Download presentation
Presentation is loading. Please wait.
1
Exercises Given the gradient ππ¦ ππ₯ , find the original function π¦. Q π
π π
π = π= 1 π₯ 5 1 6 π₯ 6 +π 2 10 π₯ 4 2 π₯ 5 +π 3 β π₯ β2 π₯ β1 +π 4 π₯ 2 3 3 5 π₯ π 5 6 π₯ 3 3 2 π₯ 4 +π 6 β5 β5π₯+π 7 6π₯ 3 π₯ 2 +π 8 β14 π₯ β8 2 π₯ β7 +π 9 2 π₯ β0.4 10 3 π₯ 0.6 +π 10 5 π₯ β 3 2 β10 π₯ β π ? ? ? ? ? ? ? ? ? ?
2
Further Examples 2 π₯ 3 β3 π₯ ππ₯ = 2 π₯ β3 β3 π₯ = β π₯ β2 β2 π₯ π 2π₯ π₯ +5 π₯ ππ₯ = π₯ 2 + π₯ β π₯ β2 ππ₯ = 4 3 π₯ 3 β 2 π₯ β 5 π₯ +π Notice that at this point, we havenβt actually integrated yet, only simplified, hence the integral symbol remains! ? ππ₯ ?
3
Working out the c Suppose that for a curve πΆ, ππ¦ ππ₯ =2π₯ and the curve passes through the point 3,10 . Work out π¦. By integrating, π= π π +π Substituting: ππ= π π +π. So π=π. π= π π +π A curve π¦=π π₯ passes through the point 4,5 and π β² π₯ = π₯ 2 β2 π₯ π β² π = π π π βπ π β π π π π = π π π π π βπ π π π +π π= π π π π π βπ π +π so π= π π Thus π= π π π π π βπ π π π + π π ? ?
4
Schoolboy ErrorsTM ? ? ππ¦ ππ₯ = 2 π₯ 2 +3 =2 π₯ β2 +3 =β2 π₯ β1 +3π₯+π
Whatβs wrong with these workings? ππ¦ ππ₯ = 2 π₯ =2 π₯ β2 +3 =β2 π₯ β1 +3π₯+π π₯ 2 ππ₯ = π₯ 3 +π ? They forgot to put π¦= on the third line. ? Theyβve integrated on the second line, so they donβt want the integral symbol!
5
Evaluating Definite Integrals
1 2 2 π₯ 3 +2π₯ ππ₯ = π₯ 4 + π₯ = 8+4 β = 21 2 β2 β1 4π₯ 3 +3 π₯ 2 ππ₯ = π₯ 4 + π₯ 3 β2 β1 = 1β1 β 16β8 =β8 ? ?
6
Harder Examples ? ? The Sketch The number crunching
Sketch the curve with equation π¦=π₯ π₯β1 π₯+3 and find the area between the curve and the π₯-axis. The Sketch The number crunching ? π¦ ? π₯ π₯β1 π₯+3 = π₯ 3 β2 π₯ 2 β3π₯ β3 0 π₯ 3 β2 π₯ 2 β3π₯ ππ₯ =11.25 0 1 π₯ 3 β2 π₯ 2 β3π₯ ππ₯ =β 7 12 Adding: =11 5 6 π₯ -3 1
7
Integration
8
Integration π¦= π₯ π π¦=π πππ₯ π¦=πππ πππ₯ ππ¦ ππ₯ =π π₯ πβ1 ππ¦ ππ₯ =πππ π₯
You need to be able to integrate standard functions You met the following in C3, in the differentiation chapter: π¦= π₯ π π¦=π πππ₯ π¦=πππ πππ₯ ππ¦ ππ₯ =π π₯ πβ1 ππ¦ ππ₯ =πππ π₯ ππ¦ ππ₯ =βπππ πππ₯πππ‘π₯ π¦= π π(π₯) π¦=πππ π₯ π¦=πππ‘π₯ ππ¦ ππ₯ = πβ²(π₯)π π(π₯) ππ¦ ππ₯ =βπ πππ₯ ππ¦ ππ₯ =βπππ π π 2 π₯ π¦=lnβ‘(π π₯ ) π¦=π‘πππ₯ π¦=π πππ₯ ππ¦ ππ₯ = πβ²(π₯) π(π₯) ππ¦ ππ₯ =π π π 2 π₯ ππ¦ ππ₯ =π πππ₯π‘πππ₯ 6A
9
Integration π₯ π = π₯ π+1 π+1 +πΆ πππ π₯= πππ πππ₯πππ‘π₯= π πππ₯+πΆ βπππ πππ₯+πΆ
You need to be able to integrate standard functions Therefore, you already can deduce the following π₯ π = π₯ π+1 π+1 +πΆ πππ π₯= πππ πππ₯πππ‘π₯= π πππ₯+πΆ βπππ πππ₯+πΆ π π₯ = π πππ₯= πππ π π 2 π₯= π π₯ +πΆ βπππ π₯+πΆ βπππ‘π₯+πΆ 1 π₯ = π π π 2 π₯= π πππ₯π‘πππ₯= lnβ‘|π₯|+πΆ π‘πππ₯+πΆ π πππ₯+πΆ The modulus sign is used here to avoid potential problems with negative numbersβ¦ (More info on the next slide!) 6A
10
Integration π= π π 1 π₯ = lnβ‘|π₯|+πΆ π= π βπ π π₯ = 1 π₯ π π₯ = 1 βπ₯ 6A
You need to be able to integrate standard functions Therefore, you already can deduce the following This is saying when we Integrate either of the following, we get the same result: As we are integrating to find the Area, you can see for any 2 points, the area will be the same for either graphβ¦ Therefore you can use either x or βx in the Integral However, you cannot find ln of a negative, just use the positive value instead! 1 π₯ = lnβ‘|π₯|+πΆ π= π βπ π π₯ = 1 π₯ π π₯ = 1 βπ₯ 6A
11
Integration 6A 2πππ π₯+ 3 π₯ β π₯ ππ₯
