INTEGRATION IKMJB 78 DUM 20022/20032 UNIT 3: INTEGRATION INTRODUCTION Every operation in mathematics has its inverse. In this chapter we learn how to reverse the process of differentiation with the process of integration. LEARNING OUTCOMES Upon completion of this unit, students should be able to: 1. find the integrals of functions by considering integration as a reverse of differentiation. 2. integrate indefinite integrals by rule of integration. 3. integrate composite polynomial functions by using substitution method. 4. integrate product of functions by part. 5. integrate quotient of functions using partial fraction method. 6. evaluate the definite integrals.
INTEGRATION IKMJB 79 DUM 20022/20032 3.1 INTEGRATION AS THE REVERSE OF DIFFERENTIATION Integration is the reverse process of differentiation. The process of obtaining dx dy from y (a function of a) is known as differentiation. Hence, the process of obtaining y from dx dy is known as integration. Integration of y with respect to x, is denoted by f(x) dx . The symbols )x(f dx denote the integral of f(x) with respect to the variable x. For example: 3.2 INDEFINITE INTEGRAL From which the derivative 5x4 was derived?. Take a look at these examples: x51x x5 dx 1x dx d and x53x x5 dx 3x dx d also x5x x5 dx x dx d 5 4 4 5 5 4 4 5 5 4 4 5 Any constant terms in the original expression becomes zero in the derivative. Therefore the presence of such constant term is replaced by adding a symbol c to the result of integration. x'f dx c)x(f c is known as the constant of integration and must always be included. Such integral is called an indefinite integral. c 3 x y 3 Differentiation Integration 2 x dx dy
INTEGRATION IKMJB 80 DUM 20022/20032 3.2.1 INTEGRATION OF POLYNOMIALS 1. a axdx awherec is constanta 2. c (n 1) (n 1) x x dx 1n n 3. c (n 1) (n 1) xa dxax 1n n c (ax b) dx d (n 1) (ax b) 4. (ax dxb) 1n n ; (n 1) 5. g(x)f(x) dx f(x) dx g(x) dx Example 3.1 : Integrating a constant Integrate the following with respect to :x 3 1 (c)5(b)2(a) Solution : cx 3 1 dx 3 1 (c) (b) 5 5xdx c 2(a) 2xdx c Example 3.2 : Integration of xn x 1 (c) x 1 (b)x(a) Integrate the following with respect to :x 3 5 Solution : c 2x 1 cx 2 1 dx x dx x 1 (b) cx 6 1 cx 15 1 x(a) dx 2 2 3 5 15 6 3
INTEGRATION IKMJB 81 DUM 20022/20032 cx 2x cx2c 1 2 1 1 dx x dx x 1 (c) 2 1 1 2 1 2 1 Example 3.3 : Integration of axn Find the following integrals: (a) -4x2 (b) 2x3 (c) 3x5 Solution : cxc x xc dx x dx cxc x xb dx x dx cxc x a x dx x dx 6 6 5 5 4 4 3 3 3 3 2 2 2 1 6 33 3 2 1 4 22 2 3 4 3 4 4 4 )( )( )( Example 3.4 : Integral of (ax + b)n Find: (a) 2x 3 dx 3(b) dx6x x1(c) 2 dx 1 4 4 Solution : cx12cx1 1 2 1 1 ( c) x1 dx 3 6x c 18 1 3 6x c 63 1 3(b) dx6x 2x c3 10 1 2x c3 25 1 (a) 2x 3 dx 2 1 2 1 2 1 4 3 3 4 5 5
INTEGRATION IKMJB 82 DUM 20022/20032 Example 3.5 : Integral of sum and differences Find the following integrals: dx x 1x dx (b) 2x 1 dx (c x 1 3x(a) 3 2 3 2 ) Solution : c 2x 1 x 1 c 2 x 1 x dx xx dx x 1x (c) x 2x cx 3 4 (b) 2x 1 4x4xdx 1 dx cx 2 1 x 3x 3 1 dx x dx 13 dx x dx x 1 (a) 3x 2 21 32 3 2 2 23 3 2 2 3 3 2
INTEGRATION BKT DUM 20132 4.2.2 INTEGRATION OF x 1 ➢ dx = ln x + c x 1 ➢ + + + = + c (ax b) dx d ln ax b dx ax b 1 Example 4.6 : + − dx 5 2x 3 dx (c) 2x 3 1 dx (b) x 5 (a) Find Solution : ln 5 2x c 2 3 dx 5 2x 3 (c) ln 2x 3 c 2 1 dx 2x 3 1 (b) dx 5ln x c x 5 (a) = − − + − = + + + = + (u) dx d u 1 ln u dx d = •
INTEGRATION BKT DUM 20132 Object Image 4.3 TECHNIQUES OF INTEGRATION To integrate functions which are not in standard form, we need to use a few other techniques to make the process of integration easier. There are three techniques: 4.3.1 Integration by Substitution 4.3.2 Integration by Part 4.3.3 Integration by Partial Fraction 4.3.1 INTEGRATION BY SUBSTITUTION f(x) dx = g(u)f g'(u) du The substitution technique is simplified as follows: a. Choose u = g(x) b. Find )x('g dx du = c. Substitute u = g(x), du = g ‘(x) dx. In this part, the integrand must be in terms of u only, meaning there is no x terms left. But if this happen, make another substitution for u. d. Solve the integration e. Back substitute u with g(x) therefore the final answer is in terms of x.
