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l– Maths Leaving Cert Higher Level – Integration Question 8 Paper 1 By Cillian Fahy and Darron Higgins Mocks.ie Maths LC HL Integration © mocks.ie Page 1 Integration Paper 1, Q8 Table of Contents: 1. Basics 2. Definite and Indefinite Integration a. Constant of Integration b. Definite Integrations 3. Integration by Substitution a. Powers b. Exponential c. Rewriting 4. Integration and Trigonometry a. Normal Trig. Functions b. Using Rules for Trig. Products c. Substitution for Trig. Functions d. Inverse Trig. Functions e. 5. Area a. Curve and the x-axis b. Two curves or a curve and a line Mocks.ie Maths LC HL Integration © mocks.ie Page 2 Integration Basics: We will begin our review of Integration with a quick overview of the basics. We will focus on clarifying the simple points which you may have overlooked. It‟s important that you are certain on the simple steps as that can make the more difficult questions easier. Basics: The basic rule of integration is: (Page 26, Table Book) N x dx x N 1 N 1 Meaning you take the power of X add 1 to it and then divide by that new power. Question 1: How is Integration the reverse of differentiation? The basic idea behind integration is the following: When you differentiate y you get . When you integrate you get y. You will find that all the rules below are very similar to Differentiation. Question 2: What if I am asked to integrate with a number in front of it? You integrate as you usually would without the number and then multiply by that number. In fact, we can do the following: ax dx a x dx where a is a constant (normal number). n n Question 3: What if I am asked to integrate a number without an x? Just like in Differentiation, you think of a number without an x as a number multiplied by because Then just follow the formula above. Question 4: What if I am asked to integrate two numbers added to each other? Again, just as in Differentiation, you integrate the terms separately and then add them together again. Question 5: How do I deal with a Square root? You just look at the square root as and integrate normally then. Question 6: How to I deal with an X below the line? Think of what you would do for Differentiation. Just use indices to bring the X above the line and integrate normally. Mocks.ie Maths LC HL Integration © mocks.ie Page 3 Definite and Indefinite Integration: We will now look at the two fundamental types of integrations: Definite and Indefinite. They are not all that different but quite easy to identify. x N dx This is an Indefinite Integral a As you can see above the only difference is that the integration sign on the Definite Integral has a x N dx This is a Definite Integral number above and below it. b Indefinite Integration: Constants of Integration The Constant of Integration may sound scary but it‟s just a “+ C”. After every indefinite integration you must add C to your answer. You‟ll recall that we claimed that Integration was the reverse of Differentiation. Now is a good time to really test that. Here are two questions: y x3 3x 2 dy 3x 2 dx y x3 y x3 15 3x 2 dy 3x 2 dx y x3 You‟ll notice that although my beginning y values are different my final answer for y is the same. This is clearly wrong. As differentiation gets rid of constants (like the 15 above) Integration cannot take account of them. So instead we add a “+C” which means any number both positive and negative could have been part of the original integration. So the example from above should read: y x3 15 3x 2 dy 3x 2 dx y x3 + C Mocks.ie Maths LC HL Integration © mocks.ie WARNING! Not including a “+C” is a sure way to lose marks. So make sure you include it for every Indefinite Integration Page 4 Definite Integration: As you‟ve seen above you can tell a Definite Integration question from