate each part

separately + 6C Rearrange Integration You can use Trigonometric Identities in Integration There are some trigonometric expressions 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: 2 Divide by 2 2=12 2 22 =1 2 1 1 2 2 2 2 2 ( 1 1 2

2 2 1 2 1 2 2 1 2 ) 1 2 4 1 1 2 + 2 4 Using the identity above, replace sin2x Integrate each part separately Use the guessing method =2

=2 2 This is 4 times what we want so divide the guess by 4 6C Integration You can use Trigonometric Identities in Integration There are some trigonometric expressions 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: 33 33 1 2 6 = 6 = 6 6 s Double angle formula s Follow the pattern Divide by 2

s 1 6 = 3 3 2 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! 1 2 6 1 6 + 12 6C Divide by cos Integration You can use Trigonometric Identities in Integration There are some trigonometric expressions 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! 2 + 2 1 2 +1 2

Subtract 1 2 2 1 2 2 ( + +2) 2 2 ( +( 1)+2 ) 2 ( 2 1+2) 2 2 1 2 Expand the bracket 2 Replace tan2x Simplify Integrate separately 2 Find: 2 +2 + 6C

What about . 32 Ex 6C C4 Method Q Split into partial fractions. If each factor of the denominator is linear, we can split like such (for constants and ): We dont like fractions in equations, so we could simplify this to: ? METHOD 1: Substitution We can easily eliminate either or by an appropriate choice of : If : If : Therefore: ? Test Your Understanding C4 Jan 2011 Q3 Let : Let : Therefore ? Notice we can move the to the front of the fraction. Note that we dont technically need this last line from the perspective of the mark scheme, but its good to just to be on the safe side

More than two fractions The principle is exactly the same if we have more than two linear factors in the denominator. Q Split into partial fractions. When : When : When : So ? Bro Tip: While substitution is generally the easier method, I sometimes compare coefficients of just the term to avoid having to deal with fractions. No need to expand; we can see by observation that: Then is easy to determine given we know and . Repeated linear factors Q Split into partial fractions. When : When : ? At this point we could substitute something else (e.g. ) but its easier to equate terms. Test Your Understanding C4 June 2011 Q1 ? Dealing with Improper Fractions The degree of a polynomial is the highest power, e.g. a quadratic has degree 2. An algebraic fraction is improper if the degree of the numerator is at least the degree of

the denominator. 2 3 +2 +1 1 3 2 +3 2 ! To split an improper fraction into partial fractions, either: 1. Divide algebraically first. 2. Or introduce a whole term and deal with identity immediately. Dealing with Improper Fractions Q Split into partial fractions. Method 1: Algebraic Division Method 2: Using One Identity (method not in your textbooks but in mark schemes) Let: Dividing algebraically gives: Turn numerator back: Let ? If : If : Comparing coefficients?of : So opinion: I personally think the second method is easier. And mark schemes present it as Method 1 implying more standard!

Test Your Understanding C4 Jan 2013 Q3 ? Integration You can use partial fractions to integrate expressions This allows you to split a fraction up it can sometimes be recombined after integration 5 + Write as two ( + 1)( 2) ( + 1) ( 2) fractions and make the denominators 5 ( 2) ( +1) equal + ( + 1)( 2) ( + 1)( 2) ( +1)( 2) Combin 5 ( 2 )+ ( +1) e ( + 1)( 2) ( +1)( 2) The numerators Let x = 2 Find:

