Document Sample

Integration It might appear to be a bit obvious but you must remember all of your C3 work on differentiation if you are to succeed with this unit. Content By the end of this unit you should be able to: 1 Integrate trigonometric, exponential functions and x Find volumes of revolution Find integrals using partial fractions and integrating by parts. Finding solutions to seperable first order differential equations. Find areas under curves and volumes generated from functions given in parametric form. Use of trapezium rule to provide approximate solution to integrals. Standard Integrals Function (f(x)) Integral f(x)dx xn xn 1 c n 1 1 ln x c x (ax + b)n (ax b)n 1 c a(n 1) 1 1 ln ax b c ax b a ex ex + c Sin x -Cos x + c Cos x Sin x + c Tan x ln|(Sec x)| + c Cot x ln|(Sin x)| + c Cosec x -ln|(Cosec x + Cot x)| + c Sec x ln|(Sec x + Tan x)| + c Sec2x Tanx + c Sec x Tan x Sec x + c -Cosec 2x Cot x + c C4 1 eax+b e ax b c a Cos (ax + b) 1 Sin(ax b) c a SinnxCosx 1 Sin x c n 1 n 1 Cosn xSinx 1 C os x c n 1 n 1 2 Sin x 1 Sin 2x 2 x 2 c 2 Cos x 1 Sin 2x 2 2 x c Integrals in red are given in formula book (but you should be able to prove them). Integration of trigonometric functions. To succeed with these types of questions you need to have a very good grasp of the trig identities from C3 and obviously the differentiation from the same unit. Integration questions involving trigonometry are either of a substitution type or areas and volumes from parametrics. The following examples give a flavour but can’t cover every eventuality. Example 1 By using the substitution u = sinx or otherwise, find 8sin x sin2xdx 2 Giving your answer in terms of x. From C3 you should remember that: sin2x = 2sinx cosx Hence: 8sin x sin2xdx 16sin3 x cos xdx 2 If u = sinx du = cosdx (part of the integral!) C4 2 Therefore 16sin x cos xdx 16u3du 3 4u4 c 4 sin 4 x c Example 2 Use the substitution x = sin θ to find the exact value of 1 4 2 0 3 dx (1 x ) 2 2 If x = sin θ then the denominator becomes (1 – sin2θ) which is cos 2θ. When this is raised to the power of 1.5 the result is cos3θ as a denominator. The x’s have now been substituted but you can’t forget about the dx. If x = sin θ then dx = cosθdθ. Therefore the integral becomes: 1 4 4 2 0 3 dx 3 cos d (1 x ) 2 2 (1 sin ) 2 2 4 4 cos cos d cos 3 2 d 4 The next thing to spot is that 4 sec2 . Therefore the question cos 2 becomes: 4sec d 4tan 2 The question isn’t quite finished as the original limits were given in terms of x. So for the lower limit of x = 0, sin θ = 0, hence θ = 0. For the upper limit x = 0.5, so sin θ = 0.5, hence θ = . The question asks for the exact value so 3 the integral is finally. 3 4 sec2 d 4 tan 0 3 0 4 4 3 3 3 C4 3 There are a few tricks in this question but practice makes perfect and you must remember all of your differentiation and trig work from C3 and C4. Use ideas outlined in the examples above to carry out the next question. Use the substitution u 2 = 1 – cos2θ and integration to find 2 0 Sin2 1 Cos2d . (You will need to use implicit differentiation when dealing with the u 2.) Volume of Revolution The basic principle here is that we take very thin strips, of width δx (delta x), underneath a curve in the xy plane and rotate them around the x axis to generate a volume. The diagram on the left shows the graph of y = x 2 in 3D. The pink section is the area under the graph between the x values of 0 and 1. This pink section has been rotated through 360 degrees about the x axis to form the green solid in the diagram on the right. If you look at the far end of the green solid it has a circular appearance. The area of the end is πy2 and therefore the volume of a very thin disc of width δx