Ch 4 Differentiation
|
|
- Eugene Flynn
- 5 years ago
- Views:
Transcription
1 Ch 1 Partial fractions Ch 6 Integration Ch 2 Coordinate geometry C4 Ch 5 Vectors Ch 3 The binomial expansion Ch 4 Differentiation
2 Chapter 1 Partial fractions We can add (or take away) two fractions only if they have the same number on the bottom (denominator). We can add (or take away) two fractions only of they have the same number on the bottom (denominator) X 3 X 2 X 4 X 3 = = = 5 6 =
3 Similarly with algebraic fractions, we can only add or take them away if they have the same thing on the bottom. We are going to make the bottoms the same on both fractions, multiply out the brackets, bring the fractions together and simplify the tops. 2 (x + 3) + 1 (x + 1) X (x + 1) X (x + 3) = 2 X (x + 1) (x + 1)(x + 3) + 1 X (x + 3) (x + 1)(x + 3) = 2x + 2 (x + 1)(x + 3) + 1x + 3 (x + 1)(x + 3) = 2x x + 3 (x + 1)(x + 3) = 3x + 5 (x + 1)(x + 3) 3
4 We can also go the other way and split a fraction up. This can be useful when we are integrating. There are two ways to solve these - Substitution. We put in a value of x that will make some of the brackets disappear. - Compare the coefficients. Both sides must have exactly the same number of everything, so look at how many x 2 s or x s or constants are on one side, the other side must have exactly the same. If we have two linear (not x 2 terms) on the bottom then we can always split it up and write it like 6x - 2 (x 3)(x + 1) A (x 3) + B (x + 1) Notice that there is a in the middle, this means that both sides are always exactly the same no matter what value of x you put in, instead of sometimes the same when we have a normal = sign. Later we will see that both sides always have exactly the same number of x s, x 2 s and constants as well. 4
5 We are going to multiply everything on both sides by the brackets on the bottom and then substitute x s that make one of the brackets disappear. 6x - 2 (x 3)(x + 1) A (x 3) + B (x + 1) X (x-3)(x+1) X (x-3)(x+1) X (x-3)(x+1) 6x 2 A(x + 1) + B(x 3) When x = -1 6x 2 A(x + 1) + B(x 3) 6(-1) 2 A(0) + B(-1 3) -8-4B B When x = +3 6x 2 A(x + 1) + B(x 3) 6(3) 2 A(3 + 1) + B(0) 16 4A A 4 5
6 We can do exactly the same thing if there are three linear (not x 2 or squared) brackets on the bottom. 6x 2 + 5x - 2 (x)(x - 1)(2x + 1) A x + B (x 1) + C (2x + 1) X (x)(x-1)(2x+1) 6x 2 + 5x 2 A(x-1)(2x+1) + B(x)(2x+1) + C(x)(x-1) Notice that each letter now multiplies all the brackets it didn t have on the bottom When x = 0 6x 2 + 5x 2 A(x-1)(2x+1) + B(x)(2x+1) + C(x)(x-1) 6(0)2 5(0) - 2 A(0-1) + B(0)( ) + C(0)( ) 2-1A A When x = +1 6x 2 + 5x 2 A(x-1)(2x+1) + B(x)(2x+1) + C(x)(x-1) 6(1) 2 + 5(1) - 2 A(0)(-1) + B(1)(3) + C(-1)(0) 9 3B B When x = 1 2 6x 2 + 5x 2 A(x-1)(2x+1) + B(x)(2x+1) + C(x)(x-1) 6( 1 2 )2 + 5( 1 2 ) 2 A( 3 2 )(0) + B( 1 2 )(0) + C( 1 2 )( ( 3 2 ) C C -4 6
7 Chapter 2 Coordinate geometry in the xy plane We can define a curve so that y and x are not defined in terms of each other anymore but a third parameter which we often call t. 7
8 Draw the curve for the parametric equation x = 2t and y = t 2 for -3 t 3 t x = 2t y = t
9 A Cartesian equation is the name for a curve in the form y = something to do with x. We can change a parametric equation back to the normal Cartesian equation. 1. Rearrange both equations until they both have the same t thing 2. Set them equal to each other so that the t s disappear 3. Rearrange so that y = something 9
10 A curve has parametric equations, x = 2t, y = t 2 Find the Cartesian equation of the curve x = 2t y = t 2 t = x 2 y = ( x 2 ) The Cartesian equation is y = x2 4 10
11 Example The diagram shows a sketch of the curve with parametric equations x = t 1 and y = 4 t 2. The curve meets the x axis at the points A and B. Find the coordinates of A and B. When it meets the x axis y = 0 A B y = 4 t 2 0 = 4 t 2 t 2 = 4 t = ± When t = +2 When t = -2 x = t 1 x = t - 1 x = 2-1 x = x = 1 x = - 3 The coordinates are (1, 0) and (-3, 0) 11
12 A curve has parametric equations x = at and y = a(t 3 + 8) where a is a constant. The curve passes through the point (2, 0). Find the value of a. y = a(t 3 + 8) both sides by a 0 = a(t 3 + 8) 0 = t t 3 = -8 We know that there is a value of t for which x = 2 when y = 0 so we can find this value of t. t = Let y = 0 and then solve for t. x = at 2 = a X (-2) a = Substitute this value of t into the x equation to find a 12 The parametric equations are x = -1t and y = -1(t 3 + 8) Substitute a into the original equations
13 A curve is given parametrically by the equations x = t 2 and y = 4t. The line x + y = - 4 meets the curve at A. Find the coordinates of A. 13
14 To find the coordinates where they meet we need to find t when they meet. Rearrange the equation of the line x + y + 4 = 0 Substitute x = t 2 and y = 4t into this equation t 2 + 4t + 4 = 0 (t + 2) 2 = 0 t = Substitute back in to find the coordinates x = t 2 = (-2) 2 = 4 y = 4t = 4 X -2 = -8 The coordinate of A is (4, -8) 14
15 Converting trigonometric parametric equations into Cartesian equations. Example - A curve has the parametric equations x = sin t + 2 y = cos t 3 Find the Cartesian equation of the curve. We need to get rid of any t s so rearrange the formulas. x = sin t + 2 y = cos t 3 sin t = x 2 cos t = y Now sin 2 t + cos 2 t = 1 so if we substitute in (x 2) 2 + (y + 3) 2 = 1 This is the Cartesian equation of the curve. It is a circle with radius 1 and centre (+2, -3). 15
16 A circle with the equation (x a) 2 + (y b) 2 = r 2 has centre (a, b) and radius r. (a, b) r 16
17 Example - A curve has the parametric equations x = sin t y = sin 2t Find the Cartesian equation of the curve. We need to get rid of any t s so open up sin 2t and substitute in x = sin t y = sin 2t y = 2 sin t cos t y = 2x cos t Now sin 2 t + cos 2 t = 1 so rearrange and substitute in x = sin t so we get t in terms of x cos t 2 t = 1 sin 2 t cos t = (1 sin 2 t) cos t = (1 x 2 ) Now substitute cos t = (1 x 2 ) into the original equation and we have y in terms of x which is what we wanted. y = 2x cos t y = 2x (1 x 2 ) 17
18 Finding the area under a parametric curve The area under a curve is given by yy dddd by using the chain rule we can rewrite this as yy dx dt dt (notice this looks a bit like we could cancel the fractions and turn it back into the original). It is very important to realise that because we are integrating with respect to t the limits must be in terms of t as well. 18
19 If x = 2t + 4 and y = 3t 1 find y dx dt. x = 2t + 4 y = 3t 1 dx dt = 2 y dx dt = 2 (3t -1) It is very important to realise that because we are integrating with respect to t the limits must be in terms of t as well. 19