5 ( +1)( 2) 2 1 (+1) ( 2) Let x = -1 5 ( 2 ) + ( +1) 3 3 1 6 3 2 5 + ( + 1)( 2) ( + 1) ( 2) 2 1 5 ( + 1)( 2) ( + 1) ( 2) must be equal Calculate A and B by choosing appropriate x values Replace A and B from the start 6D Integration You can use partial fractions to integrate

expressions 2 1 ( +1) ( 2) This allows you to split a fraction up it can sometimes be recombined after integration 2 1 ( +1) ( 2) 2ln +1 ln 2 ln Find: ln 5 ( +1)( 2) 2 ( +1 ) ln + 2 | | Integrate separately You can combine the

natural logarithms as a division 2 1 (+1) ( 2) 6D Integration You can use partial fractions to integrate expressions 1 9 2 4 9 2 3 +2 4 9 2 3 +6 This allows you to split a fraction up it can sometimes be recombined after integration 2) Multiply the answer by the whole expression youre dividing by 3) Subtract to find the remainder 4) Remember to write the remainder as a fraction of the original expression 2Find: 9 3 +2 9 2 4 1) Divide the first term by the highest power

9 2 3 + 2 2 9 4 + 3 + 6 1 9 2 4 + 6 3 1 9 2 4 Looks tidier! 6D Integration You can use partial fractions to integrate expressions This allows you to split a fraction up it can sometimes be recombined after integration 9 2 3 + 2 9 2 4 We now need to write the remainder as partial fractions 63 6 3 9 2 4 (3 +2)(3 2) + (3 +2) ( 3 2)

( 3 2 )+ (3 + 2) (3 2)(3 + 2) 2Find: 9 3 +2 9 2 4 + 6 3 1 9 2 4 ( 3 2 ) + ( 3 +2 )=6 3 4 =4 =1 4 =8 = 2 Let x = 2/3 Let x = -2/3 1+ Set the numerators equal and solve for A and B 6 3 2 1 1 + 3 +2 3 2 9 2 4 Write the final answer with the remainder broken apart!

6D Integration You can use partial fractions to integrate expressions This allows you to split a fraction up it can sometimes be recombined after integration 2Find: 9 3 +2 9 2 4 2 1 9 2 3 + 2 1 + 2 3 +2 3 2 9 4 2 1 1 3 +2 + 3 2 2 1 3 +2 1 ( 2 ) ln 3 +2 3

Integrate separately 1 3 2 1 ln 3 2 3 () () 1 ( ) ln ( 3 +2 ) 3 2 1 1 2 ln |( 3 +2 ) |+ ln |3 2|+ 3 3 1 ( 3 2) + + 3 (3 +2)2 | | You can combine the natural logarithms (be careful, the negative goes on the

bottom) 6D Ex 6D C4 Integration You can Integrate by using standard patterns You have seen how to integrate fractions of the form: 1 2 + 3 Including using partial fractions where an expression can be factorised However, this method will not work for integrals of the form: 1 2 +1 Some expressions like this can by integrated by using the standard patterns technique 6E Integration You can Integrate by using standard patterns You have seen how to integrate fractions of the form: 2 2 +1 Remember =ln ( )

Some expressions can by integrated by using the standard patterns technique ( ) = ( ) Find: =ln 2 +1 2 2 +1 Notice that the denominator would differentiate to become the numerator This is a pattern we can use to figure out what the integral is 2 = 2 + 1 So imagine starting with ln|denominator| In this case, we get straight to the answer! 2 2 +1 2 ln | +1|+ 6E =ln ( ) Integration ( )

= ( ) You can Integrate by using standard patterns You have seen how to integrate fractions of the form: Some expressions can by integrated by using the standard patterns technique Find: 3+2 3+2 =ln 3+2 2 = 3+ 2 Start by trying y = ln| denominator| Differentiate This is double what we want so multiply the guess by 1/2 3+2 1 ln |3+2 |+ 2 6E Integration You can Integrate by

using standard patterns You have seen how to integrate fractions of the form: Some expressions can by integrated by using the standard patterns technique Find: 2 3 2 3 In this case consider the power of sine. If it has been differentiated, it must have been sin3x originally = 3 = Write as a cubed bracket = 3 () 2 =3 Differentiate using the chain rule Rewrite this has given us exactly what we wanted! 2

