Calculus Integration By Norhafizah Md Sarif & Norazaliza Mohd Jamil Faculty of Instrial Science & Technology norhafizah@ump.e.my, norazaliza@ump.e.my
Description Aims This chapter is aimed to : 1. introce the concept of integration. eplain the basic properties of integral 3. compute the integral using different techniques of integration Epected Outcomes 1. Students should be able to eplain about indefinite integral and definite integral. Students should be able to know the basic properties of definite integrals 3. Student should be able to determine the appropriate techniques to solve difficult integral. References 1. Abl Wahid Md Raji, Hamisan Rahmat, Ismail Kamis, Mohd Nor Mohamad, Ong Chee Tiong. The First Course of Calculus for Science & Engineering Students, Second Edition, UTM 016.
Content 1 3 Integration by Substitution Integration by Parts Integration using Partial Fractions
Techniques of Integration 1 Integration by Substitution Integration by Partial Fraction 3 Integration by Parts
Integration by Substitution The idea of integration by substitution is to transform a difficult integral to an simpler integral by using a substitution. Theorem Integration by substitution. Let f, g and u be differentiable functions of such that f ( ) g( u) d Then f ( ) d g( u) d g( u) G( u) c d
Integration by Substitution Working Steps 01 Choose a new variable u. Let say u f ( ) 0 Differentiate u with respect to. f( ) d 03 Determine the value d. d f ( )
04 Make the substitution u f ( ) and d f ( ) 05 Evaluate the resulting integral 06 Return to the initial variable.
Eample Evaluate 5 3 7 Step 1 : Choose a substitution function u d u 53 Step : Differentiate u with respect to Step 3: From step, d Step 4: Substitute u and d into integral Step 5 : Evaluate the resulting integral Step 6 : Return to variable Therefore, the solution is 7 1 5 3 d 5 3 8 c 40 d 5 d 5 7 7 5 3 d u 1 1 5 40 7 8 u u c 1 8 1 5 3 8 u c c 40 40 5
Eample Evaluate 0 cos sin Step 1 : Choose a substitution function u d u sin Step : Differentiate u with respect to Step 3: From step, d Step 4: Substitute u and d into integral Step 5 : Evaluate the resulting integral Step 6 : Compute values of u d cos 1 3 u 1 3 3 0 cos d 1 cos sin d u cos 0 0 1 3 u u 3 1 0 0 cos Therefore, the solution is 0 cos sin d 1 3
Eample Evaluate 30 3 1 d Let u 3 1, then 6 which implies. The given d d 6 integral can be written as 30 30 3 1 d u d 1 30 u 3 31 1 u 3 31 1 3 1 c 31 c 93
Eample Evaluate 5 1 0 3 d Let u 51, then 5 which implies. The given integral d d 5 can be written as Method I : Method II : When =0, u=-1 and when =, u=9. 9 3 3 5 1 d u 0 1 4 1 u 54 5 9 1 1 4 4 9 ( 1) 0 1 6561 1 38 0 0 3 3 5 1 d u 4 1 u 54 5 c 1 5 1 0 1 6561 1 0 38 4 0
Integration by Parts Integration by parts is a technique to solve an integration in the form of proct of two functions such as: [ sin(5)] d f ( ) g( ) sin(5) The main interest in integration by parts is to transform an integral into a new integral that is easier to solve than the original.
Don t try to understand this yet. Wait for the eamples that follow u For Indefinite integrals: dv d d uv v d Definite integrals: d b a u dv d d b b uv v d a a d For convenience, this can be memorized as: u dv uv v
Integration by Parts Guideline of Selecting U Choose u by the following sequence: 1 3 4 Logarithmic (log, ln) Algebraic (, 3, ) Trigonometry (sin, cos, tan, ) Eponential (e, e 4, ) and the net function automatically becomes dv. If the new integral is more difficult than the original, change the choice of u and dv.
Eample Evaluate cos d cos d u dv Why choose as u instead of cos? is algebraic function. Meanwhile, cos is a trigonometric function. Hence, algebraic function comes first before trigonometric functions. So is chosen as u.