20 The diagram shows a sketch of the curve with parametric equations x = t 2, y = 2t(3 t), t 0. The curve meets the x axis at x = 0 and x = 9. The region R is bounded by the curve and the x axis. Find the area of R First find out what t is when x = 0 and x = 9 x = t 2 x = t 2 0 = t 2 9 = t 2 t = 0 t = 3 we ignore the negative square roots because t yy dx dt = t=3 2t(3 t)2t dt dt t=0 = t=3 t=0 12t 2 4t 3 dt = [4t 3 t 4 ] = (108 81) (0 0) = 27 20
21 Chapter 3 The Binomial Expansion We can expand brackets a lot quicker using the binomial expansion. (1 + x) n = 1 + nx + n(n 1) 2! x 2 + n(n 1)(n 2) 3! x n C r x r Remember that n C r is the number of of ways of picking a team of r from a total of n people nc r = n! (n r)! r! 5C 2 = 5! 3! 2! = 5 X 4 X 3 X 2 X 1 (3 x 2 X 1) (2 X 1) = 5 X 4 2 X 1 21
22 Use the binominal expansion to find the first four terms of (1 + 2x) -1 (1 + x) n = 1 + nx + n(n 1) 2! x 2 + n(n 1)(n 2) 3! x (1 + 2x) -1 = 1 + (-1)(2x) X X 1 (2x) (2x) X - 2 X X 2 X 1 (1 + 2x) -1 = 1-2x + 4x 2 8x Notice that if the power is negative then the expansion will go on forever so we normally stop after the first few terms. This is because it converges very quickly as if x is a small number then x 4 is a very, very small number and can be ignored. The infinite expansion is only valid if l2xl < 1 i.e. if ½<x<½. 22
23 Use the binominal expansion to find the first four terms of (1-3x) ½. (1 + x) n = 1 + nx + n(n 1) 2! x 2 + n(n 1)(n 2) 3! x (1-3x) ½ = 1 + (½)(-3x) X X 1 (-3x) X -1 2 X X 2 X 1 (-3x)3 (1-3x) ½ = x x x Again this infinite expansion would go on forever and is only valid if l-3xl < 1 i.e. if 1 3 < x <
24 If the number at the front is bigger than a one then take it out down the front. Use the binominal expansion to find the first four terms of (4 + x) ½. (4 + x) ½ = [4(1 + x)] ½ = 4 ½ (1 + x 4 )½ = 2(1 + x 4 )½ expand (1 + x 4 )½ as normal 2(1 + x 4 )½ = 2 [ x x x3 +...] = x x x valid for l x 4 l < 1 i.e l x l < 4 24
25 Use the binominal expansion to find 1 + x 1 + 3x the term in x 3. as far as 1 + x 1 + 3x = (1 + x) X (1 + 3x) Expand (1 + 3x) -1 as normal (1 + 3x) -1 = 1 3x + 9x 2 27x Multiply out the brackets (1 + x) X ( 1 3x + 9x 2 27x 3 ) Notice that you only actually need to multiply the terms that are going to result in terms less than x 4. 25
26 Use partial fractions to find the binominal expansion of as far as the term in x 3. (4 5x) (1 + x)(2 x) (4 5x) = by partial fractions (1 + x)(2 x) (1 + x) (2 x) (1 + x) = 3 x (1 + x)-1 expand (1 + x) -1 as normal = 1 x + x 2 x (1 + x) = 3 x (1 + x)-1 = 3[1 x + x 2 x 3 ] = 3 3x + 3x 2 3x (2 - x) = -2 x (2 - x)-1 expand (2 - x) -1 as normal = x x x (2 - x) = -2 x (2 - x)-1 = -2[ x x x3 ] = x x2-1 8 x (1 + x) + -2 (2 x) = [3 3x + 3x2 3x 3 ] + [ x x2-1 8 x3 ] 26
27 Chapter 4 Differentiation 27
28 We can differentiate parametric curves (ones given in terms of t) by using the fact that dy dt dx dt = dy dt dx dt =dy dt X dt dx = dy dx so to find dy we need to differentiate y with respect to t, dx differentiate x with respect to t and divide one by the other. 28
29 Find the gradient at the point P where t = 2, on the curve given parametrically by x = t 3 + t, y = t Differentiate x and y with respect to t x = t y = t dx dt = 3t2 dy dt = 2t dy dy = dt dx dx dt = 2t 3t 2 = 2 3t when t = 2 dy = 2 = 1 dx 3 X
30 Sometimes we need to find dy dx for curves that can t be rearranged into y = something like x 2 + y 2 = 8x. These are called implicit relationships. The differential with respect to x of anything to do with y is what it would be if you differentiated with respect to y times by dy. dx The differential of y 2 = 2y dy dx d(y 2 ) dx = 2y dy dx Sometimes we may need to use the product rule. The differential of x 2 y 3 = x 2 3y 2 dy dx + 2xy3 d(x 2 y 3 ) dx = x 2 3y 2 dy dx + 2xy3 30
31 To find a gradient of a curve, differentiate everything with respect to x and then rearrange for dy. dx Find the gradient of the curve x 2 + y 3 = 4x at the point (1, 2). Differentiate everything with respect to x and rearrange to get dy = something dx 2x + 3y 2 dy dx = 4 3y 2 dy dx = 4 2x dy dx = 4 2x 3y Substitute x = 1, y = 2 into dy substitute both x and y now. dx. Notice that we need to dy dx = 4 2x 3y 2 = 4 2(1) 3(2) 2 =
32 We can differentiate something like y = 2 x where x is a power rather than a multiple. In general if y = a x then dy dx = ax Ln a. If y = 2 x find dy dx. y = 2 x Ln both sides Ln y = Ln 2 x Bring x down the front by log laws Ln y = x Ln 2 Differentiate both sides 1 dy = Ln 2 y dx Multiply y up dy = y Ln 2 dx Substitute y = 2 x in from the beginning dy = dx 2x Ln 2 32
33 The radius of a circle is expanding at the rate of 5 cm per sec. Find the rate at which its Area is expanding when the radius is 3. Rate means with respect to time, we want the rate of change of its Area (with respect to time) which is da. dt We know the rate of change of the radius with respect to time which is dr dt = 5. and by the chain rule da = da X dr dt dr dt so we need to work out da dr For a circle A = πr 2 so da dr = 2πr da = da X dr dt dr dt da dt = 2πr X 5 when r = 3 da dt = 30π cm2 /sec When the radius is 3cm the area of the circle is increasing by 30π cm2 per second. 33
34 We can set up Differential equations for many real life situations but they especially occur in the cases of radioactive decay, population growth and cooling objects. In the decay of radioactive particles the rate at which particles decay is proportional to how many particles are left. dn dt α N which can be rewritten as dn dt = -kn the minus sign arises because the number of particles is going down. 34
35 The population of a town is growing at a rate proportional to the size of the population. dp dt α P which can be rewritten as dp dt = kp Newton s Law of Cooling says that an object loses temperature at a rate that is proportional to the difference between the object and the surrounding air temperature. dθ dt α (Θ Θ 0) where Θ 0 is the surrounding air temperature dθ dt = (Θ Θ 0) 35
36 36
37 Chapter 5 Vectors 37
38 If we want to describe where an ant walking on a classroom floor is we need to use 2 coordinates, (x, y), that tell us that how far along and how far across the ant is y x This ant is at the point (4, 3). 38
39 If we want to describe where a fly is in a classroom we need to use three numbers, (x, y, z) to tell us how far along, how far across and how far up it is. z y x This fly is at the point (2, 3, 4). 39
40 To find out how far a point (x, y) is from the origin in two dimensions we use Pythagoras in two dimensions. Distance = (x 2 + y 2 ) Find the distance of the point (4, 3) from the origin. Distance = (x 2 + y 2 ) Distance = ( Distance = ( Distance = (25 Distance = 5 40
41 To find out how far a point (x, y, z) is from the origin in three dimensions we use Pythagoras in three dimensions. Distance = (x 2 + y 2 + z 2 ) Find the distance of the point (6, 7, 8) from the origin. Distance = (x 2 + y 2 + z 2 ) = ( ) = ( ) = (149) = 12.2 (3sf) 41