an Indefinite Integration Question because a Definite Integration sign will have a number above and below it. But there are other differences two: Definite Integration does not have a “+C” ever! With Definite Integration you will get an answer of numbers A Definite Integration works something like this: a b f ( x)dx f (a) f (b) In short, you integrate, sub in the top value into that integration and then take the bottom value subbed into the integration away from it. These values are called „limits‟. Example: Definite Integration 1 1 x 2 dx Simplify first 0 1 2 1 x x dx 0 1 1 1 2 x 2 x dx Now integrate 0 1 3 2 2 2x x x 3 2 0 2 1 2 3 4 x x2 x 3 2 0 Substitute x 1 minus x 0 2 2 2 3 3 4 1 1 4 0 02 1 0 3 2 3 2 4 1 1 3 2 0 1 6 Mocks.ie Maths LC HL Integration © mocks.ie Page 5 Integration by Substitution: Integration by Substitution is a method of integrating both Definite and Indefinite Integrals. The idea is quite similar to Differentiation Rules where you let U and V stand for part of the sum. The Substitution method for Integration is just slightly more advanced. Your focus is to turn complicated sums into equations which can be computed by rules from the table book. We will look at: Dealing with powers Exponential Function ( Manipulation and Substitution Example: Dealing with Powers Leaving Cert 2006 Q8 b(i) Evaluate x(1 x ) dx 2 3 U = 1+x 2 du 2x dx 3 (U ) xdx U3 U4 2 du 8 (1 x 2 )4 C 8 dx du xdx 2 With powers, select whichever part has the highest power. Then let U equal the numbers inside the bracket of highest power Sub your U value back into the final equation WARNING! When we differentiate U above our objective is to replace dx with du. We also need to remove all other traces of x from the question. We cannot integrate unless it is all in terms of one variable only! Example: Exponential Function and Substitution ( Leaving Cert 2008 Q8 b(i) Evaluate 3x e 2 x3 dx u 2 e 3 x dx Let u = x 3 du = 3x 2 ==> du=3x 2 dx dx u u e du e WARNING! Don’t forget to sub the value back in e e +C u Mocks.ie x3 Maths LC HL Integration © mocks.ie Page 6 Example: Manipulation and Substitution Integrate x x3 dx x Let u x 3 dx x3 x u 12 dx u3 u u u 3 3 2 1 1 2 x u 3 du 1 dx dx du du 2 3u u 2 1 1 1 2 du 2 2 3 2 x 3 2 6 x 3 c 3 WARNING! The only element that is different here is that you needed to manipulate your U value to get rid of the X. If you ever get to a situation where you can’t find a way to remove an X from a sum then look closely at this. Mocks.ie Maths LC HL Integration © mocks.ie Page 7 Trigonometry: Trigonometry is a big section of the Integration Paper. Take a look at the 2010 question and you might think it should have been on Paper 2! But don’t let that put you off, even if you are not a Trig. Fan you’ll find the Trigonometry in Integration quite manageable as it usually follows set rules and is mainly concerned with Sin and Cos. We will focus on: Normal Trig. Functions Using Rules for Products of Trig. Functions Substitution for Trig. Functions Inverse Trig. Functions Normal Trigonometry Function: Integrating Trig Functions can basically be put down to two rules: Sin( ax ) a Cos ( ax l ) Sin(ax l )dx a Cos(ax)dx Or this could be seen in another way as: , basically you differentiate whatever the function (Cos or Sin) is acting on and divide by the result. Leaving Cert 2008, Q8 (a) Find (2 x cos3x)dx 2 xdx cos3xdx x2 {Split and Integrate separately} sin 3x C 3 Mocks.ie Maths LC HL Integration © mocks.ie Page 8 Products of Trigonometry Functions: To integrate products of Sin and Cos, then you must follow certain formulae: 2 cos A cos B cos( A B) cos( A B) 2sin A cos B sin( A B) sin( A B) 2sin A sin B cos( A B) cos( A B) 2 cos A sin B sin( A B) sin( A B) Table Book, page 15 WARNING! Notice in particular how important the order of Sin and Cos are in the