3 3 + Dont forget the + C! 6E Integration You can Integrate by using standard patterns You have seen how to integrate fractions of the form: = Some expressions can by integrated by using the standard patterns technique Find: Consider the power on the bracket = 4 (2 ) =8 As it is a power 3, it must have been a power 4 before differentiation 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 1 8 Dont forget to add C! 6E Integration You can Integrate by using standard patterns You have seen how to integrate fractions of the form: Some expressions can by integrated by using the standard patterns technique Find: 2 2 Write using powers 2 ( ) = Imagine how we could end

up with a -3 as a power... = 2 ( 2 ) 2 = 2 2 2 = Use the chain rule Rewrite This is double what we want so multiply the guess by 1/2 2 1 2 6E Integration You can Integrate by using standard patterns You have seen how to integrate fractions of the form: Some expressions can by integrated by using the standard patterns technique Find:

4 5 4 5 = 5 = Consider using a power 5 Write as a bracket to the power 5 =5 () 5 =5 Differentiate using the chain rule We 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! 6E Integration You can Integrate by

using standard patterns You have seen how to integrate fractions of the form: Some expressions can by integrated by using the standard patterns technique Find: 4 5 4 5 = 4 = Consider using a power 4 Write as a bracket to the power 4 = 4 () Differentiate using the chain rule 4 =4 This is what we want, but 4/5 of the amount Multiply by the guess by 5/4 4

5 5 4 = + 4 6E Ex 6E C4 Integration It is sometimes possible to simplify an integral by changing the variable. This is known as integration by substitution. 2+5 =2 +5 Differentiate Rearrange to get dx 5=2 1 ( 5)= 2 = 2 To find: 2+5 Rearrange to find x =2

Use the substitution: =2 +5 To integrate this, you need to replace the x terms with equivalent u terms, and replace the dx with an equivalent du 2+5 1 2 (5) 2 1 4 ( 5 ) 1 3 2 5 1 2 Replace each x term with an equivalent u term Rearrange you should leave du at the end Combine terms including the square root, changed to a power 1/2

6F Integration 3 It is sometimes possible to simplify an integral by changing the variable. This is known as integration by substitution. Use the substitution: =2 +5 To find: 2+5 1 1 2 5 2 4 4 5 5 1 2 4 3 4 5 2 2 ( ) ( ) 5 Differentiate terms separately 3

2 + 3 1 2 2 5 2 2 + 4 5 4 3 ( ) ( ) 5 3 1 5 2 2 + 5 6 5 3 Flip the dividing fractions Calculate the fraction parts 1 5 (2 +5) 2 (2 +5) 2+ 5 6 Finally,

replace with u with its equivalent from the start! 6F Integration It is sometimes possible to simplify an integral by changing the variable. This is known as integration by substitution. Use the substitution: =+1 To find: =+1 Differentiate Rearrange to get dx = Rearrange to find Sinx 1= =

3 ) (1 4 3 Replace each x term with an equivalent u term ( 1)3 Cancel the Cosx terms Multiply out 1 5 1 4 + 5 4 Integrate 1 1 5 4 (+ 1) ( +1) + 5 4 Replace u with x terms again! 6F Integration 2

It is sometimes possible to simplify an integral by changing the variable. This is known as integration by substitution. 0 Differentiate = +1 =1 Rearrange to get dx 1= = You also need to recalculate limits in terms of u Use integration 2 by substitution to find: 0 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: = +1

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! 2 3 ( 1)3 1 3 Multiply out bracket 4 3 1 [ ( 5 Integrate 4 3 5 4 ] 1

4 5 4 1 1 3 3 5 4 5 4 28.4 )( = +1 =2, =3 =0, =1 Replace x limits with u limits and the x terms with u terms 0 5 Rearrange to find x ) Sub in limits and calculate

6F Ex 6F C4 = Integration You can use integration by parts to integrate some expressions You may need the following Integrals, which you are given in the formula booklet =ln||+ =ln||+ =ln|+|+ =ln|+|+ 6G Integration You can use integration by parts to integrate some expressions This is the differential of two functions multiplied together You could think of it as: () =