will be πy 2δx. We use integration to add up the volumes of all of the very thin discs to give the volume of revolution. ∑ πy2δx = y2dx C4 4 Curves can be rotated about the x or y axis and therefore there are two formulae to calculate volumes. The volume generated by a curve as it rotates around the x axis is given by: y2dx The volume generated by a curve as it rotates around the y axis is given by: x2dy The example outlined below is very standard but hopefully the pitfalls are well signposted. Example 3 The finite region bounded by the curve with the equation y = 8x – x2 and the x axis is rotated through 360º about the x axis. Using integration find, in terms of π, the volume of its solid form. The volume generated by a curve as it rotates around the x axis is given by: y2dx The mistake a number of candidates make is to state that y 2 in this case is 64x2 + x4 WRONG Y2 = (8x – x2)( 8x – x2) = 64x2 – 16x3 + x4 Therefore volume is given by : (64x2 - 16x3 + x 4 )dx 64 3 x5 x 4x 4 3 5 The quadratic, y = 8x – x2, has solutions of x = 0 and x = 8 and therefore these become the limits on the integral. C4 5 8 64 x5 x3 4x 4 3 5 0 16380 15 Example 4 Find the volume generated when the region bounded by the curve with 5 1 equation y 2 , the x axis and the lines x = and x = 3 is rotated x 3 through 360 degrees about the x axis. Volume = y2dx 2 3 5 3 20 25 1 2 dx 1 4 2 dx 3 x 3 x x 3 25 4x 20ln x x 1 3 25 4 1 12 20ln3 20ln 75 3 3 3 232 20ln 9 3 Take care with the negative powers when integrating and the rules of logs. In most cases the question will specifically ask for answers in terms of a + bln9 for example. A calculator answer will loose you the final answer mark. Integrating by Parts. This concept is guaranteed to appear on a C4 paper and is not too tricky. Integration by parts is used to integrate functions that are made up of products of functions. Not surprisingly the process is the reverse of differentiating a product. The idea is best explained through an example. C4 6 Example 5 3 Use integration by parts to find the exact value of 1 x5 ln xdx The rule for integrating by parts is: dv du u dx dx uv v dxdx The thing to remember is that one part of the question becomes u and the dv other part becomes . Since we can’t integrate lnx, it must be u and dx dv du therefore x5 must equal . On the right of the rule we need v and this dx dx can be achieved by integrating and differentiating respectively. Therefore: 3 1 x5 ln xdx dv Let u = lnx and x5 dx du 1 x6 v dx x 6 So by substituting into the rule: dv du u dx dx uv v dxdx C4 7 3 x6 3 1 x6 1 x5 ln xdx 6 dx 1 x 6 3 x6 3x 5 x6 x6 6 1 6 dx 6 36 1 3 5x6 405 5 4 36 36 1 910 9 I was telling a little lie before. lnx can be integrated and you should know how!! Differential Equations Most of the differential equation questions will require a number of integration techniques. Integration of trig functions, use of partial fractions or integration by parts could be used. The following example uses integration by parts to find the general solution. Example 6 Given that y = 7 at x = π, solve the differential equation dy 2yx2 cos x, y>0 dx Differential equations in C4 are solved by seperating variables (x’s and dx to one side, y’s and dy to the other). So the question becomes: dy 2y x cos xdx 2 The left hand side is simply a natural log answer. The right hand side, however, will require two uses of integration by parts and a question of this difficultly has been on recent exam papers. Here goes! C4 8 If we start by integrating the right hand side by parts: 1 dv ln y x2 cos xdx u x2 cosx 2 dx We have chosen the x2 to be u because it is made simpler by differentiation. 