42 To find out the distance between two points (x 1, y 1 ) (x 2, y 2 ) we use Pythagoras. Distance = { (x 2 x 1 ) 2 + (y 2 y 1 ) 2 } Find the distance between the points (9, 3) and (6, -1). Distance = { (x 2 x 1 ) 2 + (y 2 y 1 ) 2 } = { (9 6) 2 + (3-1) 2 } = { (3) 2 + (4) 2 } = { } = { 25 } Distance = 5 42
43 To find out the distance between two points (x 1, y 1, z 1 ) (x 2, y 2, z 2 ) we use Pythagoras in three dimensions. Distance = { (x 2 x 1 ) 2 + (y 2 y 1 ) 2 + (z 2 z 1 ) 2 Find the distance between the points (2, 7, 4) and (9, - 3, 8). Distance = { (x 2 x 1 ) 2 + (y 2 y 1 ) 2 + (z 2 z 1 ) 2 } = { (9-2) 2 + (-3-7) 2 + (8-4) 2 } = { (7) 2 + (-10) 2 + (4) 2 } = { } = { 165 } = 12.8 (3sf) 43
44 We can also use a position vector to say where a fly is. We use i, j and k where i, j and k are unit vectors (i.e. of length one) pointing in the x, y and z directions. Vectors are normally written in bold. This is nearly the same idea as coordinates but a position vector has a direction as well as a length now. We can write a position vector in terms of i, j and k or as a column vector. Instead of saying the fly is at coordinates (7, 3, 8) we 7 would say it has position vector 7i + 3j + 8k or
45 We can find the modulus (or magnitude) of a vector v = xi + yj + zk. lvl = (x 2 + y 2 + z 2 ) Find the modulus of the vector v = 3i + 8j - 5k. lvl = (x 2 + y 2 + z 2 ) lvl = ( (-5) 2 ) lvl = ( ) lvl = 98 lvl =
46 Exactly the same as in two dimensions if we know the position vectors a and b for points A and B we can find the vector AB by calculating b a. A OA AB a b - a O OB B b The points A and B have position vectors 4i + 2j + 7k and 3i + 4j 1k. Find the vector AB. AB = b a = (4i - 2j + 7k) (3i + 4j -1k) = 1i 6j + 8k 46
47 The points A and B have position vectors a = t 5 2t and b = t. t 1 3 a. Find AB. b. By differentiating labl 2, find the value of t for which labl is at a minimum. c. Find the minimum value of labl. AB = b a 2t = t t 5. 3 t 1 t = t 5 4 t If labl = (x 2 + y 2 + z 2 ) then labl 2 = (x 2 + y 2 + z 2 ) so labl 2 = t 2 + (t 5) 2 + (4 t) 2 multiplying out brackets and simplifying labl 2 = 3t 2 18t + 41 If we call labl 2 y then y = 3t 2 18t + 41 dddd dddd = 6t 18 The minimum of labl 2 occurs when dy = 0 and this is exactly the same as the minimum dt of labl so labl is at a minimum at the point t = Putting t = 3 back in to find out the actual magnitude of labl labl = (3t 2 18t + 41) = (3X X ) = 41 47
48 If we have two vectors a = ui + vj + wk b = xi + yj + zk then the scalar product = a.b = ux + vy + wz Find the scalar product of the vectors a = 8i - 5j - 4k b = 5i + 4j - k a.b = (8 X 5) + (-5 X 4) + (-4 Xa -1) = = 24 48
49 49
50 We can use the scalar product to find the angle between two vectors. cos (angle) = a.b lal lbl where a.b is the scalar product and lal is the magnitude of the vector a. a O b 50
51 Find the angle between the vectors a = - i + j + 3k b = 7i - 2j + 2k a.b = (-1 X 7) + (1 X -2) + (3 X 2) = = lal = [(-1) 2 + (1) 2 + (3) 2 ] = lbl = cos (angle) = = X 57 angle = 96.9 a.b lal lbl 51
52 We can use just the scalar product to prove that two vectors are perpendicular. Because cos (angle) = a.b lal lbl and cos 90 = 0 We only need to prove that the top is = 0 because 0 divided by anything is sill zero. So if the scalar product, a.b = 0 then the two vectors cross at right angles. Given that the vectors a = 2i - 6j + 1k b = 5i + 2j + λk are perpendicular find the value of λ a.b = (2 X 5) + (-6 X 2) + (1 X λ) = -2 + λ to be perpendicular a.b = 0 so λ = 2. 52
53 Given the vectors a = - 2i + 5j - 4k b = 4i - 8j + 5k Find a vector that is perpendicular to both a and b. Call the vector perpendicular to a and b v = xi + yj + zk then a.v = 0 and b.v = 0 because they are perpendicular If a.v = 0 then -2x + 5y 4z = If b.v = 0 then 4x 8y + 5z = Now we can pick any one dimension and choose it to be a constant. So if we pick the z dimension and let z = 1 so if if we substitute z = 1 we get the simultaneous equation -2x + 5y = 4 4x - 8y = -5 solving simultaneous equations x = 7 4 and y = 3 2 and we have already chosen z to be = One possible vector that is perpendicular to a and b is v = 7 4 i j + 1k. we can find another one by just multiplying by 4 to get v = 7i + 7j + 4k 53
54 We can write the equation of a straight line in which passes through a point a and is parallel to a vector b. r = a + tb t is the parameter. You can think about this as saying go to point a and continue in the direction b. This is similar to the equation of a line y = mx + c. b r = a + tb a O 54
55 Find a vector equation of the straight line which passes through the point A, with position vector 3i 5j + 4k, and is parallel to the vector 7i 3k. r = a + tb or or r = 3i 5j + 4k +t(7i 3j + 0k) r = (3 +7t)i + (-5-3t)j + (4 +0t)k 3 + 7t r = 5 3t 4 55
56 We can also write the equation of a straight line which passes through two points c and d. r = c + t(d c) Where t is the parameter. You can think about this as saying start at point c and continue in the direction (d c) which is the same as vector CD. Again the form of this is similar to the equation of a line y = m(x 2 x 1 )+ y 1 in two dimensions. (d c) = CD r = c + t(d c) c d O We do not have to use c as our start point we could use d or indeed any point on the line so the line r = d + t(d c) 56
57 Find a vector equation of the straight line which passes through the points c = 4i + 5j - 1k and d = 6i + 3j + 2k r = c + t(d c) (d c) = 2i - 2j + 3k r = 4i + 5j -1k + t(2i - 2j +3k) or r = (4-2t)i + (5-2t)j + (-1 +3t)k or 4 2t r = 5 2t 1 + 3t or 4 2 r = 5 + t
58 In two dimensions two lines will definitely cross over one another unless they are parallel. In three dimensions two lines will generally not intersect because there are so many more ways that they can miss each other rather than intersect. If they do intersect then each of the three coordinates must be equal in both lines at exactly the same time. 58
59 Show that the lines with vector equations r = (3i + 8j 2k) + μ(2i - 1j + 3k) r = (7i + 4j +3k) + λ(2i +1j + 4k) intersect and find the position vector of the point of intersection. If the lines intersect then each of the three coordinates must be the same at exactly the same time so taking simultaneous equations we get 3 + 2μ = 7 + 2λ 8 - μ = 4 + λ μ = 3 + 4λ We can choose any two of these but choosing the top two and solving simultaneously we get μ = 3 λ = Check that this also works for the z coordinate X3 = 3 + 4X1 It does work! So subbing back into all three coordinates we get the point of intersection p = 9i + 5j +7k or 9 p =
60 We can find the angle between two straight lines by using cos (angle) = a.b lal lbl where a and b are only the direction bit of the lines. 60
61 Find the acute angle between the two straight lines with vector equations a = (2i + 1j + 1k) + μ(3i - 8j - 1k) b = (7i + 4j + k) + λ(2i + 2j + 3k) We only need the direction bits of the lines! lal = [(3) 2 + (-8) 2 + (-1) 2 ] = 74 lbl = [(2)2 + (2)2 + (3)2 ] = a.b = 3X2 + -8X2 + -1X3 = cos (angle) = a.b lal lbl cos (angle) = angle = X 17 61
