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. When dealing with indefinite integrals you need to add a constant of Integration .

2 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=. 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.

6 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. 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.

7 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. +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.

8 () +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. ()() +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.

9 () 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. +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.

10 +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. + +10dx2x11x1= ()()[]102xln1xln+ + = {(ln(2) ln(3)) (ln(1) ln(2))} = ln(2) ln(3) + ln(2) = 2ln(2) ln(3) = ln(22) ln(3) = ln(4) ln(3) = 34ln