1 dv ln y x2 cos xdx u x2 cosx 2 dx du 2x v sinx dx Substituting the values into the equation for integrating by parts gives: 1 du ln y uv v dx 2 dx 1 ln y x2 sin x 2x sin xdx 2 Unfortunately the integral on the right is still too complex so we have to integrate by parts again. At least this time the 2x will differentiate to 2. 1 ln y x2 sin x 2x sin xdx 2 1 dv ln y x2 sin x .... u 2x sin x 2 dx du 2 v - cosx dx 1 2 ln y x2 sin x 2x cos x 2 cos xdx 1 ln y x2 sin x 2x cos x 2 sin x c 2 Great care must be taken with the signs in a question of this complexity. C4 9 The following example tests the concept of partial fractions but then uses the solution in a differential equation. Remember that separation of the variables is the key. Example 7 13 2x (a) Express in partial fractions. (2x 3) (x 1) (b) Given that y = 4 at x = 2, use your answer to part (a) to find the solution of the differential equation dy y (13 2x ) = , x > 1.5 dx (2x 3) (x 1) Express your answer in the form y = f(x). 13 2x (a) Express in partial fractions. (2x 3) (x 1) 13 2x A B (2x 3)(x 1) 2x 3 x 1 13 2x A(x 1) B(2x 3) By letting x = -1 and x = 0 we find that A = 4 and B = -3. Therefore: 13 2x 4 3 (2x 3)(x 1) 2x 3 x 1 (b) Given that y = 4 at x = 2, use your answer to part (a) to find the solution of the differential equation dy y (13 2x ) = , x > 1.5 dx (2x 3) (x 1) Express your answer in the form y = f(x). C4 10 As mentioned earlier we solve first order differential equations by separating the variables and then integrate with respect to x and y. dy y (13 2x) dx (2x 3)(x 1) Seperate variables and int egrate : dy (13 2x)dx y (2x 3)(x 1) Usin g part (a) and integrating the left we get: 4 4 3 3 lny= ln y 2x-3 dxdx 1 dx 2x 3 x x 1 ln y 2ln(2x 3) 3ln(x 1) c ln y 2ln(2x 3) 3ln(x 1) c If we use the initial condition of y = 4 and x = 2 to find c. ln 4 2ln(1) 3ln(3) c Re member that ln1 0 and 3ln3=ln27 Therefore : c = ln4 +ln27=ln108 Therefore: ln y 2ln(2x 3) 3ln(x 1) ln108 Usin g rules of log s ln y ln(2x 3)2 ln(x 1)3 ln108 108(2x 3)2 ln y ln (x 1) 3 We can now remove the logarithms to give the answer in the form y = f(x): 108(2x 3)2 y (x 1) 3 C4 11 The following example deals with exponential decay, a regular differential equation question. Example 8 A radioactive substance has a half life of 1200 years (the time taken for half of the substance to decay) and at time t = 0, 6kg are present. If the substance decays at a rate proportional to the amount, x, present at a time t find an equation for x in terms of t. Find also the amount of substance left after 800yrs. So the rate of change is proportional to the amount of the substance hence: dx x dt Therefore: dx kx dt The negative is used to highlight decay. By separating the variables we get: dx x kdt And integrating gives: lnx + lnC = -kt lnCx = -kt Removing the logs: 1 Cx = e-kt let A C So finally x = Ae-kt Using the initial conditions of t = 0 and x = 6 A=6 C4 12 By definition after 1200 yrs the mass of substance should be 3kg. Therefore: x = 6e-1200k 3 = 6e-1200k Taking ln of both sides: ln0.5 = -1200k k = 0.000578 Hence x = 6e-0.000578t If t = 800 the amount of substance left is: x = 6e-0.000578 × 800 x = 3.78kg More complex integrations by substitution. Example 9 Use the substitution u 2 = 2x – 1, or otherwise, find the exact value of 25 6x 13 2x 1 dx From an earlier example you should remember that there are three parts that need dealing with, 6x, the dx and the square root in the denominator If u2 = 2x – 1 then 2udu = 2dx ( by implicit differentiation). So dx = udu. If u2 = 2x – 1 then x = 0.5u 2 + 0.5. The integral becomes: C4 13 25 6x 3u 2 3 13 2x 1 dx u udu (3u2 3)du u3 3u And now for the limits. If x = 13 then u = 5 and if x = 25 then u = 7. 