As the terms are separate you can integrate them separately 2πππ π₯+ 3 π₯ β π₯ ππ₯ You need to be able to integrate standard functions Find the following integral: 2πππ π₯ 3 π₯ π₯ =2π πππ₯ =3lnβ‘|π₯| 2πππ π₯+ 3 π₯ β π₯ ππ₯ Rewrite this term as a power = π₯ 1 2 = π₯ = 2 3 π₯ 3 2 =2π πππ₯+3 ln π₯ β 2 3 π₯ πΆ Remember the + C! 6A
12
Integration 6A πππ π₯ π π π 2 π₯ β2 π π₯ ππ₯
As the terms are separate you can integrate them separately πππ π₯ π π π 2 π₯ β2 π π₯ ππ₯ You need to be able to integrate standard functions Find the following integral: πππ π₯ π π π 2 π₯ 2 π π₯ πππ π₯ π π π 2 π₯ β2 π π₯ ππ₯ = πππ π₯ π πππ₯ 1 π πππ₯ =2 π π₯ = πππ‘π₯πππ πππ₯ Try to rewrite as an integral you βknowβ =βπππ πππ₯ Remember the + C! =βπππ πππ₯β2 π π₯ +πΆ 6A
13
Ex 6A C4
14
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: π¦=sinβ‘(2π₯+3) Consider starting with sin(2x + 3) and the answer that would give ππ¦ ππ₯ =2cosβ‘(2π₯+3) This is double what we are wanting to integrate ο Therefore, we must βstartβ with half the amountβ¦ cos 2π₯+3 ππ₯ cos 2π₯+3 ππ₯ Divide the original βguessβ by 2 = 1 2 sin 2π₯+3 +πΆ This is a VERY common method of integration β considering what we might start with that would differentiate to our answerβ¦ 6B
15
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: π¦= π 4π₯+1 Consider starting with e4x + 1 and the answer that would give ππ¦ ππ₯ =4 π 4π₯+1 This is four times what we are wanting to integrate ο Therefore, we must βstartβ with a quarter of the amountβ¦ π 4π₯+1 ππ₯ π 4π₯+1 ππ₯ Divide the original βguessβ by 4 = 1 4 π 4π₯+1 +πΆ 6B
16
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: π¦=π‘ππ3π₯ Consider starting with tan3x and the answer that would give ππ¦ ππ₯ =3π π π 2 3π₯ This is three times what we are wanting to integrate ο Therefore, we must βstartβ with a third of the amountβ¦ π π π 2 3π₯ ππ₯ π π π 2 3π₯ ππ₯ Divide the original βguessβ by 3 = 1 3 π‘ππ3π₯+πΆ 6B
17
Integration π β² ππ₯+π = 1 π π ππ₯+π +πΆ 6B cos 2π₯+3 ππ₯ π 4π₯+1 ππ₯
π 4π₯+1 ππ₯ π π π 2 3π₯ ππ₯ 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) These three answers illustrate a rule: = 1 2 sin 2π₯+3 +πΆ = 1 4 π 4π₯+1 +πΆ = 1 3 π‘ππ3π₯+πΆ 1) Integrate the function using what you know from C3 π β² ππ₯+π = 1 π π ππ₯+π +πΆ 2) Divide by the coefficient of x 3) Simplify if possible and add C 6B
18
π β² ππ₯+π = 1 π π ππ₯+π +πΆ Integration 1 3π₯+2 ππ₯ 1) Integrate the function using what you know from C3 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) Divide by the coefficient of x = 1 3 ln 3π₯+2 +πΆ 3) Simplify if possible and add C 1 3π₯+2 ππ₯ 6B
19
As this is 10 times what we want, we need to divide our βguessβ by 10
π β² ππ₯+π = π β² ππ₯+π = 1 π π ππ₯+π +πΆ Integration π¦=(2π₯+3 ) 5 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: Consider a function that would leave you with a power 4 and the same bracket ππ¦ ππ₯ =5(2π₯+3 ) 4 (2) Simplify after using the Chain rule ππ¦ ππ₯ =10(2π₯+3 ) 4 As this is 10 times what we want, we need to divide our βguessβ by 10 (2π₯+3 ) 4 ππ₯ (2π₯+3 ) 4 ππ₯ = 1 10 (2π₯+3 ) 5 +πΆ 6B
20
Ex 6B C4
21
π π π 2 π₯+ππ π 2 π₯β‘1 Integration Divide by cos π‘π π 2 π₯+1β‘π π π 2 π₯ Subtract 1 π‘π π 2 π₯β‘π π π 2 π₯β1 π‘π π 2 π₯ ππ₯ 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 integrateβ¦ Trigonometric Identities are invaluable in this! Find: Using the identity above, replace tan2x = (π π π 2 π₯β1) ππ₯ Integrate each part separately π π π 2 π₯ 1 =π‘πππ₯ =π₯ =π‘πππ₯βπ₯+πΆ π‘π π 2 π₯ ππ₯ 6C
22