62 62
63 Chapter 6 Integration x n = xn+1 n C ex = ex + C 1 x = Ln lxl + C cos x = sin x + C sin x = - cos x + C sec 2 x = tan x + C cosec x cot x = - cosec x + C cosec 2 x = - cot x + C sec x tan x = sec x + C 63
64 Integrate the following with respect to x 2 cos x + 3 x - x Separately 2 cos x dx= 2 sin x 3 x dx = 3 Ln lxl x dx = x ½ = 2 3 x3/ so (2 cos x + 3 x - x )dx = 2 sin x + 3 Ln lxl x3/2 + C 64
65 Find ( cos x sin 2 x - 2ex ) dx cos x = cos x X 1 sin 2 x sin x sin x = cot x cosec x cot x cosec x dx = - cosec x 2 e x dx = 2e x so ( cos x sin 2 x - 2ex ) dx = - cosec x 2e x + C 65
66 f (ax +b) dx = 1 a f(ax + b) + C (ax +b) n dx = 1 a (ax +b) n+1 n C e ax +b dx = 1 a eax +b + C 1 ax +b dx = 1 a Ln lax +bl + C cos (ax +b) dx = 1 a sin (ax + b) + C sin (ax +b) dx = 1 a cos (ax + b) + C sec 2 (ax +b) dx = 1 a tan (ax + b) + C cosec (ax +b) cot (ax + b) dx = 1 a cosec (ax + b) + C cosec 2 (ax +b) dx = 1 a cot (ax + b) + C sec (ax +b) tan (ax +b) dx= 1 sec (ax + b) + C a 66
67 (3x +4) 5 dx = (3x + 4)6 + C = 1 18 (3x + 4)6 + C sin (2x +1) dx = 1 2 cos (2x +1) + C 1 dx = 1. Ln l 5x +4 l + C 5x cosec (3x +2) cot (3x +2) dx = 1 3 cosec (3x +2) + C 67
68 We may not be able to integrate it in the form that we are given. We may need to change it using trigonometric identities before it can be integrated. Find (tan 2 x) dx (tan 2 x) dx tan 2 x + 1 = sec 2 x = (sec 2 x 1) dx = tan x x + C 68
69 Find sin 2 x dx (sin 2 x) dx cos 2x = 1 2sin 2 x sin 2 x = 1-1 cos 2x 2 2 = ( 1 1 cos 2x) dx 2 2 = 1 x - 1 sin 2x + C
70 Sometimes it can be useful to use partial fractions. Find x - 5 (x +1)(x 2) dx x - 5 (x +1)(x 2) = A + B (x +1) (x 2) solving partial fractions as normal we find that A = 2 and B = - 1 x - 5 (x +1)(x 2) = 2-1 (x +1) (x 2) 2-1 (x +1) (x 2) = 2 Ln lx + 1l Ln lx 2l 70
71 If the power of the top is bigger than the bottom then we may need to do long division on the improper fraction. Find 9x2 3x + 2 9x 2-4 dx (9x 2 4) (9x 2 3x + 4) = 1 rem ( 3x + 6) this means that 9x 2 3x + 2 9x 2-4 = x (9x 2 4) next we split 6 3x (9x 2 4) using partial fractions 6 3x = 1-2 so (9x 2 4) (3x + 2) (3x 2) 9x 2 3x + 2 9x 2-4 = (3x + 2) (3x 2) 9x2 3x + 2 9x 2-4 = (3x + 2) (3x 2) = x + 1 Ln l3x 2l - 2 Ln l3x + 2l + C
72 Sometimes we can simplify an integral by changing the variable by choosing a good substitution. Use the substitution u = 2x + 5 to find x (2x +5) dx If u = 2x + 5 then x = u If u = 2x +5 then (2x +5) = u = u ½ If u = 2x + 5 then du dx = 2 so rearranging dx = 1 2 du x (2x +5) dx = (u 5) 2 X u ½ X 1 2 du this multiplies out and simplifies to 1 2 u3/2 5 u 1/2 du = 1 10 u5/2-5 6 u3/2 + C Don t forget to substitute back in to give the answer in x s. = 1 10 (2x + 5)5/2-5 6 (2x + 5)3/2 + C 72
73 Use the substitution u = 1 + sin x to find cos x sin x (1 + sin x) 3 dx If u = 1 + sin x then du dx = cos x i.e. cos x dx = du If u = 1 + sin x then sin x = u cos x sin x (1 + sin x) 3 dx = (u 1)u 3 du this multiplies out and simplifies to u 4 u 3 du = 1 5 u5-1 4 u4 + C Substitute back in to give the answer in terms of the original x s. = 1 5 (1 + sin x)5-1 4 (1 + sin x)4 + C 73
74 Use the substitution u 2 = 2x + 5 to find x (2x + 5) dx If u 2 = 2x + 5 then x = u If u 2 = 2x + 5 then (2x + 5) = u If u 2 = 2x + 5 then 2u du i.e. dx = u du dx = 2 by implicit differentiation x (2x + 5) dx = ( u2-5 2 ) X u X u du this multiplies out and simplifies to 1 2 u4 5 2 u2 du = 1 10 u5-5 6 u3 + C Substitute back in to give the answer in terms of the original x s. = 1 10 (2x + 5)5/2-5 6 (2x + 5)3/2 + C 74
75 If we have limits then we need to change these as well. 2 Use the substitution u = x + 1 to find x(x 0 +1)3 dx If u = x + 1 then when x = 0 u = 1 x = 2 u = If u = x + 1 then x = u If u = x + 1 then (x + 1) 3 = u If u = x + 1 then du = 1 i.e. dx = du dx xx= 2 xx = 0 x(x + 1)3 dx = uu = 3 uu =1 (u 1)u3 du this multiplies out and simplifies to uu=3 uu= 1 u4 u 3 du = [ 1 5 u5-1 4 u4 ] u = 1 = [ ] - [ ] = =
76 We can use integration by parts to integrate some expressions. u dv dx dx = uv - v du dx dx We use integration by parts to change an integral that is hard to do into one that is easier. We would look to use this for an expression that has one bit that would be easier to integrate if it was differentiated times by another bit that would not be any harder to integrate if it is integrated. These will often be of the form x cos x dx because the x differentiates to 1 which is easier and cos integrates to sin which is no harder. Another example is x e x dx x differentiates to 1 and e x integrated is just e x which is no harder. again 76
77 Use integration by parts to find x cos x dx Choose u = x dv dx = cos x then du dx = 1 v = sin x u dv dx dx = uv - v du dx dx x cos x dx = x sin x - sin x. 1 dx = x sin x - (- cos x) = x sin x + cos x + C 77
78 Usually if we have an x or a power of x at the front (x 2,x 3 etc) we will choose that to be the u as this will (possibly eventually) turn into 1 which is easier to integrate. The only time we don t choose x to be the u is when we have an Ln term in which case we let u = the Ln term. Use integration by parts to find x Ln x d x Choose u = Ln x dv dx = x then du = 1 dx x v = 1 2 x u dv du dx = uv - v dx dx dx x Ln x dx = 1 2 x2 Ln x x2. 1 x dx = 1 2 x2 Ln x x dx = 1 2 x2 Ln x x2 + C 78
79 If we have a power of x bigger than one, i.e. x 2, x 3 etc, we may need to do integration by parts twice in a row before we get rid of the x term. Use integration by parts to find x 2 e x dx Choose u = x 2 dv dx = ex then du dx = 2x v = ex u dv dx = uv - dx v du dx dx x 2 e x dx = x 2 e x - 2x e x dx Now 2x e x dx still can t be integrated so we need to do integration by parts again on just this bit Choose u = 2x dv dx = ex then du dx = 2 v = ex u dv du dx = uv - v dx dx dx = 2x e x - 2 e x dx = 2x e x - 2 e x Substituting this back into our original integral we end up with x 2 e x dx = x 2 e x - 2x e x + e x 79
80 tan x dx = Ln lsec xl +C sec x dx = Ln lsec x + tan xl +C cot x dx = Ln lsin xl +C tan x dx = - Ln lcosec x + cot xl +C 80
81 We can use standard patterns to integrate some expressions. Find 2x x dx We notice that the top is the differential of the bottom 2x x dx = Ln lx2 + 1l + C 81
82 Find cos x sin 2 x dx Try y = sin 3 x = (sin x) 3 then dy dx = 3 cos x sin2 x it is 3 times too big so if y = 1 3 sin3 x then dy dx = cos x sin2 x so 1 3 sin3 x dx = cos x sin 2 x 82
83 Find x(x 2 + 5) 3 dx Try y = (x 2 + 5) 4 then dy dx = 4 X 2x X (x2 + 5)3 = 8x(x 2 + 5) 3 which is 8 times too big so if y = 1 8 (x2 + 5) 4 then dy dx = x(x2 + 5) 3 so x(x 2 + 5) 3 x dx = 1 8 (x2 + 5) 4 83
84 Remember the Trapezium rule y dx 1 2 h X [y 0 + y n + 2(y 1 + y y n-1 ] where h = b - a n 84
85 ππ/3 Find 0 sec x dx using the chain rules with four strips. h = π/3-0 4 = π 12 X 0 π 12 π 6 π 4 π 3 Y I 1 2 h X [y 0 + y n + 2(y 1 + y y n-1 ] 1 2 X π 12 X [ ( )] 1.34 (3sf) 85