25 6x 7 13 2x 1 dx u3 3u 5 224 Integration of Parametrics to Find Areas under curves. Example 10 The diagram shows a sketch of the curve C with parametric equations x = 7tsint, y = 5sect, 0≤t< 2 The point P(a, 10) lies on C. a) Find the exact value of a. The region R is enclosed by the curve C, the axes and the line x = a. b) Show that the area of R is given by 35 3 tan t t dt 0 C4 14 c) Find the exact value of the area R. a) Find the exact value of a. The point P has y coordinate 10. Therefore: 5sect = 10 If t = 3 7 sect =2 x= sin 3 3 7 3 cost = 0.5 x= 6 7 3 t= a= 3 6 b) When integrating to find the area under a parametric curve you need to integrate the following: dx y dt dt The x part of the parametric is a product and therefore we need to differentiate the first (7t) and multiply it second (sint) then add the first multiplied by the differential of the second. This gives: dx 7 sin t 7tcos t dt So to find the area R: dx y dt dt 5 sec t 7 sin t 7t cos t 1 sint Re membering that sect = and that tan t cost cost 35 3 tan t t dt 0 C4 15 The limits are defined from the work in part (a). c) The integral from part (b) becomes: sin t 35 3 t dt 0 cos t The integral of tant should be in your notes. The denominator nearly differentiates to give the numerator. Therefore the integral is given by the negative ln of cost. sin t 35 3 t dt 35 3 tan t t dt 0 cos t 0 35 ln cos t 0.5t2 3 0 8.45 = 43.5 If you are asked to find the volume generated by a parametric as it rotates about the x axis then the integral to evaluate is: dx y2 dt dt Example 11 The diagram below is the graph of the function y = Sec x C4 16 Use the trapezium rule with five equally spaced ordinates to estimate the area of the region bounded by the curve with equation y = sec x, the x-axis and the lines x = - and x = , 3 3 giving your answer to two decimal places. π π π π The five x coordinates are , ,0, and . The best way to 3 6 6 3 succeed with these questions is to put the information into a table. x π π 0 π π 3 6 6 3 Y 2 1.155 1 1.155 2 Trapezium rule is introduced in AS. 1 h xo 2 (x1 x2 x3 ...xn 1 ) xn b a f( x)dx 2 π 1 π 3 π Secxdx 2 2 1.155 2 1 2 1.155 2 3 2 6 2.78(2dp) Finally an integration question that has an exponential substitution! Example 12 Use the substitution u = 3x to find the exact value of 2 3x dx 0 3x 1 The integral very quickly becomes C4 17 udx u1 but we haven’t dealt with the dx. From example 9 of the C4 differentiation notes the differential of u is: du 3x ln 3 dx du Therefore dx and the integral becomes: u ln 3 9 9 du 1 1 ln 3(u 1) ln 3 ln(u 1) 1 ln 5 ln 3 Remember that the question said exact answer so leave it in terms of logs. C4 18

DOCUMENT INFO

Shared By:

Categories:

Tags:
Data Integration, application integration, Integration Solutions, integration software, SOA governance, Data Migration, White Paper, online dictionary, enterprise integration, how to

Stats:

views: | 311 |

posted: | 3/6/2010 |

language: | English |

pages: | 18 |

OTHER DOCS BY malj

How are you planning on using Docstoc?
BUSINESS
PERSONAL

By registering with docstoc.com you agree to our
privacy policy and
terms of service, and to receive content and offer notifications.

Docstoc is the premier online destination to start and grow small businesses. It hosts the best quality and widest selection of professional documents (over 20 million) and resources including expert videos, articles and productivity tools to make every small business better.

Search or Browse for any specific document or resource you need for your business. Or explore our curated resources for Starting a Business, Growing a Business or for Professional Development.

Feel free to Contact Us with any questions you might have.