πππ 2π₯=1β2π π π 2 π₯ Integration Rearrange 2π π π 2 π₯=1βπππ 2π₯ Divide by 2 π π π 2 π₯β‘ 1 2 β 1 2 πππ 2π₯ π π π 2 π₯ ππ₯ 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 integrateβ¦ Trigonometric Identities are invaluable in this! Find: Using the identity above, replace sin2x = β 1 2 πππ 2π₯ ππ₯ Integrate each part separately 1 2 1 2 πππ 2π₯ = 1 2 π₯ = 1 4 π ππ2π₯ π¦=π ππ2π₯ Use the βguessingβ method ππ¦ ππ₯ =2πππ 2π₯ = 1 2 π₯β 1 4 π ππ2π₯+πΆ π π π 2 π₯ ππ₯ This is 4 times what we want so divide the βguessβ by 4 6C
23
Integration 6C π ππ3π₯πππ 3π₯ ππ₯
sππ2π₯=2π πππ₯πππ π₯ Integration Double angle formula sππ4π₯=2π ππ2π₯πππ 2π₯ Follow the patternβ¦ sππ6π₯=2π ππ3π₯πππ 3π₯ Divide by 2 1 2 π ππ6π₯=π ππ3π₯πππ 3π₯ π ππ3π₯πππ 3π₯ ππ₯ 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 integrateβ¦ Trigonometric Identities are invaluable in this! Find: Replace with the aboveβ¦ 1 2 π ππ6π₯ ππ₯ This will give us the sin 6x when differentiating, but is negative and 12 times too big! ο Make the guess negative and divide by 12! π¦=πππ 6π₯ ππ¦ ππ₯ =β6π ππ6π₯ 1 2 π ππ6π₯ ππ₯ π ππ3π₯πππ 3π₯ ππ₯ =β 1 12 πππ 6π₯+πΆ 6C
24
π π π 2 π₯+ππ π 2 π₯β‘1 Integration Divide by cos π‘π π 2 π₯+1β‘π π π 2 π₯ Subtract 1 π‘π π 2 π₯β‘π π π 2 π₯β1 (π πππ₯+π‘πππ₯ ) 2 ππ₯ 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 integrateβ¦ Trigonometric Identities are invaluable in this! Find: Expand the bracket π π π 2 π₯+π‘π π 2 π₯+2π πππ₯π‘πππ₯ ππ₯ Replace tan2x π π π 2 π₯+(π π π 2 π₯β1)+2π πππ₯π‘πππ₯ ππ₯ Simplify 2π π π 2 π₯β1+2π πππ₯π‘πππ₯ ππ₯ Integrate separately 2π π π 2 π₯ 1 2π πππ₯π‘πππ₯ =2π‘πππ₯ =π₯ =2π πππ₯ (π πππ₯+π‘πππ₯ ) 2 ππ₯ =2π‘πππ₯βπ₯+2π πππ₯+πΆ 6C
25
What about β¦. π ππ3π₯πππ 2π₯ ππ₯
26
Ex 6C C4
27
Method ? ? Split 6π₯β2 π₯β3 π₯+1 into partial fractions. Q
If each factor of the denominator is linear, we can split like such (for constants π΄ and π΅): 6π₯β2 π₯β3 π₯+1 β‘ π΄ π₯β3 + π΅ π₯+1 We donβt like fractions in equations, so we could simplify this to: ππβπβ‘π¨ π+π +π© πβπ ? METHOD 1: Substitution ? We can easily eliminate either π΄ or π΅ by an appropriate choice of π₯: If π₯=β1: β8=β4π΅ β π΅=2 If π₯=3: 16=4π΄ β π΄=4 Therefore: 6π₯β2 π₯β3 π₯+1 β‘ 4 π₯β3 + 2 π₯+1
28
Test Your Understanding
C4 Jan 2011 Q3 ? 5 π₯β1 3π₯+2 β‘ π΄ π₯β1 + π΅ 3π₯+2 5=π΄ 3π₯+2 +π΅ π₯β1 Let π₯=1: 5=5π΄ β π΄=1 Let π₯=β 2 3 : 5=β 5 3 π΅ β π΅=β3 Therefore 5 π₯β1 3π₯+2 β‘ 1 π₯β1 β 3 3π₯+2 Notice we can move the βββ to the front of the fraction. Note that we donβt technically need this last line from the perspective of the mark scheme, but itβs good to just to be on the safe side
29
More than two fractions
The principle is exactly the same if we have more than two linear factors in the denominator. Q Split 6 π₯ 2 +5π₯β2 π₯ π₯β1 2π₯+1 into partial fractions. ? 6 π₯ 2 +5π₯β2 π₯ π₯β1 2π₯+1 β‘ π΄ π₯ + π΅ π₯β1 + πΆ 2π₯+1 6 π₯ 2 +5π₯β2β‘π΄ π₯β1 2π₯+1 +π΅π₯ 2π₯+1 +πΆπ₯ π₯β1 When π₯=0: β2=βπ΄ β π΄=2 When π₯=1: 9=3π΅ β π΅=3 When π₯=β 1 2 : β3= 3 4 πΆ β πΆ=β4 So 6 π₯ 2 +5π₯β2 π₯ π₯β1 2π₯+1 β‘ 2 π₯ + 3 π₯β1 β 4 2π₯+1 Bro Tip: While substitution is generally the easier method, I sometimes compare coefficients of just the π₯ 2 term to avoid having to deal with fractions. No need to expand; we can see by observation that: 6=2π΄+2π΅+2πΆ Then πΆ is easy to determine given we know π΄ and π΅.
30
Repeated linear factors
Q Split 11 π₯ 2 +14π₯+5 π₯ π₯+1 into partial fractions. ? 11 π₯ 2 +14π₯+5β‘π΄ π₯+1 2π₯+1 +π΅ 2π₯+1 +πΆ π₯+1 2 When π₯=β1: 2=βπ΅ β π΅=β2 When π₯=β 1 2 : = 1 4 πΆ β πΆ=3 At this point we could substitute something else (e.g. π₯=1) but itβs easier to equate π₯ 2 terms. 11=2π΄+πΆ π΄=4
31
Test Your Understanding
C4 June 2011 Q1 ?