Differentiate u u d, dv cos d 1, v sin (omit c) Integrate dv Plug everything into the formula udv uv v sin sin d Integrating sin d cos c. Therefore, cos d sin sin d sin ( cos ) c sin cos c
Eample Evaluate 3 ln d ln 3 d u dv Why ln is a u? In guideline of choosing u, we refer LATE in which Logarithmic (L) function comes first in the list. Hence, ln is chosen as u
Differentiate u u ln, d dv 3 d 1, 3 v (omit c) Integrate dv Plug everything into the formula udv uv v ln d 3 3 1 Integrating 1 3 3 d c, we obtained, 3 ln ln 3 d d 3 sin c 3
Eample Evaluate e d u, dv e d d, ve (omit c) By using integration by parts udv uv v e e d e e c
Sometimes, integration by parts must be applied several times to evaluate a given integral. See eample below. Eample Evaluate e cos d u cos, dv e d d sin, v e By using integration by parts udv uv v e cos e sin d e cos e sin d c 1
Another integration by parts applied to the last integral, i.e. e sin d will complete the solution. Hence, by doing by parts once again, we obtain u sin, dv e d d Substitute into the formula cos, v e e sin dv uv v e sin e cos d Substitute result in () into Equation (1)
e cos d e cos e sin d e cos e sin e cos d Notice that the last term is similar to the original problem. Hence, by moving the last term into the left hand side equation, we get e cos d e cos d e cos e sin c e cos d e cos e sin c We want to find e cos d, therefore e cos e sin e cos d c
Integration by Partial Fractions If the integrand is in the form of an algebraic fraction and the integral cannot be evaluated by simple methods, the fraction needs to be epressed in partial fraction. Definition Proper Fraction Any rational function of, P ( ) Q ( ) where the P ( ) is less than the degree of Q ( ) could be epressed as sum of relatively simpler rational functions, called partial fractions.
1. Linear Factor Q( ) ( a b )( a b ) ( a b ) 1 1 n n Partial fraction : A1 A An ( a b ) ( a b ) ( a b ) 1 1 n n. Repeated Linear Factor Q( ) ( a b) n Partial fraction : A A 1 n ( a b) ( a b) n ( a b) A 3. Quadratic Factor Q( ) a b c Partial fraction : A B ( a b c)
Eample Evaluate 6 3 ( 1) d Step 1 Factor the denominator 4 3 ( 3)( 1) Step Break up the fraction into sum of partial fractions 6 A B ( 3)( 1) 3 1 Step 3 Multiply both sides of the equation by the left side denominator 6 A( 1) B( 3)
Step 4 Take the roots of the linear factors and plug them, one at a time, into on the equation from step 3, and solve If 1, B 3 3, A 3 Split up the original integral and integrate Step 5 6 d 3 3 d 3 ( 1) 3 1 3ln 3 3ln 1 c
Eample 7 6 3 Evaluate 3 d The function can be written as 76 76 3 ( 3) 3 The denominator is a combination of linear and repeated linear case, therefore we have 7 6 A B C 3 3 3 We have three unknown A, B and C. Multiply both sides by the left side of denominator,
7 6 A B 3 C A 3A B 3B C A C ( 3 A B) 3B Equating the coefficient of the polynomial, where A C 0 : 0 : 3A B 7 : 3B6 Solving the simultaneous equation, we obtain A 3, B, C 3. Alternatively, A, B and C can be found using the another approach. If 3, 9C 7 C 3 0, 6 3B B 1, 13 ( A ) 3 A 3
Hence, 7 6 3 3 3 3 3 The integration becomes 7 6 3 3 d 3 d 3 3 3 d d d 3 3 Simplify whenever necessary and then integrate 7 6 3 3 3 3 3ln 3ln 3 c d d d d 3
Conclusion In integration by substitution, making appropriate choices for u will come with eperience. Selecting u for by part techniques should follow the LATE guideline. If the power of denominator is less than the power of numerator, then the fraction is called proper fraction
Author Information Norhafizah Binti Md Sarif Email: norhafizah@ump.e.my Google Scholar: Norhafizah Md Sarif Scopus ID : 571905369 UmpIR ID: 3479 Norazaliza Binti Mohd Jamil Email: norazaliza@ump.e.my Google Scholar: Norazaliza Mohd Jamil Scopus ID : 4061495500