86 We can create a solid shape by revolving a curve, between the limits a and b, 360 degrees (2π radians) around the x axis. This is called a Volume of revolution. We can find the volume of this curve by integrating y squared and multiplying by pi (notice this is very similar to πr 2 the area of a circle). bb Volume = π aa y 2 dx a b 86
87 The region R is bounded by the curve with equation y = sin 2x, the x axis and the 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. b Volume of revolution = π a y2 dx Remember that sin 2 x = 1 2 (1 cos 2x) so sin 2 2x = 1 2 (1 cos 4x) b π a y2 dx π/2 = π 0 sin 2 2x dx π/2 = π (1 cos 4x) dx = π [ 1 2 x sin 4x ]π/2 = π [ ( π 0) (0-0) ] 4 87
88 The curve C has parametric equations x = t(1 + t) y = t where t is the parameter and t 0. The region R is bounded by the curve C, the x axis and the lines x = 0 and x = 2. Find the volume of the solid formed when the region R is rotated 2π radians about the x axis. b Volume of revolution = π a y2 dx If x = t + t 2 then when x = 2 t = 1 dx dt = 1 + 2t When x = 0 t = 0 dx = (1 + 2t) dt b π a y2 dx 1 = π 0 1 (1 + t) 2 (1 + 2t) dt 1 = π 0 2 (1 + t) - 1 (1 + t) 2 dt by partial fractions = π [ 2 Ln l 1 + t l + 1 (1 + t) ]1 = π [ ( 2 Ln ) (0 + 1) ] 2 = π ( 2 Ln ) 88
89 We can use integration to solve differential equations. We always get all the y things on one side, the x things on the other side and then integrate. Find a general equation to the differential equation (1 + x 2 ) dy dx = x tan y Get all the x things on one side and the y things on the other and then integrate. 1 dy = x dx tan y 1 + x 2 cot y dy = 1 2 2x 1 + x 2 dx Ln lsin yl = 1 2 Ln l1 +x2 l + C Another way we could have finished off is by choosing our constant to be = Ln k instead (it doesn t change anything it s still a constant!) Ln lsin yl = 1 2 Ln l1 +x2 l + Ln k Ln lsin yl = Ln l k (1 + x 2 ) l so sin y = k (1 + x 2 ) this is just an alternative way which makes the final answer neater. 89
90 Sometimes we are given boundary conditions which allow us to calculate exactly what the equation is without any need for C. Find a particular solution to the differential equation dy dx = -3(y 2) (2x + 1)(x + 2) if x = 1 when y = 4. Get all the x things on one side and the y things on the other and then integrate. 1 (y 2) dy = -3 (2x + 1)(x + 2) dx taking partial fractions 1 (y 2) dy = 1 (x + 2) - 2 (2x + 1) dx Ln ly 2l = Ln lx + 2l - Ln l2x + 1l + Ln k Ln ly -2l = Ln l k xx+2 2xx + 1 l remove Ln s y 2 = k xx+2 2xx + 1 Substitute x = 1 when y = = k k = 2 y = xx+2 2xx + 1 which can be simplified to 90 y = xx + 1
91 91
92 92
Week beginning Videos Page
1 M Week beginning Videos Page June/July C3 Algebraic Fractions 3 June/July C3 Algebraic Division 4 June/July C3 Reciprocal Trig Functions 5 June/July C3 Pythagorean Identities 6 June/July C3 Trig Consolidation
More informationHigher Mathematics Course Notes
Higher Mathematics Course Notes Equation of a Line (i) Collinearity: (ii) Gradient: If points are collinear then they lie on the same straight line. i.e. to show that A, B and C are collinear, show that
More informationMathsGeeks. Everything You Need to Know A Level Edexcel C4. March 2014 MathsGeeks Copyright 2014 Elite Learning Limited
Everything You Need to Know A Level Edexcel C4 March 4 Copyright 4 Elite Learning Limited Page of 4 Further Binomial Expansion: Make sure it starts with a e.g. for ( x) ( x ) then use ( + x) n + nx + n(n
More informationAlgebra and functions; coordinate geometry in the (x, y) plane; sequences and series; differentiation; integration; vectors.
Revision Checklist Unit C4: Core Mathematics 4 Unit description Assessment information Algebra and functions; coordinate geometry in the (x, y) plane; sequences and series; differentiation; integration;
More informationCore A-level mathematics reproduced from the QCA s Subject criteria for Mathematics document
Core A-level mathematics reproduced from the QCA s Subject criteria for Mathematics document Background knowledge: (a) The arithmetic of integers (including HCFs and LCMs), of fractions, and of real numbers.
More information1.4 Techniques of Integration
.4 Techniques of Integration Recall the following strategy for evaluating definite integrals, which arose from the Fundamental Theorem of Calculus (see Section.3). To calculate b a f(x) dx. Find a function
More informationYEAR 13 - Mathematics Pure (C3) Term 1 plan
Week Topic YEAR 13 - Mathematics Pure (C3) Term 1 plan 2016-2017 1-2 Algebra and functions Simplification of rational expressions including factorising and cancelling. Definition of a function. Domain
More informationPossible C4 questions from past papers P1 P3
Possible C4 questions from past papers P1 P3 Source of the original question is given in brackets, e.g. [P January 001 Question 1]; a question which has been edited is indicated with an asterisk, e.g.
More information(x 3)(x + 5) = (x 3)(x 1) = x + 5. sin 2 x e ax bx 1 = 1 2. lim
SMT Calculus Test Solutions February, x + x 5 Compute x x x + Answer: Solution: Note that x + x 5 x x + x )x + 5) = x )x ) = x + 5 x x + 5 Then x x = + 5 = Compute all real values of b such that, for fx)
More informationPractice Exam 1 Solutions
Practice Exam 1 Solutions 1a. Let S be the region bounded by y = x 3, y = 1, and x. Find the area of S. What is the volume of the solid obtained by rotating S about the line y = 1? Area A = Volume 1 1
More informationa Write down the coordinates of the point on the curve where t = 2. b Find the value of t at the point on the curve with coordinates ( 5 4, 8).
Worksheet A 1 A curve is given by the parametric equations x = t + 1, y = 4 t. a Write down the coordinates of the point on the curve where t =. b Find the value of t at the point on the curve with coordinates
More informationIntegration by parts Integration by parts is a direct reversal of the product rule. By integrating both sides, we get:
Integration by parts Integration by parts is a direct reversal of the proct rule. By integrating both sides, we get: u dv dx x n sin mx dx (make u = x n ) dx = uv v dx dx When to use integration by parts
More informationMath Final Exam Review
Math - Final Exam Review. Find dx x + 6x +. Name: Solution: We complete the square to see if this function has a nice form. Note we have: x + 6x + (x + + dx x + 6x + dx (x + + Note that this looks a lot
More informationC4 "International A-level" (150 minute) papers: June 2014 and Specimen 1. C4 INTERNATIONAL A LEVEL PAPER JUNE 2014
C4 "International A-level" (150 minute) papers: June 2014 and Specimen 1. C4 INTERNATIONAL A LEVEL PAPER JUNE 2014 1. f(x) = 2x 3 + x 10 (a) Show that the equation f(x) = 0 has a root in the interval [1.5,
More informationPartial Fractions. June 27, In this section, we will learn to integrate another class of functions: the rational functions.