32
Dealing with Improper Fractions
The βdegreeβ 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. π₯ 3 β π₯ π₯ 2 βπ₯ π₯ 2 β3 π₯+2 π₯+1 π₯β1 ! To split an improper fraction into partial fractions, either: Divide algebraically first. Or introduce a whole term π΄+β¦ and deal with identity immediately.
33
Dealing with Improper Fractions
Q Split 3 π₯ 2 β3π₯β2 π₯β1 π₯β2 into partial fractions. Method 1: Algebraic Division Method 2: Using One Identity (method not in your textbooks but in mark schemes) ? 3 π₯ 2 β3π₯β2 π₯β1 π₯β2 β‘ 3 π₯ 2 β3π₯β2 π₯ 2 β3π₯+2 Dividing algebraically gives: 3+ 6π₯β8 π₯ 2 β3π₯+2 Turn numerator back: =3+ 6π₯β8 π₯β1 π₯β2 Let 6π₯β8 π₯β1 π₯β2 β‘ π΄ π₯β1 + π΅ π₯β2 π΄=2 π΅=4 So π π π βππβπ πβπ πβπ β‘π+ π πβπ + π πβπ ? Let: 3 π₯ 2 β3π₯β2 π₯β1 π₯β2 β‘π΄+ π΅ π₯β1 + πΆ π₯β2 3 π₯ 2 β3π₯β2β‘π΄ π₯β1 π₯β2 +π΅ π₯β2 +πΆ π₯β1 If π₯=2: 4=πΆ If π₯=1: β2=βπ΅ β π΅=2 Comparing coefficients of π₯ 2 : 3=π΄ opinion: I personally think the second method is easier. And mark schemes present it as βMethod 1β β implying more standard!
34
Test Your Understanding
C4 Jan 2013 Q3 ?
35
Integration π₯β5 (π₯+1)(π₯β2) = π΄ (π₯+1) + π΅ (π₯β2) You can use partial fractions to integrate expressions This allows you to split a fraction up β it can sometimes be recombined after integrationβ¦ Find: Write as two fractions and make the denominators equal π₯β5 (π₯+1)(π₯β2) = π΄(π₯β2) (π₯+1)(π₯β2) + π΅(π₯+1) (π₯+1)(π₯β2) Combine π₯β5 (π₯+1)(π₯β2) = π΄ π₯β2 +π΅(π₯+1) (π₯+1)(π₯β2) The numerators must be equal π₯β5 =π΄ π₯β2 +π΅(π₯+1) Let x = 2 β3 =3π΅ Calculate A and B by choosing appropriate x values β1 =π΅ π₯β5 (π₯+1)(π₯β2) Let x = -1 β6 =β3π΄ 2 =π΄ 2 (π₯+1) β 1 (π₯β2) π₯β5 (π₯+1)(π₯β2) = π΄ (π₯+1) + π΅ (π₯β2) Replace A and B from the start π₯β5 (π₯+1)(π₯β2) = 2 (π₯+1) β 1 (π₯β2) 6D
36
You can combine the natural logarithms as a division
Integration 2 (π₯+1) β 1 (π₯β2) You can use partial fractions to integrate expressions This allows you to split a fraction up β it can sometimes be recombined after integrationβ¦ Find: Integrate separately 2 (π₯+1) 1 (π₯β2) =2lnβ‘|π₯+1| =lnβ‘|π₯β2| =lnβ‘|(π₯+1 ) 2 | =lnβ‘|(π₯+1 ) 2 |β ln π₯β2 +πΆ π₯β5 (π₯+1)(π₯β2) You can combine the natural logarithms as a division =lnβ‘ π₯ π₯β2 +πΆ 2 (π₯+1) β 1 (π₯β2) 6D
37
Integration 1 1) Divide the first term by the highest power You can use partial fractions to integrate expressions This allows you to split a fraction up β it can sometimes be recombined after integrationβ¦ Find: 9 π₯ 2 β4 9 π₯ 2 β3π₯+2 9 π₯ 2 β 4 β 2) Multiply the answer by the whole expression youβre dividing by β3π₯ + 6 3) Subtract to find the remainder 4) Remember to write the remainder as a fraction of the original expression 9 π₯ 2 β3π₯+2 9 π₯ 2 β4 9 π₯ 2 β3π₯+2 9 π₯ 2 β4 + β3π₯+6 9 π₯ 2 β4 = 1 + 6β3π₯ 9 π₯ 2 β4 Looks tidier! = 1 6D
38
Integration 6D 9 π₯ 2 β3π₯+2 9 π₯ 2 β4 + 6β3π₯ 9 π₯ 2 β4 = 1
+ 6β3π₯ 9 π₯ 2 β4 We now need to write the remainder as partial fractions = 1 You can use partial fractions to integrate expressions This allows you to split a fraction up β it can sometimes be recombined after integrationβ¦ Find: 6β3π₯ 9 π₯ 2 β4 = 6β3π₯ (3π₯+2)(3π₯β2) = π΄ (3π₯+2) + π΅ (3π₯β2) = π΄ 3π₯β2 +π΅(3π₯+2) (3π₯β2)(3π₯+2) 9 π₯ 2 β3π₯+2 9 π₯ 2 β4 Set the numerators equal and solve for A and B π΄ 3π₯β2 +π΅ 3π₯+2 =6β3π₯ Let x = 2/3 4π΅=4 π΅=1 Let x = -2/3 β4π΄=8 π΄=β2 Write the final answer with the remainder broken apart! 1 + 6β3π₯ 9 π₯ 2 β4 = 1β 2 3π₯ π₯β2 6D
39
Integration 6D 9 π₯ 2 β3π₯+2 9 π₯ 2 β4 = 1β 2 3π₯+2 + 1 3π₯β2