This is the formula used for Integration by parts! You get given this in the booklet In C3 you met the following: (the product rule) ( ) = + ( ) = ( ) = = 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 As a general rule, it is easiest to let u = anything of the form x n. The exception is when there is a lnx term, in which case this should be used as 6G u = Integration You can use integration by parts to integrate some expressions = Find: You can recognise that Integration by parts is needed as we have two functions multiplied together Unlike when using the product rule, we now have one function to differentiate, and one to integrate

= = =1 = Differentiate Integrate Now replace the relevant parts to find the integral ( )() ()(1) () ++ The integral here is simpler! Be careful with negatives here! As a general rule, it is easiest to let u = anything of the form x n. The exception is when there is a lnx term, in which case this should be used as u 6G = Integration

You can use integration by parts to integrate some expressions = = 1 = Find: 2 Differentiate Integrate Now replace the relevant parts to find the integral 3 You can recognise that Integration by parts is needed as we have two functions multiplied together Let u be lnx! 3 = 3 2 = ( )

3 3 3 1 ( ) ( )( ) 3 1 2 3 3 3 3 3 9 Simplify terms Integrate the second part +

6G = Integration You can use integration by parts to integrate some expressions Find: 2 You can recognise that Integration by parts is needed as we have two functions multiplied together = = =2 = Differentiate

Integrate Now replace the relevant parts to find the integral ( )(2) 2 ( 2 )( ) 2 =2 = =2 = Differentiate Integrate Sometimes you will have to use the process twice! This happens if the new integral still has two functions multiplied together = 2

2 [ 2 2 ] 2 [ 2 2 ] 2 2 +2 + Work out the square bracket which is the second integration by parts Careful with negatives!! 6G = When integrating lnx, you MUST think of it as lnx times 1, and use lnx as u and 1 as dv/dx Integration You can use integration by parts to integrate some expressions

= 2Evaluate: = = 1 = =1 Differentiate Integrate 1 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!! Now replace the relevant parts to find the integral ()( ) () 1 2 [ ] 1

1 () Integrate and use a square bracket with limits ( 2 2 2 (11 ) 1) 2 2 1 Simplify terms Sub in the limits Calculate and leave in terms of ln2 6G Integrating and definite integration Q Find , leaving your answer in terms of natural logarithms. = ln ? =1 Q Find , leaving your answer in terms of natural logarithms. If we were doing it from scratch: ? In general: Ex 6G C4 Overview

Integration by standard result (Theres certain expressions youre expected to know straight off.) 2 sec =tan + Integration by substitution (We make a substitution to hopefully make the expression easier to integrate) Let Integration by reverse chain rule (We imagine what would have differentiated to get the expression.) 1 4 sin cos = 4 sin + 3 Integration by parts (Allows us to integrate a product, just as the product rule allowed us to differentiate one) Test Your Understanding Q Find = =sin ? Summary of Functions How to deal with it Standard Standard result result Standard result In formularesult booklet, but use which is

Standard of the form For both and use identities for ? ? (+constant) Formula booklet? No No No Yes No ? No Yes No No Yes No Yes ? Yes Would use substitution , but too hard for exam. Would use substitution , but too hard for exam. which is of the form No ? No ?

Yes ? Yes ? ? Yes Summary of Functions How to deal with it By observation. By observation. (+constant) ? ? ? For any product of sin and cos with same coefficient of , use double angle. ? Use IBP, where Use algebraic division. Use partial fractions. ? ? ? Use partial fractions. ? Formula booklet? No! Yes

(but memorise) No No No No No No Summary of Functions How to deal with it (+constant) Reverse chain rule. Of form Power around denominator so NOT ? of form . Rewrite as product. Reverse chain rule (i.e. Consider and differentiate) ? For any function where inner function is linear expression, divide by coefficient of Use sensible substitution. or even better, . Reverse chain rule. ? ? Reverse chain rule. ?

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