Partial Fractions June 7, 04 In this section, we will learn to integrate another class of functions: the rational functions. Definition. A rational function is a fraction of two polynomials. For example,
More informationCh1 Algebra and functions. Ch 2 Sine and Cosine rule. Ch 10 Integration. Ch 9. Ch 3 Exponentials and Logarithms. Trigonometric.
Ch1 Algebra and functions Ch 10 Integration Ch 2 Sine and Cosine rule Ch 9 Trigonometric Identities Ch 3 Exponentials and Logarithms C2 Ch 8 Differentiation Ch 4 Coordinate geometry Ch 7 Trigonometric
More informationDIFFERENTIATION AND INTEGRATION PART 1. Mr C s IB Standard Notes
DIFFERENTIATION AND INTEGRATION PART 1 Mr C s IB Standard Notes In this PDF you can find the following: 1. Notation 2. Keywords Make sure you read through everything and the try examples for yourself before
More informationCore Mathematics 3 Differentiation
http://kumarmaths.weebly.com/ Core Mathematics Differentiation C differentiation Page Differentiation C Specifications. By the end of this unit you should be able to : Use chain rule to find the derivative
More informationA-Level Notes CORE 1
A-Level Notes CORE 1 Basic algebra Glossary Coefficient For example, in the expression x³ 3x² x + 4, the coefficient of x³ is, the coefficient of x² is 3, and the coefficient of x is 1. (The final 4 is
More informationChapter 3. Integration. 3.1 Indefinite Integration
Chapter 3 Integration 3. Indefinite Integration Integration is the reverse of differentiation. Consider a function f(x) and suppose that there exists another function F (x) such that df f(x). (3.) For
More informationAS PURE MATHS REVISION NOTES
AS PURE MATHS REVISION NOTES 1 SURDS A root such as 3 that cannot be written exactly as a fraction is IRRATIONAL An expression that involves irrational roots is in SURD FORM e.g. 2 3 3 + 2 and 3-2 are
More informationNewbattle Community High School Higher Mathematics. Key Facts Q&A
Key Facts Q&A Ways of using this booklet: 1) Write the questions on cards with the answers on the back and test yourself. ) Work with a friend who is also doing to take turns reading a random question
More informationReview for the First Midterm Exam
Review for the First Midterm Exam Thomas Morrell 5 pm, Sunday, 4 April 9 B9 Van Vleck Hall For the purpose of creating questions for this review session, I did not make an effort to make any of the numbers
More informationTechniques of Integration
Chapter 8 Techniques of Integration 8. Trigonometric Integrals Summary (a) Integrals of the form sin m x cos n x. () sin k+ x cos n x = ( cos x) k cos n x (sin x ), then apply the substitution u = cos
More informationMathematics 1 Lecture Notes Chapter 1 Algebra Review
Mathematics 1 Lecture Notes Chapter 1 Algebra Review c Trinity College 1 A note to the students from the lecturer: This course will be moving rather quickly, and it will be in your own best interests to
More informationFall 2013 Hour Exam 2 11/08/13 Time Limit: 50 Minutes
Math 8 Fall Hour Exam /8/ Time Limit: 5 Minutes Name (Print): This exam contains 9 pages (including this cover page) and 7 problems. Check to see if any pages are missing. Enter all requested information
More informationEdexcel past paper questions. Core Mathematics 4. Parametric Equations
Edexcel past paper questions Core Mathematics 4 Parametric Equations Edited by: K V Kumaran Email: kvkumaran@gmail.com C4 Maths Parametric equations Page 1 Co-ordinate Geometry A parametric equation of
More informationLevel 3, Calculus
Level, 4 Calculus Differentiate and use derivatives to solve problems (965) Integrate functions and solve problems by integration, differential equations or numerical methods (966) Manipulate real and
More informationMEI Core 1. Basic Algebra. Section 1: Basic algebraic manipulation and solving simple equations. Manipulating algebraic expressions
MEI Core Basic Algebra Section : Basic algebraic manipulation and solving simple equations Notes and Examples These notes contain subsections on Manipulating algebraic expressions Collecting like terms
More information3.4 Conic sections. Such type of curves are called conics, because they arise from different slices through a cone
3.4 Conic sections Next we consider the objects resulting from ax 2 + bxy + cy 2 + + ey + f = 0. Such type of curves are called conics, because they arise from different slices through a cone Circles belong
More informationCore Mathematics C4. You must have: Mathematical Formulae and Statistical Tables (Pink)
Write your name here Surname Other names Pearson Edexcel GCE Centre Number Core Mathematics C4 Advanced Candidate Number Friday 23 June 2017 Morning Time: 1 hour 30 minutes Paper Reference 6666/01 You
More informationSTEP Support Programme. Pure STEP 1 Questions
STEP Support Programme Pure STEP 1 Questions 2012 S1 Q4 1 Preparation Find the equation of the tangent to the curve y = x at the point where x = 4. Recall that x means the positive square root. Solve the
More information4.5 Integration of Rational Functions by Partial Fractions
4.5 Integration of Rational Functions by Partial Fractions From algebra, we learned how to find common denominators so we can do something like this, 2 x + 1 + 3 x 3 = 2(x 3) (x + 1)(x 3) + 3(x + 1) (x
More informationSolutions for the Practice Final - Math 23B, 2016
olutions for the Practice Final - Math B, 6 a. True. The area of a surface is given by the expression d, and since we have a parametrization φ x, y x, y, f x, y with φ, this expands as d T x T y da xy
More informationMathematics 104 Fall Term 2006 Solutions to Final Exam. sin(ln t) dt = e x sin(x) dx.
Mathematics 14 Fall Term 26 Solutions to Final Exam 1. Evaluate sin(ln t) dt. Solution. We first make the substitution t = e x, for which dt = e x. This gives sin(ln t) dt = e x sin(x). To evaluate the
More informationWeek 2 Techniques of Integration
Week Techniques of Integration Richard Earl Mathematical Institute, Oxford, OX LB, October Abstract Integration by Parts. Substitution. Rational Functions. Partial Fractions. Trigonometric Substitutions.
More informationCoordinate goemetry in the (x, y) plane
Coordinate goemetr in the (x, ) plane In this chapter ou will learn how to solve problems involving parametric equations.. You can define the coordinates of a point on a curve using parametric equations.
More information1 Exponential Functions Limit Derivative Integral... 5
Contents Eponential Functions 3. Limit................................................. 3. Derivative.............................................. 4.3 Integral................................................
More informationUpdated: January 16, 2016 Calculus II 7.4. Math 230. Calculus II. Brian Veitch Fall 2015 Northern Illinois University
Math 30 Calculus II Brian Veitch Fall 015 Northern Illinois University Integration of Rational Functions by Partial Fractions From algebra, we learned how to find common denominators so we can do something
More informationMathematics Revision Questions for the University of Bristol School of Physics
Mathematics Revision Questions for the University of Bristol School of Physics You will not be surprised to find you have to use a lot of maths in your stu of physics at university! You need to be completely
More informationSTEP Support Programme. Pure STEP 3 Solutions
STEP Support Programme Pure STEP 3 Solutions S3 Q6 Preparation Completing the square on gives + + y, so the centre is at, and the radius is. First draw a sketch of y 4 3. This has roots at and, and you
More informationAnswers for NSSH exam paper 2 type of questions, based on the syllabus part 2 (includes 16)
Answers for NSSH eam paper type of questions, based on the syllabus part (includes 6) Section Integration dy 6 6. (a) Integrate with respect to : d y c ( )d or d The curve passes through P(,) so = 6/ +
More informationIntegrals. D. DeTurck. January 1, University of Pennsylvania. D. DeTurck Math A: Integrals 1 / 61
Integrals D. DeTurck University of Pennsylvania January 1, 2018 D. DeTurck Math 104 002 2018A: Integrals 1 / 61 Integrals Start with dx this means a little bit of x or a little change in x If we add up
More informationFormulas to remember
Complex numbers Let z = x + iy be a complex number The conjugate z = x iy Formulas to remember The real part Re(z) = x = z+z The imaginary part Im(z) = y = z z i The norm z = zz = x + y The reciprocal
More informationThe definite integral gives the area under the curve. Simplest use of FTC1: derivative of integral is original function.