= 1β 2 3π₯ π₯β2 You can use partial fractions to integrate expressions This allows you to split a fraction up β it can sometimes be recombined after integrationβ¦ Find: 1β 2 3π₯ π₯β2 Integrate separately 1 2 3π₯+2 1 3π₯β2 = lnβ‘|3π₯+2| = lnβ‘|3π₯β2| =π₯ 9 π₯ 2 β3π₯+2 9 π₯ 2 β4 = lnβ‘| 3π₯+2 2 | =π₯β 1 3 ln 3π₯ ln 3π₯β2 +πΆ You can combine the natural logarithms (be careful, the negative goes on the bottomβ¦) =π₯ (3π₯β2) (3π₯+2) 2 +πΆ 6D
40
Ex 6D C4
41
Integration 6E You can Integrate by using standard patterns 1 2π₯+3
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 βstandard patternsβ technique 1 2π₯+3 1 π₯ 2 +1 6E
42
Integration Notice that the denominator would differentiate to become the numerator ο This is a pattern we can use to figure out what the integral isβ¦ 2π₯ π₯ ππ₯ 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 βstandard patternsβ technique Find: Rememberβ¦ ππ π¦=lnβ‘|π π₯ | ππ¦ ππ₯ = πβ²(π₯) π(π₯) So imagine starting with ln|denominator| π¦=lnβ‘| π₯ 2 +1| 2π₯ π₯ ππ₯ In this case, we get straight to the answer! ππ¦ ππ₯ = 2π₯ π₯ 2 +1 2π₯ π₯ ππ₯ = ln π₯ πΆ 6E
43
Integration 6E ππ π¦=lnβ‘|π π₯ | ππ¦ ππ₯ = πβ²(π₯) π(π₯) πππ π₯ 3+2π πππ₯ ππ₯
πππ π₯ 3+2π πππ₯ ππ₯ 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 βstandard patternsβ technique Find: Start by trying y = ln|denominator| π¦=lnβ‘|3+2π πππ₯| Differentiate ππ¦ ππ₯ = 2πππ π₯ 3+2π πππ₯ This is double what we want so multiply the βguessβ by 1/2 πππ π₯ 3+2π πππ₯ ππ₯ πππ π₯ 3+2π πππ₯ ππ₯ = 1 2 ln 3+2π πππ₯ +πΆ 6E
44
Integration 6E 3πππ π₯π π π 2 π₯ ππ₯
In this case consider the power of sine. ο If it has been differentiated, it must have been sin3x originallyβ¦ 3πππ π₯π π π 2 π₯ ππ₯ 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 βstandard patternsβ technique Find: π¦=π π π 3 π₯ Write as a cubed bracket π¦=(π πππ₯ ) 3 Differentiate using the chain rule ππ¦ ππ₯ =3(π πππ₯ ) 2 (πππ π₯) Rewrite β this has given us exactly what we wanted! 3πππ π₯π π π 2 π₯ ππ₯ ππ¦ ππ₯ =3πππ π₯π π π 2 π₯ 3πππ π₯π π π 2 π₯ ππ₯ Donβt forget the + C! =π π π 3 π₯+πΆ 6E
45
Integration Consider the power on the bracket ο As it is a power 3, it must have been a power 4 before differentiation π₯( π₯ 2 +5 ) 3 ππ₯ 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 βstandard patternsβ technique Find: π¦=( π₯ 2 +5 ) 4 Differentiate the bracket to the power 4 using the chain rule ππ¦ ππ₯ =4( π₯ 2 +5 ) 3 (2π₯) Simplify ππ¦ ππ₯ =8π₯( π₯ 2 +5 ) 3 π₯( π₯ 2 +5 ) 3 ππ₯ This is 8 times too big so multiply the βguessβ by 1/8 π₯( π₯ 2 +5 ) 3 ππ₯ Donβt forget to add C! = 1 8 ( π₯ 2 +5 ) 4 +πΆ 6E
46
Integration 6E πππ π π 2 π₯ (2+πππ‘π₯ ) 3 ππ₯
πππ π π 2 π₯ (2+πππ‘π₯ ) ππ₯ 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 βstandard patternsβ technique Find: Write using powers (πππ π π 2 π₯)(2+πππ‘π₯ ) β3 ππ₯ Imagine how we could end up with a -3 as a power... π¦=(2+πππ‘π₯ ) β2 Use the chain rule ππ¦ ππ₯ =β2(2+πππ‘π₯ ) β3 (βπππ π π 2 π₯) Rewrite ππ¦ ππ₯ =2πππ π π 2 π₯(2+πππ‘π₯ ) β3 πππ π π 2 π₯ (2+πππ‘π₯ ) ππ₯ This is double what we want so multiply the βguessβ by 1/2 ππ¦ ππ₯ = 2πππ π π 2 π₯ (2+πππ‘π₯ ) 3 πππ π π 2 π₯ (2+πππ‘π₯ ) ππ₯ = 1 2 (2+πππ‘π₯ ) β2 +πΆ 6E
47
Integration 6E 5π‘πππ₯π π π 4 π₯ ππ₯
5π‘πππ₯π π π 4 π₯ ππ₯ 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 βstandard patternsβ technique Find: Consider using a power 5 π¦=π π π 5 π₯ Write as a bracket to the power 5 π¦=(π πππ₯ ) 5 Differentiate using the chain rule ππ¦ ππ₯ =5(π πππ₯ ) 4 (π πππ₯π‘πππ₯) We have an extra secx ππ¦ ππ₯ =5π‘πππ₯π π π 5 π₯ 5π‘πππ₯π π π 4 π₯ ππ₯ HOWEVER: We cannot just add this to our βguessβ 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