formula. Leaving Cert 2006 Q8, b (ii) 2sin A cos B sin( A B) sin( A B) sin( A B) sin( A B) {Adjust the formula to suit the question} 2 sin(8 ) sin(2 ) 4 sin5 cos3 d = d 0 04 2 sinAcosB= sin(8 ) sin(2 ) -cos(8 ) cos(2 ) 4 d {Integrate as usual} 04 2 4 0 16 -cos(2 ) cos( 2 ) -cos(0) cos(0) 4 16 4 16 1 1 1 1 0 16 16 4 4 TOP TIP: It’s always a good idea to put the larger angle first when completing these sums. It saves you a little bit of extra work and ensures that you will not get something like sin( ). Mocks.ie Maths LC HL Integration © mocks.ie Page 9 Example: Substitution for Trig Functions These questions are quite straightforward once you know what to substitute for but if you don’t you will find them quite difficult to deal with. Leaving Cert 2004 Q8 b(ii) 3 0 For these questions always let U equal the function which is to the highest power sin x cos3 xdx U=cosx (U )3 sin xdx -du sin xdx 3 0 3 0 U4 (U ) du = 4 3 4 (cos ) (cos x) 4 3 (cos 0) 4 3 4 0 4 4 1 1 15 64 4 64 Inverse Trig. Functions: Table Book, page 26 1 sin 1 x a a2 x2 1 1 1 x tan x2 a2 a a Integrating Inverse Trig functions is quite similar to that of Differentiation. Again you have a formula and your focus is to manipulate your question to suit that formula. Mocks.ie Maths LC HL Integration © mocks.ie Page 10 Example: Inverse Function with Manipulation 1 3 dx 81 9 x 2 dx 9(9 x ) 2 dx (3) x 2 2 3 dx 9 x 2 {X 2 can only be multiplied by 1} 1 x sin 1 C 3 3 Completing the Square: These questions increase in difficulty if you are given something along the lines of: x 2 dx 6 x 14 This is a problem as you cannot immediately factorise this into anything workable. You then must complete the square, as in turn this into a squared number. x 2 6 x 13 ( x 3) 2 a Divide the number multiplied by x by 2 ( . This is how you know what to square. ( x 3) 2 a x 2 6 x 9 a x 2 6 x 13 a 4 (2) 2 x 2 6 x 13 ( x 3) 2 (2) 2 WARNING! It’s important that both numbers are squared as that is the only way that it will fit into the formula. If you do this method and the remainder (a) can’t be squared easily then you shouldn’t do this method. Mocks.ie Maths LC HL Integration © mocks.ie Page 11 Example: Inverse Function with Completing the dx x 2 4 x 20 x 2 4 x 20 ( x 2) 2 a Again, just divide 4 by 2 and you’ll get 2. Therefore it must be ( x 2) 2 a x 2 4 x 4 a x 2 4 x 16 Square a 16 (4) 2 x 2 4 x 20 ( x 2) 2 (4) 2 dx 1 1 x 2 tan C (4)2 ( x 2)2 4 4 : This is a nasty equation which can be quite difficult to tackle unless you know precisely what to do. You need to always use the following substitution: Let x a sin x2 a 2 sin 2 Example: Integrate 3 0 9 x 2 dx STEP ONE: Simplify 9 x 2 dx = (3) 2 x 2 dx Let x =3sin (3) 2 9sin 2 dx x a sin dx=-3cos d 9 9sin 2 (-3cos d ) 9(1 sin 2 ) (-3cos d ) (1 sin 2 ) cos 2 Table book, p.13 9 cos 2 (-3cos d ) 3cos (-3cos d ) = -9cos 2 d Changing Limits: Mocks.ie Maths LC HL Integration © mocks.ie Page 12 In this sum you are required to change your limits. This is simply because when you get to the final part your answer is too complicated to sub x back into the equation. You original limits are in terms of x and you want to get them in terms of . Use your original link: to find this. STEP TWO: Change Limits x =3sin 3 {Use the first limit to gain a value for } 3sin 3 sin 1 = 90= x=3sin 0 sin 0 2 =0 STEP THREE: Finding the value 2 0 -9cos d 9 2 cos 2 d {Take the -9 out} 2 0 1 9 2 cos 2 d 9 2 (1+cos2 ) d 0 0 2 Table Book, p. 14 9 2 9 sin 2 2 (1+cos2 ) d 0 2 2 2 0 9 sin 2 2 9 sin 9 sin 0 0 2 2 0 2 2 2 2 2 9 9 9 ( 0) (0 0) 2 2 2 4 Mocks.ie Maths LC HL Integration © mocks.ie Page 13 Area: You can be asked to find the area using integration methods. This can usually feature as a c part but not always. There are some clear rules which if you follow should guide you through the question. Area between the