5.3: The Fundamental Theorem of Calculus EX. Given the graph of f, sketch the graph of x 0 f(t) dt. The definite integral gives the area under the curve. EX 2. Find the derivative of g(x) = x 0 + t 2 dt.
More informationDerivatives and Integrals
Derivatives and Integrals Definition 1: Derivative Formulas d dx (c) = 0 d dx (f ± g) = f ± g d dx (kx) = k d dx (xn ) = nx n 1 (f g) = f g + fg ( ) f = f g fg g g 2 (f(g(x))) = f (g(x)) g (x) d dx (ax
More informationabc Mathematics Pure Core General Certificate of Education SPECIMEN UNITS AND MARK SCHEMES
abc General Certificate of Education Mathematics Pure Core SPECIMEN UNITS AND MARK SCHEMES ADVANCED SUBSIDIARY MATHEMATICS (56) ADVANCED SUBSIDIARY PURE MATHEMATICS (566) ADVANCED SUBSIDIARY FURTHER MATHEMATICS
More informationChapter 1 Review of Equations and Inequalities
Chapter 1 Review of Equations and Inequalities Part I Review of Basic Equations Recall that an equation is an expression with an equal sign in the middle. Also recall that, if a question asks you to solve
More informationChapter 5: Integrals
Chapter 5: Integrals Section 5.5 The Substitution Rule (u-substitution) Sec. 5.5: The Substitution Rule We know how to find the derivative of any combination of functions Sum rule Difference rule Constant
More informationEdexcel Core Mathematics 4 Parametric equations.
Edexcel Core Mathematics 4 Parametric equations. Edited by: K V Kumaran kumarmaths.weebly.com 1 Co-ordinate Geometry A parametric equation of a curve is one which does not give the relationship between
More informationMATH 250 TOPIC 13 INTEGRATION. 13B. Constant, Sum, and Difference Rules
Math 5 Integration Topic 3 Page MATH 5 TOPIC 3 INTEGRATION 3A. Integration of Common Functions Practice Problems 3B. Constant, Sum, and Difference Rules Practice Problems 3C. Substitution Practice Problems
More informationMAS153/MAS159. MAS153/MAS159 1 Turn Over SCHOOL OF MATHEMATICS AND STATISTICS hours. Mathematics (Materials) Mathematics For Chemists
Data provided: Formula sheet MAS53/MAS59 SCHOOL OF MATHEMATICS AND STATISTICS Mathematics (Materials Mathematics For Chemists Spring Semester 203 204 3 hours All questions are compulsory. The marks awarded
More informationTime: 1 hour 30 minutes
Paper Reference (complete below) Centre No. Surname Initial(s) Candidate No. Signature Paper Reference(s) 6663 Edexcel GCE Pure Mathematics C Advanced Subsidiary Specimen Paper Time: hour 30 minutes Examiner
More informationC6-2 Differentiation 3
chain, product and quotient rules C6- Differentiation Pre-requisites: C6- Estimate Time: 8 hours Summary Learn Solve Revise Answers Summary The chain rule is used to differentiate a function of a function.
More information*P46958A0244* IAL PAPER JANUARY 2016 DO NOT WRITE IN THIS AREA DO NOT WRITE IN THIS AREA DO NOT WRITE IN THIS AREA. 1. f(x) = (3 2x) 4, x 3 2
Edexcel "International A level" "C3/4" papers from 016 and 015 IAL PAPER JANUARY 016 Please use extra loose-leaf sheets of paper where you run out of space in this booklet. 1. f(x) = (3 x) 4, x 3 Find
More information2.2 The derivative as a Function
2.2 The derivative as a Function Recall: The derivative of a function f at a fixed number a: f a f a+h f(a) = lim h 0 h Definition (Derivative of f) For any number x, the derivative of f is f x f x+h f(x)
More informationVectors Part 1: Two Dimensions
Vectors Part 1: Two Dimensions Last modified: 20/02/2018 Links Scalars Vectors Definition Notation Polar Form Compass Directions Basic Vector Maths Multiply a Vector by a Scalar Unit Vectors Example Vectors
More informationOrdinary Differential Equations (ODEs)
c01.tex 8/10/2010 22: 55 Page 1 PART A Ordinary Differential Equations (ODEs) Chap. 1 First-Order ODEs Sec. 1.1 Basic Concepts. Modeling To get a good start into this chapter and this section, quickly
More informationPure Core 2. Revision Notes
Pure Core Revision Notes June 06 Pure Core Algebra... Polynomials... Factorising... Standard results... Long division... Remainder theorem... 4 Factor theorem... 5 Choosing a suitable factor... 6 Cubic
More informationTABLE OF CONTENTS 2 CHAPTER 1
TABLE OF CONTENTS CHAPTER 1 Quadratics CHAPTER Functions 3 CHAPTER 3 Coordinate Geometry 3 CHAPTER 4 Circular Measure 4 CHAPTER 5 Trigonometry 4 CHAPTER 6 Vectors 5 CHAPTER 7 Series 6 CHAPTER 8 Differentiation
More informationCHAPTER 7: TECHNIQUES OF INTEGRATION
CHAPTER 7: TECHNIQUES OF INTEGRATION DAVID GLICKENSTEIN. Introduction This semester we will be looking deep into the recesses of calculus. Some of the main topics will be: Integration: we will learn how
More informationC-1. Snezana Lawrence
C-1 Snezana Lawrence These materials have been written by Dr. Snezana Lawrence made possible by funding from Gatsby Technical Education projects (GTEP) as part of a Gatsby Teacher Fellowship ad-hoc bursary
More informationEssex County College Division of Mathematics MTH-122 Assessments. Honor Code
Essex County College Division of Mathematics MTH-22 Assessments Last Name: First Name: Phone or email: Honor Code The Honor Code is a statement on academic integrity, it articulates reasonable expectations
More informationQuestions Q1. The function f is defined by. (a) Show that (5) The function g is defined by. (b) Differentiate g(x) to show that g '(x) = (3)
Questions Q1. The function f is defined by (a) Show that The function g is defined by (b) Differentiate g(x) to show that g '(x) = (c) Find the exact values of x for which g '(x) = 1 (Total 12 marks) Q2.
More information2t t dt.. So the distance is (t2 +6) 3/2
Math 8, Solutions to Review for the Final Exam Question : The distance is 5 t t + dt To work that out, integrate by parts with u t +, so that t dt du The integral is t t + dt u du u 3/ (t +) 3/ So the
More informationMAC 2311 Calculus I Spring 2004
MAC 2 Calculus I Spring 2004 Homework # Some Solutions.#. Since f (x) = d dx (ln x) =, the linearization at a = is x L(x) = f() + f ()(x ) = ln + (x ) = x. The answer is L(x) = x..#4. Since e 0 =, and
More informationReview Problems for the Final
Review Problems for the Final Math -3 5 7 These problems are provided to help you study. The presence of a problem on this handout does not imply that there will be a similar problem on the test. And the
More informationChapter 9 Overview: Parametric and Polar Coordinates
Chapter 9 Overview: Parametric and Polar Coordinates As we saw briefly last year, there are axis systems other than the Cartesian System for graphing (vector coordinates, polar coordinates, rectangular
More informationMark Scheme (Results) January 2007
Mark Scheme (Results) January 007 GCE GCE Mathematics Core Mathematics C (666) Edexcel Limited. Registered in England and Wales No. 96750 Registered Office: One90 High Holborn, London WCV 7BH January 007
More informationSOUTH AFRICAN TERTIARY MATHEMATICS OLYMPIAD
SOUTH AFRICAN TERTIARY MATHEMATICS OLYMPIAD. Determine the following value: 7 August 6 Solutions π + π. Solution: Since π
More informationA polynomial expression is the addition or subtraction of many algebraic terms with positive integer powers.