48
Integration 6E 5π‘πππ₯π π π 4 π₯ ππ₯
5π‘πππ₯π π π 4 π₯ ππ₯ 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 βstandard patternsβ technique Find: Consider using a power 4 π¦=π π π 4 π₯ Write as a bracket to the power 4 π¦=(π πππ₯ ) 4 Differentiate using the chain rule ππ¦ ππ₯ =4(π πππ₯ ) 3 (π πππ₯π‘πππ₯) ππ¦ ππ₯ =4π‘πππ₯π π π 4 π₯ 5π‘πππ₯π π π 4 π₯ ππ₯ This is what we want, but 4/5 of the amount ο Multiply by the guess by 5/4 5π‘πππ₯π π π 4 π₯ ππ₯ π¦= 5 4 π π π 4 π₯+πΆ 6E
49
Ex 6E C4
50
Integration To integrate this, you need to replace the x terms with equivalent u terms, and replace the dx with an equivalent du π₯ 2π₯+5 ππ₯ It is sometimes possible to simplify an integral by changing the variable. This is known as integration by substitution. Use the substitution: To find: π’=2π₯+5 Differentiate Rearrange to find x ππ’ ππ₯ =2 π’β5=2π₯ π’=2π₯+5 Rearrange to get dx 1 2 (π’β5)=π₯ ππ’ 2 =ππ₯ π₯ 2π₯+5 ππ₯ π₯ 2π₯+5 ππ₯ Replace each βxβ term with an equivalent βuβ term 1 2 (π’β5) ππ’ 2 π’ Rearrange β you should leave βduβ at the end 1 4 π’β5 π’ ππ’ Combine terms including the square root, changed to a power β1/2β 1 4 π’ β 5 4 π’ ππ’ 6F
51
Integration 1 4 π’ β 5 4 π’ ππ’ It is sometimes possible to simplify an integral by changing the variable. This is known as integration by substitution. Use the substitution: To find: Differentiate terms separately = π’ β π’ + πΆ Flip the dividing fractions = π’ 5 2 β π’ 3 2 π’=2π₯+5 + πΆ Calculate the fraction parts π₯ 2π₯+5 ππ₯ = 1 5 π’ 5 2 β π’ 3 2 + πΆ Finally, replace with u with its equivalent from the start! = 1 5 (2π₯+5) 5 2 β (2π₯+5) 3 2 + πΆ 6F
52
Integration πππ π₯π πππ₯(1+π πππ₯ ) 3 ππ₯ It is sometimes possible to simplify an integral by changing the variable. This is known as integration by substitution. Use the substitution: To find: π’=π πππ₯+1 Differentiate Rearrange to find Sinx ππ’ ππ₯ =πππ π₯ π’β1=π πππ₯ Rearrange to get dx ππ’ πππ π₯ =ππ₯ π’=π πππ₯+1 πππ π₯π πππ₯(1+π πππ₯ ) 3 ππ₯ Replace each βxβ term with an equivalent βuβ term πππ π₯π πππ₯(1+π πππ₯ ) 3 ππ₯ πππ π₯ ππ’ πππ π₯ (π’β1) π’ 3 Cancel the Cosx terms π’β1 π’ 3 ππ’ Multiply out π’ 4 β π’ 3 ππ’ Integrate = 1 5 π’ 5 β 1 4 π’ 4 +πΆ Replace u with x terms again! = 1 5 (π πππ₯+1) 5 β 1 4 (π πππ₯+1) 4 +πΆ 6F
53
Integration 0 2 π₯(π₯+1 ) 3 ππ₯ It is sometimes possible to 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: π’=π₯+1 Rearrange to find x Differentiate ππ’ ππ₯ =1 π’β1=π₯ Rearrange to get dx ππ’=ππ₯ π’=π₯+1 You also need to recalculate limits in terms of u 0 2 π₯(π₯+1 ) 3 ππ₯ π₯=2, π’=3 π₯=0, π’=1 0 2 π₯(π₯+1 ) 3 ππ₯ Replace x limits with u limits and the x terms with u terms 1 3 (π’β1) π’ 3 ππ’ Multiply out bracket 1 3 π’ 4 β π’ 3 ππ’ Integrate = π’ 5 5 β π’ π’=π₯+1 An alternative method is to replace the βuβ terms with x terms at the end and then just use the original βxβ limits β either way is fine! Sub in limits and calculate = β β β 6F =28.4
54
Ex 6F C4
55
Integration π‘πππ₯ ππ₯= ln π πππ₯ +πΆ πππ‘π₯ ππ₯= ln π πππ₯ +πΆ
π’ ππ£ ππ₯ =π’π£ β π£ ππ’ ππ₯ Integration You can use integration by parts to integrate some expressions You may need the following Integrals, which you are given in the formula bookletβ¦ π‘πππ₯ ππ₯= ln π πππ₯ +πΆ πππ‘π₯ ππ₯= ln π πππ₯ +πΆ πππ πππ₯ ππ₯= βln πππ πππ₯+πππ‘π₯ +πΆ π πππ₯ ππ₯= ln π πππ₯+π‘πππ₯ +πΆ 6G
56
You can use integration by parts to integrate some expressions
