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Integration Integration It might appear to be a bit obvious

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					                                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  Cos2d .
(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
                   u1                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

				
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