LEAVING CERT Honours Maths notes on Algebra. A polynomial expression is the addition or subtraction of many algebraic terms with positive integer powers. The degree is the highest power of x. 3x 2 + 2x
More informationCalculus I Review Solutions
Calculus I Review Solutions. Compare and contrast the three Value Theorems of the course. When you would typically use each. The three value theorems are the Intermediate, Mean and Extreme value theorems.
More information2 Analogies between addition and multiplication
Problem Analysis The problem Start out with 99% water. Some of the water evaporates, end up with 98% water. How much of the water evaporates? Guesses Solution: Guesses: Not %. 2%. 5%. Not 00%. 3%..0%..5%.
More informationN13/5/MATHL/HP2/ENG/TZ0/XX/M MARKSCHEME. November 2013 MATHEMATICS. Higher Level. Paper pages
N/5/MATHL/HP/ENG/TZ0/XX/M MARKSCHEME November 0 MATHEMATICS Higher Level Paper 0 pages N/5/MATHL/HP/ENG/TZ0/XX/M This markscheme is confidential and for the exclusive use of examiners in this examination
More informationMath 265H: Calculus III Practice Midterm II: Fall 2014
Name: Section #: Math 65H: alculus III Practice Midterm II: Fall 14 Instructions: This exam has 7 problems. The number of points awarded for each question is indicated in the problem. Answer each question
More information2 2xdx. Craigmount High School Mathematics Department
Π 5 3 xdx 5 cosx 4 6 3 8 Help Your Child With Higher Maths Introduction We ve designed this booklet so that you can use it with your child throughout the session, as he/she moves through the Higher course,
More informationIndefinite Integration
Indefinite Integration 1 An antiderivative of a function y = f(x) defined on some interval (a, b) is called any function F(x) whose derivative at any point of this interval is equal to f(x): F'(x) = f(x)
More informationFunctions and their Graphs
Chapter One Due Monday, December 12 Functions and their Graphs Functions Domain and Range Composition and Inverses Calculator Input and Output Transformations Quadratics Functions A function yields a specific
More informationUNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education
UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/02 Paper 2 Examination from 2013 SPECIMEN PAPER 2 hours Candidates
More informationChapter 8A. Recall. where. DeMoivre: then. Now lets do something NEW some groovy algebra. therefore, So clearly, we get:
Chapter 8A We are revisiting this so we can use Euler Form later in the Chapter to assist us Integrate different functions (amongst other fun things). But first lets go over some old ground: Recall where
More informationTime: 1 hour 30 minutes
Paper Reference(s) 6666/ Edexcel GCE Core Mathematics C4 Gold Level (Harder) G3 Time: hour 3 minutes Materials required for examination Mathematical Formulae (Green) Items included with question papers
More informationTime: 1 hour 30 minutes
Paper Reference(s) 6666/0 Edexcel GCE Core Mathematics C4 Silver Level S Time: hour 0 minutes Materials required for examination papers Mathematical Formulae (Green) Items included with question Nil Candidates
More informationCore Mathematics C34
Write your name here Surname Other names Pearson Edexcel International Advanced Level Centre Number Candidate Number Core Mathematics C34 Advanced Monday 16 June 2014 Morning Time: 2 hours 30 minutes You
More informationTable of Contents. Module 1
Table of Contents Module 1 1.1 Order of Operations 1.6 Signed Numbers 1. Factorization of Integers 1.7 Further Signed Numbers 1.3 Fractions 1.8 Power Laws 1.4 Fractions and Decimals 1.9 Introduction to
More informationPearson Edexcel Level 3 Advanced Subsidiary GCE in Mathematics (8MA0) Pearson Edexcel Level 3 Advanced GCE in Mathematics (9MA0)
Pearson Edexcel Level 3 Advanced Subsidiary GCE in Mathematics (8MA0) Pearson Edexcel Level 3 Advanced GCE in Mathematics (9MA0) First teaching from September 2017 First certification from June 2018 2
More informationIES Parque Lineal - 2º ESO
UNIT5. ALGEBRA Contenido 1. Algebraic expressions.... 1 Worksheet: algebraic expressions.... 2 2. Monomials.... 3 Worksheet: monomials.... 5 3. Polynomials... 6 Worksheet: polynomials... 9 4. Factorising....
More informationHSC Marking Feedback 2017
HSC Marking Feedback 017 Mathematics Extension 1 Written Examination Question 11 Part (a) The large majority of responses showed correct substitution into the formula x = kx +lx 1 k+l given on the Reference
More informationWeek Topics of study Home/Independent Learning Assessment (If in addition to homework) 7 th September 2015
Week Topics of study Home/Independent Learning Assessment (If in addition to homework) 7 th September Functions: define the terms range and domain (PLC 1A) and identify the range and domain of given functions
More informationFINAL EXAM STUDY GUIDE
FINAL EXAM STUDY GUIDE The Final Exam takes place on Wednesday, June 13, 2018, from 10:30 AM to 12:30 PM in 1100 Donald Bren Hall (not the usual lecture room!!!) NO books/notes/calculators/cheat sheets
More informationPreliminary algebra. Polynomial equations. and three real roots altogether. Continue an investigation of its properties as follows.
978-0-51-67973- - Student Solutions Manual for Mathematical Methods for Physics and Engineering: 1 Preliminary algebra Polynomial equations 1.1 It can be shown that the polynomial g(x) =4x 3 +3x 6x 1 has
More informationTable of Contents. Module 1
Table of Contents Module Order of operations 6 Signed Numbers Factorization of Integers 7 Further Signed Numbers 3 Fractions 8 Power Laws 4 Fractions and Decimals 9 Introduction to Algebra 5 Percentages
More informationUnit 1 PreCalculus Review & Limits
1 Unit 1 PreCalculus Review & Limits Factoring: Remove common factors first Terms - Difference of Squares a b a b a b - Sum of Cubes ( )( ) a b a b a ab b 3 3 - Difference of Cubes a b a b a ab b 3 3 3
More informationMathematics (MEI) Advanced Subsidiary GCE Core 1 (4751) May 2010
Link to past paper on OCR website: http://www.mei.org.uk/files/papers/c110ju_ergh.pdf These solutions are for your personal use only. DO NOT photocopy or pass on to third parties. If you are a school or
More informationa b = a a a and that has been used here. ( )
Review Eercise ( i j+ k) ( i+ j k) i j k = = i j+ k (( ) ( ) ) (( ) ( ) ) (( ) ( ) ) = i j+ k = ( ) i ( ( )) j+ ( ) k = j k Hence ( ) ( i j+ k) ( i+ j k) = ( ) + ( ) = 8 = Formulae for finding the vector
More information7.5 Partial Fractions and Integration
650 CHPTER 7. DVNCED INTEGRTION TECHNIQUES 7.5 Partial Fractions and Integration In this section we are interested in techniques for computing integrals of the form P(x) dx, (7.49) Q(x) where P(x) and
More informationTo factor an expression means to write it as a product of factors instead of a sum of terms. The expression 3x
Factoring trinomials In general, we are factoring ax + bx + c where a, b, and c are real numbers. To factor an expression means to write it as a product of factors instead of a sum of terms. The expression
More informationCandidates are expected to have available a calculator. Only division by (x + a) or (x a) will be required.
Revision Checklist Unit C2: Core Mathematics 2 Unit description Algebra and functions; coordinate geometry in the (x, y) plane; sequences and series; trigonometry; exponentials and logarithms; differentiation;
More informationΠ xdx cos 2 x
Π 5 3 xdx 5 4 6 3 8 cos x Help Your Child with Higher Maths Introduction We ve designed this booklet so that you can use it with your child throughout the session, as he/she moves through the Higher course,
More informationChapter 5: Integrals
Chapter 5: Integrals Section 5.3 The Fundamental Theorem of Calculus Sec. 5.3: The Fundamental Theorem of Calculus Fundamental Theorem of Calculus: Sec. 5.3: The Fundamental Theorem of Calculus Fundamental
More informationSection 6-5 : Stokes' Theorem
ection 6-5 : tokes' Theorem In this section we are going to take a look at a theorem that is a higher dimensional version of Green s Theorem. In Green s Theorem we related a line integral to a double integral
More information