In C3 you met the following: (the product rule) You can use integration by parts to integrate some expressions π ππ₯ π’π£ = π’ ππ£ ππ₯ +π£ ππ’ ππ₯ This is the differential of two functions multiplied together ο You could think of it as: Rearrange by subtracting vdu/dx π’ ππ£ ππ₯ = π ππ₯ π’π£ β π£ ππ’ ππ₯ π(π’π£) ππ₯ Integrate each term with respect to x π’ ππ£ ππ₯ = π ππ₯ π’π£ β π£ ππ’ ππ₯ π’ ππ£ ππ₯ =π’π£ β π£ ππ’ ππ₯ The middle term is just a differential ο Integrating a differential cancels them both out! π’ ππ£ ππ₯ =π’π£ β π£ ππ’ ππ₯ This is the formula used for Integration by parts! ο You get given this in the booklet The other terms do not cancel as only part of them are differentiatedβ¦ 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 6G
57
π’ ππ£ ππ₯ =π’π£ β π£ ππ’ ππ₯ Integration Unlike when using the product rule, we now have one function to differentiate, and one to integrateβ¦ π’ ππ£ ππ₯ =π’π£ β π£ ππ’ ππ₯ 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 togetherβ¦ π’=π₯ π£=π πππ₯ ππ’ ππ₯ =1 ππ£ ππ₯ =πππ π₯ π₯πππ π₯ ππ₯ Differentiate Integrate Now replace the relevant parts to find the integralβ¦ β (π πππ₯)(1) =(π₯)(π πππ₯) The integral here is simpler! =π₯π πππ₯ β (βπππ π₯) Be careful with negatives here! =π₯π πππ₯ + πππ π₯ + πΆ 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 6G
58
Integration 6G π’ ππ£ ππ₯ =π’π£ β π£ ππ’ ππ₯ π₯ 2 πππ₯ ππ₯ π’ ππ£ ππ₯ =π’π£ β π£ ππ’ ππ₯
π’ ππ£ ππ₯ =π’π£ β π£ ππ’ ππ₯ Integration π’ ππ£ ππ₯ =π’π£ β π£ ππ’ ππ₯ π£= π₯ 3 3 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 togetherβ¦ π’=πππ₯ ππ’ ππ₯ = 1 π₯ ππ£ ππ₯ = π₯ 2 π₯ 2 πππ₯ ππ₯ Differentiate Integrate Now replace the relevant parts to find the integralβ¦ =(πππ₯) π₯ 3 3 β π₯ π₯ Simplify terms = π₯ 3 3 πππ₯ β π₯ 2 Let u be lnx! Integrate the second part = π₯ 3 3 πππ₯ β π₯ 3 9 + πΆ 6G
59
Integration 6G π’ ππ£ ππ₯ =π’π£ β π£ ππ’ ππ₯ π₯ 2 π π₯ ππ₯ π’ ππ£ ππ₯ =π’π£ β π£ ππ’ ππ₯
π’ ππ£ ππ₯ =π’π£ β π£ ππ’ ππ₯ Integration π’ ππ£ ππ₯ =π’π£ β π£ ππ’ ππ₯ π’= π₯ 2 π£= π π₯ 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 togetherβ¦ Sometimes you will have to use the process twice! This happens if the new integral still has two functions multiplied togetherβ¦ ππ’ ππ₯ =2π₯ ππ£ ππ₯ = π π₯ Differentiate Integrate Now replace the relevant parts to find the integralβ¦ π₯ 2 π π₯ ππ₯ β ( π π₯ )(2π₯) =( π₯ 2 )( π π₯ ) β 2π₯ π π₯ = π₯ 2 π π₯ π’=2π₯ π£= π π₯ ππ’ ππ₯ =2 ππ£ ππ₯ = π π₯ Differentiate Integrate β 2π₯ π π₯ β 2 π π₯ = π₯ 2 π π₯ Work out the square bracket which is the second integration by parts = π₯ 2 π π₯ β 2π₯ π π₯ β2 π π₯ Careful with negatives!! = π₯ 2 π π₯ β 2π₯ π π₯ + 2 π π₯ + πΆ 6G
60
You can use integration by parts to integrate some expressions
π’ ππ£ ππ₯ =π’π£ β π£ ππ’ ππ₯ Integration When integrating lnx, you MUST think of it as βlnx times 1, and use lnx as βuβ and 1 as βdv/dxβ π’ ππ£ ππ₯ =π’π£ β π£ ππ’ ππ₯ You can use integration by parts to integrate some expressions Evaluate: Leave your answer in terms of natural logarithmsβ¦ You will be asked to leave exact answers a lot so make sure you know your log laws!! π’=πππ₯ π£=π₯ ππ’ ππ₯ = 1 π₯ ππ£ ππ₯ =1 1 2 πππ₯ ππ₯ Differentiate Integrate Now replace the relevant parts to find the integralβ¦ β (π₯) 1 π₯ =(πππ₯)(π₯) Simplify terms β 1 =π₯πππ₯ Integrate and use a square bracket with limits = π₯πππ₯βπ₯ 1 2 Sub in the limits = 2ππ2β2 β (1ππ1β1) Calculate and leave in terms of ln2 =2ππ2β1 6G
61
Integrating ln π₯ and definite integration
