Integration webversion - Learnhigher

1 Integration - the basics Dr. Mundeep Gill Brunel University 1 Integration Integration is used to find areas under curves. Integration is the reversal of differentiation hence functions can be integrated by indentifying the anti-derivative. However, we will learn the process of Integration as a set of rules rather than identifying anti-derivatives. Terminology Indefinite and Definite integrals There are two types of integrals: Indefinite and Definite. Indefinite integrals are those with no limits and definite integrals have limits.

2 When dealing with indefinite integrals you need to add a constant of Integration . For example, if integrating the function f(x) with respect to x: () dxxf= g(x) + C where g(x) is the integrated function. C is an arbitrary constant called the constant of Integration . dx indicates the variable with respect to which we are integrating, in this case, x. The function being integrated, f(x), is called the integrand. Integration - the basics 2 The rules The Power Rule dxxn = C1nx1n+++ provided that n -1 Examples: dxx5= 6x6+ C dxx-4= 3-x-3+ C When n = -1 dxx-1 = dxx1 = ln x + C Constant rule dxk= kx + C where k is a constant Example: dx2= 2x + C Exponentials dxekx = Cek1kx+ Example: Ce91dxe9x9x+= Cedxexx+= Trig functions - Cos ()= dxxcossin(x) + C ()= dxkxcos()kxsink1 + C where k is a constant Example.

3 ()= dx12xcos()12xsin121 + C Integration - the basics 3 - Sin ()= dxxsin-cos(x) + C ()= dxkxsin()kxcosk1 + C where k is a constant Example: ()= dx10xsin()10xcos101 + C ()= dx5x-sin()5x-cos51 + C Linearity Suppose f(x) and g(x) are two functions in terms of x, then: ()()[]()() = dxxgdxxfdxxgxf Additionally, if A and B are constants, then ()()[]()() = dxxgBdxxfAdxxBgxAf Examples: () +dx3x2x54= dx2x4+ dx3x5 = dxx24+ dxx35 = 5x25+ 6x36 + C = 2x52x65+ + C ()() dx3e3x5cos7x = () dx3e-dx3x5cos7x =() dxe3-dx3xcos57x = () 7xe7133xsin315 = ()7xe733xsin35 Integration - the basics 4 Questions (General rules): Integrate the following functions: 1.

4 () + dxxx523x16 2. () +dx5x3x8 3. () dx3x9x-12 4. ()() +dxe4xsin3x 5. ()() +dx7x7xcos2 (Solutions on page 8) Definite Integrals Earlier we saw that () dxxf= g(x) + C Suppose now we are given limits, ( ) badxxf= g(x) + C This can be interpreted as: (value of g(x) + C at x = b) (value of g(x) + C at x = a) In other words, since C will cancel out: ( ) badxxf= g(b) g(a) The full calculation of definite integrals is usually written out as: ( ) badxxf= ()[]baxg = g(b) g(a) integrate the function first (find g(x)) then substitute in the given limits (always substitute the upper limit first).

5 (where a is the lower limit and b is the upper limit) Integration - the basics 5 Examples 1. 102dxx= 103x31 = []103x31 = 31{(1)3 (0)3} = 31(1 0) = 31 2. ()dx12x31 += 312x22x += []312xx+ = {(32 + 3) (12 + 1)} = {(9 + 3) (1 + 1)} = 12 2 = 10 3. ( )dxxcos2 0 = ( )[]2 0xsin= {(sin(2 )) (sin(0))} = 1 0 = 1 Questions (Definite integrals): Integrate the following functions: 1. ()dx52x3x212 + 2. dxe107x 3. () 0dx2xsin 4. () +414xdxx412e (Solutions on page 9) Integration that leads to log functions We know that if we differentiate y = ln(x) we find x1dxdy=.