INTEGRATION BKT DUM 20132 Example 4.9 : By using a suitable substitution, find the following integration: ( ) ( ) (a) 5x − 4 dx (b) 3x −1x dx 6 32 (c) − 3 (1 x) dx (d) x9 (1 )x dx 2 1 3 2 − Solution : ( ) ( ) ( ) c 35 5x 4 c 7 u 5 1 du 5 1 5x 4 dx u du 5 1 and5 dx dx du uLet 5x 4 (a) 5x 4 dx 7 7 6 6 6 + − = + = =− −= = = − ( ) c 2 (x 1) c 2 u 3x (x dx1) u du uLet (x dx3xdu1) (b) 3x 1x dx 23 2 332 3 2 32 + − = += =− =−= −
INTEGRATION BKT DUM 20132 (c) − 3 (1 x) dx c 2(1 x) 1 c 2 (1 x) c 2 u u ( du) (1 x) dx (1 x) dx du dx Let u (1 x) du dx 2 2 2 3 3 3 + − = + − = + − = − = − = − − − = = − = − − − − − (d) x9 (1 )x dx 2 1 3 2 − 2(1 x ) c c 3 2 u 3 3 u du 9x (1 x ) dx u ( 3du) 3du 9x dx Let u 1 x du 3x dx 2 3 3 2 3 2 1 2 1 2 3 2 3 2 = − − + + = − = − − = − − = = − = −
INTEGRATION BKT DUM 20132 4.3.2 INTEGRATION BY PART This is a method of integrating the product of two functions. The product were either function not the derivative of the other, for example: x lnx dx, xe dx 2 x , xsine dx x etc. Consider the product rule for differentiation: ( ) , dx du dx v dx dv uv u dx d += u and v are functions of x. Integrate both sides with respect to x, dx, dx du dx v dx dv uv u = + Rearranging the term, dx, dx du dx uv v dx dv u −= This is known as integration by part udv −= vvu du The procedures: a. Split the function into two simpler functions, one called u and the other dx dv . b. To decide which one will be u and dv. (i) u should be a function which becomes a simpler function after differentiation. There is a simple acronym to remember which function to equate u: L – P – E – T 1. L = Logarithm 2. P = Polynomial 3. E = Exponential 4. T = Trigonometry WHEN & HOW TO USE * If substitution doesn’t work. * f(x).g(x) dx match with udv . * choose u & dv and find du v&
INTEGRATION BKT DUM 20132 (ii) dx dv must be a function that is possible to integrate to obtain v. Example 4.10 : Find the following integrals using the integration by part techniques. xcosx(a) dx x(b) ln x dx ex(c) dx x2 Solution : ( ) ( ) c 4 x ln x 2 x cx 2 1 2 1 ln x 2 x dx x 1 2 x ln x 2 x x ln x dx Substituti ng uinto uvdv v du gives theomit constantnintegratio 2 x xv, dx x 1 dx du uLet(b) ln x dv xdx cxcosxsinx cxcosxsinx xcosx dx 1xsinxsinx dx uintogubstitutinS uvdv v du gives xcosv,1 dx sin theomit constantnintegratio dx du xuLet(a) dv xcos dx 2 2 2 2 2 2 2 +−= + −= −= −= = == = = ++= +−−= −= −= = = = = = x