Q Find ln π₯ ππ₯ , leaving your answer in terms of natural logarithms. π’= ln π₯ ππ£ ππ₯ =1 ππ’ ππ₯ = 1 π₯ π£=π₯ ln π₯ ππ₯ =π₯ ln π₯ β 1 ππ₯ =π₯ ln π₯ βπ₯+πΆ ? Q Find ln π₯ ππ₯ , leaving your answer in terms of natural logarithms. 1 2 ln π₯ ππ₯ = π₯ ln π₯ βπ₯ 1 2 =π π₯π§ π βπ If we were doing it from scratch: π’= ln π₯ ππ£ ππ₯ =1 ππ’ ππ₯ = 1 π₯ π£=π₯ ln π₯ ππ₯ = π ππ π π π β ππ₯ =2 ln 2 β1 ln 1 β π₯ =2 ln 2 β(2β1) =2 ln 2 β1 ? In general: π π π’ ππ£ ππ₯ ππ₯ = π’π£ π π β π π π£ ππ’ ππ₯ ππ₯
62
Ex 6G C4
63
Overview Integration by βreverse chainβ rule
(We imagine what would have differentiated to get the expression.) Integration by standard result (Thereβs certain expressions youβre expected to know straight off.) sec 2 π₯ ππ₯= tan π₯ +πΆ sin 3 π₯ cos π₯ ππ₯= sin 4 π₯ +πΆ Integration by parts (Allows us to integrate a product, just as the product rule allowed us to differentiate one) Integration by substitution (We make a substitution to hopefully make the expression easier to integrate) π₯ cos π₯ ππ₯ π’=π₯ ππ£ ππ₯ = cos π₯ ππ’ ππ₯ = π£= sin π₯ π₯ cos π₯ ππ₯=π₯ sin π₯ β sin π₯ ππ₯ =π₯ sin π₯ + cos π₯ π₯ 2π₯+5 ππ₯ Let π’=2π₯ β π₯= π’β5 2 ππ’ ππ₯ =2 β ππ₯= 1 2 π’ π₯ 2π₯+5 ππ₯= π’β ππ’=β¦
64
Test Your Understanding
Q Find 0 π 2 π₯ sin π₯ ππ₯ ? π’=π₯ ππ£ ππ₯ = sin π₯ ππ’ ππ₯ = π£=β cos π₯ π₯ sin π₯ ππ₯ =βπ₯ cos π₯ β β cos π₯ ππ₯ =βπ₯ cos π₯ + cos π₯ ππ₯ =βπ₯ cos π₯ + sin π₯ ππ₯ β΄ 0 π 2 π₯ sin π₯ ππ₯ = βπ₯ cos π₯ + sin π₯ 0 π 2 = β π 2 cos π 2 + sin π 2 β 0+ sin 0 =1
65
Summary of Functions ? ? ? ? ? ? ? ? ? π(π) How to deal with it
π π π
π (+constant) Formula booklet? πππ π Standard result β cos π₯ No πππ π sin π₯ πππ π In formula booklet, but use sin π₯ cos π₯ ππ₯ which is of the form π π β² π₯ π π₯ ππ₯ ln sec π₯ Yes ππ π π π For both sin 2 π₯ and cos 2 π₯ use identities for cos 2π₯ cos 2π₯ =1β2 sin 2 π₯ sin 2 π₯ = 1 2 β 1 2 cos 2π₯ 1 2 π₯β 1 4 sin 2π₯ ππ π π π cos 2π₯ =2 cos 2 π₯ β1 cos 2 π₯ = cos 2π₯ 1 2 π₯ sin 2π₯ ππ π π π 1+ tan 2 π₯ β‘ sec 2 π₯ tan 2 π₯ β‘ sec 2 π₯ β1 tan π₯ βπ₯ πππππ π Would use substitution π’=πππ ππ π₯+ cot π₯ , but too hard for exam. βln πππ ππ π₯+ cot π₯ πππ π Would use substitution π’= sec π₯ + tan π₯ , but too hard for exam. ln sec π₯ + tan π₯ πππ π cos π₯ sin π₯ ππ₯ which is of the form π β² π₯ π π₯ ππ₯ ln π ππ π₯ ? ? ? ? ? ? ? ? ?
66
Summary of Functions ? ? ? ? ? ? ? ? π(π) How to deal with it
π π π
π (+constant) Formula booklet? ππππ π π π By observation. β ππ¨π π No! ππ π π π tan π₯ Yes (but memorise) ππ π π π 1+ cot 2 π₯ β‘πππ π π 2 π₯ β cot π₯ βπ₯ No πππ ππ πππ ππ For any product of sin and cos with same coefficient of π₯, use double angle. sin 2π₯ cos 2π₯ β‘ 1 2 sin 4π₯ β 1 8 cos 4π₯ π π ln π₯ π₯π§ π Use IBP, where π’= ln π₯ , ππ£ ππ₯ = ln π₯ π₯ ln π₯ βπ₯ π π+π Use algebraic division. π₯ π₯+1 β‘1β 1 π₯+1 π₯β ln π₯+1 π π π+π Use partial fractions. ln π₯ β ln π₯+1 ? ? ? ? ? ? ? ?
67
Summary of Functions ? ? ? ? ? ππ π π +π
π(π) How to deal with it π π π
π (+constant) ππ π π +π Reverse chain rule. Of form π π β² π₯ π π₯ 2 ln π₯ 2 +1 π π π +π π Power around denominator so NOT of form π π β² π₯ π π₯ . Rewrite as product. π₯ π₯ β2 Reverse chain rule (i.e. βConsider π¦= π₯ β1 " and differentiate) β π₯ β1 π ππ+π π πβππ For any function where βinner functionβ is linear expression, divide by coefficient of π₯ 1 2 π 2π₯+1 β 1 3 ln 1β3π₯ π ππ+π Use sensible substitution. π’=2π₯+1 or even better, π’ 2 =2π₯+1. π₯ π₯β1 π¬π’π§ π π πππ π Reverse chain rule. 1 6 sin 6 π₯ ? ? ? ? ?
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.