Integration After completing this chapter you should be able to integrate: A number of standard mathematical functions Using the reverse of the chain rule Using trigonometrical identities Using partial fractions By substitution By parts To find areas and volumes To solve differential equations In addition some functions are too difficult to integrate and hence you should be able to Use the trapezium rule to approximate their value 6.1 You need to be able to integrate standard functions
There are 13 of these so please look at the sheet and then practise with exercise 6A page 89 6.2 You can integrate some functions using the reverse of the chain rule Basically you look to see what might have differentiated to the function you want to integrate! 1 ( + ) = ( + ) + Lets look at some examples to see have this is applied Find Let
I= Then from (x) on our sheet we can see I= Find Let I= Now consider y = ln Then This is x the function we want to integrate So I= Exercise 6B page 92
6.3 You can use trigonometric identities in integration It is sometimes easier to change an expression, using trigonometrical identities, into one you know how to integrate. For example you could change sin x into Or Or sin3x cos 3x into (sin2A = 2 sinA cosA) (sec x + tan x) = secx + 2 sec x tan x + tanx = secx + 2 sec x tan x + secx 1 = 2secx + 2 sec x tan x 1 All these are in the list, so you can take it from here !
Example Find Use cosecA = cotA + 1 So I= This is on the list and gives us I= Exercise 6C page 94 6.4 You can use partial fractions to integrate expressions It can be useful to split some expressions into partial fractions before integrating. +8 3 2
( 1)( +2) = 1 +2 1 1 | | 1 | +2|+ 3 2 Exercise 6D page 98 6.5 You can use standard patterns to integrate some expressions To integrate an expression is of the form try lnand differentiate to check and adjust any constants To integrate an expression of the form Try and differentiate to check and adjust any constants
Example Find Let I = This looks like it could be of the form So we try y = = So I = + C Example Find Let I = This looks like it could be of the form So we try y = sin(4x 1) = 12 So I=
Exercise 6E page 100 6.6 Sometimes you can simplify an integral by changing the variable. This process is similar to using the chain rule in differentiation and is called integration by substitution. When you try to find it is easier to do if we substitute for 2x + 5 Let u = 2x + 5 then x = and this gives dx = Replace these elements into the original integral I= = which we can hopefully do and get I= +C Example Find by using the substitution u = tanx
Let I = dx u = tanx giving du = secx dx Substituting this information in gives I= I= So I= The substitutions suggested by the question writers should make life easier, if they dont youre doing something wrong! Example Use the substitution u = 1 + sin x to integrate I= u = 1 + sin x
sinx = u - 1 so du = cosx dx Substituting these in gives I= I= I= - we can rearrange this to give I = Now substitute for U I= I= Exercise 6F page 105
HINT if you see u -1 in any shape or form factorise it to (u + 1)(u 1) 6.7 You can use integration by parts to integrate some expressions The product rule gives ( ) = + Rearranging this gives us
Integrating each term with respect to x gives ( ) = This simplifies to give us the integration by parts formula = Example 1: Find x 2 xe dx .
This is a suitable candidate for integration by parts with u 2 x and du 2 dx dv e x v e x dx u 2 x Substitute these into the formula: x x x x x 2 xe
dx 2 xe 2 e dx 2 xe 2 e c dv
e x : dx : Find x ln xdx . This can be found using integration by parts if we take u ln x and du 1 dx 2 x dv x x v dx 2 u ln x
Substitute these into the formula: x2 1 x2 1 2 x ln xdx ln x dx x ln x 2
2 2 x 12 x 2 ln x 14 x 2 c 1 2 xdx dv x . dx Find ln xdx . This can be thought of as 1ln xdx and so can be integrated by parts with u ln x and du 1
dx x dv 1 v x dx u ln x 1 x x dx x ln x 1dx = x ln x x c ln xdx x ln x dv 1 dx Find
x cos xdx . dv Here we take u x and cos x : dx du u x 1 dx dv cos x v sin x dx Substitute these into the formula: x cos xdx x sin x sin xdx x sin x ( cos x) c x sin x cos x c Use integration by parts to find the exact value of
u=x =1 = cosecx v = -cotx substituting all of this back into the formula 6 3 3 1
= + 18 2 2 ( ) ( ( ) ( )) 3 = + 3 18 6.8 You can use numerical integration You may be asked to use the trapezium rule for integrals involving some of the new functions we have met in C3 and C4 Remember the trapezium rule looks like this:
1 = 2 h [ 0+2 ( 1 + 2 +. . .+ 1 ) + ] Where and 1 = h [ 0+2 ( 1 + 2 +. . .+ 1 ) + ] 2 Where and
Example a) Complete the table below, giving your answers to 3 significant figures x 1 + ln 2x 1 1.5 2.10 2 2.5 3 2.61 2.79
b) Use the trapezium rule, with 4 intervals, to estimate the value of c) Find the exact value of I d) Calculate the percentage error in using your answer from part (b) to estimate the value of I 1 = h [ 0+2 ( 1 + 2 +. . .+ 1 ) + ] 2 Where and a)
x 1 1.5 2 2.5 3 1 + ln 2x 1.69 2.10 2.39
2.61 2.79 b) Use the trapezium rule, with 4 intervals, to estimate the value of So I= I = 4.67 Find the exact value of I 1 Use integration by parts to deal with = ln 2 1
= 3 =1 = 3 1 ln 2 = [ ln 2 ] 1 1 3 1 6 3 =2+
2 2 () I = ln 108 Calculate the percentage error in using your answer from part (b) to estimate the value of I Percentage error = 100 x Exercise 6H page 110 6.9 You can use integration to find areas and volumes Area of a region between y = f(x), the x-axis and x = a and x = b is given by Area =
Volume of revolution formed when y = f(x) is rotated about the x-axis between x + a and x + b is given by: Volume Now we use these with all the nice new integrals we have . . . . . . Example the region R is bounded by the curve y = , the x-axis and the lines with equations x = 2 and x = 4. This region is rotated through 2 radians about the x-axis. Find the exact value of the volume of the solid generated. 4 2 = 2 =16 2 48+ 36 2
2 So 3 = 16 48 36 1 3 [ = [( =211 ] 4
2 1024 128 192 9 96 18 3 3 2 3 ) ( )]
Exercise 6I page 113 6.10 You can use integration to solve differential equations We can solve first order differential equations by the process called separation of variables When you can write Sometimes boundary conditions may be given in a question which means you can find an particular solution. In this case you take the general solution and use the boundary conditions to find the constant of integration. Example Obtain the solution of for which y = 1 when x = . Give your answer in the form y = f(x). Separate the variables and integrate
1 1 ( +1) = Use partial fractions to simplify the integral in y and remember ( 1 1 = +1 ) This gives us ln y ln (y + 1) = ln sec x + C use log laws to combine the y terms and remember + C can be written as ln k
getting out of log notation gives us use the given conditions, y = 1 when x = , to find k so k = This gives us 4y cos x = y + 1 rearrange to get in the required form y(4 cos x 1) = 1 so Exercise 6J page 116 6.11 Sometimes the differential equation will arise out of a context and the solution may need interpreting in terms of that context Example The rate of increase of a population of P organisms at time t is given
by , where k is a positive constant. Given that at t = 0 the population was of size 8, and at t + 1 the population is 56, find the size of the population at time t = 2. start by separating the variables ln ln 8 = C so ln find C using t = 0 P = 8 simplify using log laws use t = 1 and P = 56 to find out that k = ln7 this gives us
= 7 8 || now use t = 2 =2 ln 7 8 || ln | |= ln 49 8
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