INTEGRATION BKT DUM 20132 ( ) ( ) e (x 2x 2) c x e 2xe 2e c x e 2xe e 2 dx 2 and v e dx e dx du u 2x and dv e x e 2x e dx again by part x e dx x e e 2x dx Substituti ng into u dv uv v du gives 2x v e dx e omit the integratio n constant dx du (c) Let u x dv e dx x 2 2 x x x 2 x x x x x x 2 x x 2 x 2 x x x x 2 x = − + + = − − + + = − − = = = = = = − = − = − = = = = =
INTEGRATION BKT DUM 20132 4.3.3 INTEGRATION BY PARTIAL FRACTION In general, the integration in the form of ( ) ( ) dx xg xf where f(x) and g(x) are polynomials in terms of x, we have to express ( ) ( ) xg xf as a partial fraction before integration is attempted. Only proper fractions can be converted directly into partial fraction. ➢ nonrepeated linear factor ( ) ax + b (x c) C (x b) B (x a) A (x b)(xa)(x c) f(x) + + + + + = +++ Example 4.11 : Find: −− dx (x 2)(x 3) 1 (a) ( ) ( )( ) +− + dx 2x 2x3 7x b Solution : Substitute B1,3x 1B Substitute 1AA1,2x So A(x1 3) B(x 2) 3x B 2x A (x 2)(x 3) 1 Let(a). = = = = −=−= −+−= − + − = −− dx g(x) f(x) STEPS * determine the shape of the partial fractions ( numbers of factor xg )( ) * find value A, B & C * evaluate the integral
INTEGRATION BKT DUM 20132 ln x 2 ln x 3) c dx x 3 1 x 2 1 (x 2)(x 3) 1 Hence x 3 1 x 2 1 (x 2)(x 3) 1 Thus = − + − + − + − − = − − − − − − = − − ( )( ) ( ) ( ) ( )( ) ( )( ) ( ) ln (x 2) c 14 17 ln 2x 3 ln x 2 c 2 ln 2x 3 7 17 ) dx x 2 1 2x 3 7 17 dx ( 2x 3 x 2 x 7 Hence x 2 1 2x 3 7 17 2x 3 x 2 x 7 Thus 7 17 A A 2 7 2 17 2 3 x Substitute x 2 5 5B B 1 x 7 A x 2 B 2x 3 x 2 B 2x 3 A dx 2x 3 x 2 x 7 (b) Let − + + − = − + + − = + − − = − + + + − − − + + = = = = − = − = − + + + − + + − − + +
INTEGRATION BKT DUM 20132 4.4 THE DEFINITE INTEGRAL An integral with limits is called a definite integral ( ) xf dx ( ) ( ) aFbF)x(F b a b a −== b is known as the upper limit and a is known as the lower limit of the integral. In the definite integral there is no constant of integration. Example 4.12 : Evaluate. ( ) dx x 1 dx (e) x x 1 (d) x (a) dx6x 1(b) 2x x dx (c) x dx 1 2 3 2 2 2 2 25 1 2 1 2 2 0 2 − − − − − ++ Solution : ( ) ( ) ( ) ( ) 3 2 82 15 3 2 25 1 3 2 x 3 2 (c) x dx x dx 9 3 1 11 3 8 42 3 x 1(b) 2x x dx xx 82022x2 16 3 6x (a) dx6x 3 2 3 25 1 2 25 3 1 2 25 1 1 2 1 3 2 2 1 2 3 2 0 3 2 0 2 3 0 2 = −= = − = = = −+−− ++= ++=++ ==−== = − −
INTEGRATION BKT DUM 20132 ( ) ( ) ( ) 6 5 6 5 1 3 2 2 2 1 4 3 8 2 1 3 1 x 1 2x 3 x 1 x 2x 3 x dx x 2 x dx x 1 (e) x 6 1 6 2 1 3 8 3 1 27 3 1 x 1 3 x 1 x 3 x dx x x dx x 1 (d) x 1 2 3 1 2 1 3 1 2 2 2 1 2 2 3 2 3 3 2 3 3 1 2 2 2 3 2 2 2 = = − − − + + = − + + = − − − = − + = − + − = − + = + = + − = − = − − − − − − − − − − − − − −