x-axis and a curve Area between curves/curve and a line Area, the curve and the x-axis: b Area f ( x)dx a Where f(x) is the curve. b, a are the points you want the area between. Important Formulae: Not in the Table Book! Points to watch out for: If you get a negative value then you just take the absolute value (remove the minus). Why? Well area cannot be negative. If the graph drops under the other side of the x-axis, then you must calculate this area separately. (As below) Leaving Cert 2010 Q8 b The curve y = crosses the x -axis at x=0, x=1 and x=3, as shown. Calculate the total area of the shaded regions enclosed by the curve and the x-axis. Rough Graph (Given) Mocks.ie Maths LC HL Integration © mocks.ie Page 14 y = 12x 3 -48x 2 +36x x=0, x=1,x=3 1 3 2 4 3 2 12x -48x +36x 3 x 16 x 18 x 1 0 0 | 3(1)4 16(1)3 18(1)2 3(0)4 16(0)3 18(0) 2 | 5 3 1 12x 3 -48x 2 +36x 3 x 4 16 x3 18 x 2 3 1 3(3)4 16(3)3 18(3) 2 3(1) 4 16(1)3 18(1)2 | (243 432 162) (5) || 32 | 32m2 Area below the graph gives negative value. Note the absolute value signs. Area = 32+5 =37 WARNING! You can also be asked about the curve and the y-axis. It is the exact same ideas as the question above with some slight changes (explained below). It does not feature regularly and the xaxis is a much more common question but you should still now how to approach it in case they surprise you in the exam. b Area f ( y )dx a Where f(y) is the curve. b, a are the points you want the area between. As it is on the y-axis, y=b, y=a Important Formula: Not in the Table Book! The important point to note about the formula above is that it is , rather than and as such all your calculations will be in terms of y. When dealing with the y-axis the rules are very similar to that of the x-axis but converted to the yaxis. Again; If you get a negative value then you just take the absolute value (remove the minus). Why? Well area cannot be negative. If the graph drops under the other side of the y-axis, then you must calculate this area separately. (As below) Otherwise you just approach the question as you did above. Mocks.ie Maths LC HL Integration © mocks.ie Page 15 Area between Curves/Curve and a Line: b b a a Area= f ( x)dx g ( x)dx Where f(x), g(x) are the two curves/curve and a line b, a are the points where they intersect This is the more difficult type of question as you need to find a lot of the points yourself. Again the same rules apply in that area must be positive, therefore negative values become positives ones and that area below the axis should be calculated separately from area above it. Leaving Cert 2008, Q8 c The diagram shows the curve y = 4 and the line 2x = 0. Calculate the area of the shaded region enclosed by the curve and the line Mocks.ie Maths LC HL Integration © mocks.ie Page 16 STEP ONE: Cuts x-axis 2x+y-1=0 y=0 {Cuts the x-axis} 2x-1=0 x= 1 2 y 4 x2 0 4 x2 y=0 {Again, cuts x-axis} x2 4 x 2 x 2, {Judge this by the graph} STEP TWO: Find intersection points y=x 2 4 2x y 1 0 y -2 x +1 2 x 1 4 x x2 2x 3 0 (x-3)(x+1)=0 Mocks.ie 2 x=3, x=-1 Maths LC HL Integration © mocks.ie Page 17 STEP THREE: Area above x-axis 2 1 1 2 1 4 x dx 1 2 x 2 x3 4 x 3 2 x3 4 x 3 2 x x 2 1 1 2 1 8 1 16 11 (8 ) (4 ) 9 3 3 3 3 1 1 2 1 1 1 ( ) (1 1) 2 1 2 4 4 1 3 92 6 4 4 x x 2 STEP FOUR: Area below x-axis 3 3 2 1 2 xdx 4 x dx 1 2 2 3 2 3 x x x x 2 1 2 3 1 2 x3 4 x 32 1 1 1 (3 9) ( ) 6 2 4 4 WARNING! Absolute value signs should not go in here. They are only put in when you are adding these numbers together in the last line! 3 x3 8 9 16 7 4 x (12 9) (8 ) 3 2 3 3 3 3 1 7 75 28 47 47 | (6 ) ( ) | || | 4 3 12 12 12 12 STEP FIVE: Add Area together 3 47 81 47 128 6 4 12 12 12 12 Mocks.ie {WARNING! This is the step that is easy forget. Don't!} Maths LC HL Integration © mocks.ie Page 18 Mocks.ie Maths LC HL Integration © mocks.ie Page 19