6 We also know that if y = ln f(x), this differentiates as: ()( )xfx'fdxdy= If we can recognise that the function we are trying to integrate is the derivative of another function, we can simply reverse the above process. So if the function we are trying to integrate is a quotient, and if the numerator is the derivative of the denominator, then the integral will involve a logarithm, ()( ) dxxfx'f= ln (f(x)) + C Example: y = ln(2x2 + 5) t = 2x2 + 5 y = ln t dxdt= 4x dtdy=t1 dxdy = 4x x t1=t4x=5+22x4x Integration - the basics 6 Example 1: +dx5x35 The derivative of the denominator is 5 which is the same as the numerator, hence +dx5x35= ln (3 + 5x) + C Example 2: +dxx1x2 The derivative of the denominator is 2x.

7 This is not the same as the numerator but we can make it the same by re-writing the function 2x1x+ as 2x12x21+ , therefore +dxx1x2= +dxx12x212= 21ln (1 + x2) + C Example 3: ( ) dxxxln1 The derivative of ln x is x1, so we can rewrite the function as: ( )xlnx1. Hence ( ) dxxxln1= ( ) dxxlnx1= ln(ln(x)) + C Example 4: + 21dx1x3x3 + 21dx1x3x3= +21dx1x1- x13 = ()()[]211x3lnx3ln+ = {(3ln(2) 3ln(3)) (3ln(1) 3ln(2))} = 3ln(2) 3ln(3) + 3ln(2) = 6ln(2) 3ln(3) = ln(26) ln(33) = ln(64) ln(27) = 2764ln Integration - the basics 7 Questions ( Integration that leads to log functions): Integrate the following functions: 1.

8 +dx3x23 2. +dx2x1x2 3. +dx1ee2x2x 4. +dx4xx2--3 5. + +10dx2x11x1 (Solutions on page 10) Integration - the basics 8 Solutions (General rules): 1. () + dxxx523x16= +dxdxx-dxx523x16 = +dxxdxx-dxx5623 = ()4xx7x425725 + + C = 474x152x7x25 + C 2. () +dx5x3x8= dx3x8+ dxx - dx5 = dxx38+ dxx - dx5 = 5x2x93x29 + + C = 5x2x3x29 + + C 3. () dx3x9x-12= dx9x2- dx3x-1 = dxx92- dxx3-1 = 39x3- ()x3ln + C = 33x- ()x3ln + C 4.

9 ()() +dxe4xsin3x = () dx4xsin + dxe3x = ()3xe314xcos41+ + C 5. ()() +dx7x7xcos2= () dx7xcos + dxx72 = ()3x377xsin71+ + C Integration - the basics 9 Solutions (Definite integrals): 1. ()dx52x3x212 + = 21235x22x33x + = []21235xxx+ = {(23 22 + 5(2)) (13 12 + 5(1))} = {(8 4 + 10) (1 1 + 5)} = 14 5 = 9 2. []107x107x107xe71e71dxe= = = 71{e7 e0} = 71(e7 1) 3. () 0dx2xsin= () 02xcos21 = ()[] 0212xcos = -21{cos(2 ) cos (0)} = -21{1 1} = 0 4. () +414xdxx412e= () +414xdx4x12e21 = 41234x234x412e += 414x38x3e23 + = ( ) + +383e3483e41623 = () + +383e3883e416 = 383e3643e416 + = 3563e3e416+ Integration - the basics 10 Solutions ( Integration that leads to log functions): 1.

10 +dx3x23= ln (2 + 3x) + C 2. +dx2x1x2 Differentiating the denominator gives 4x Therefore rewrite the function: 22x1x+ = 22x14x41+ Hence, +dx2x1x2 = + dx2x14x412= +dx2x14x241= 41ln (1 + 2x2) + C 3. +dx1ee2x2x Differentiating the denominator gives 2e2x hence we can rewrite the function as: 1ee2x2x+ = 1e2e212x2x+ + dx1ee212x2x= 21ln (e2x + 1) + C 4. +dx4xx2--3 Differentiating the denominator gives -2x-3, hence the function can be rewritten as: 4xx2--3+ = 4x2x212--3+ +dx4xx2--3= + dx4x2x212--3= -21ln(x-2 + 4) + C Integration - the basics 11 5.