Chapter 4 : Integration 177 CHAPTER 4 INTEGRATION List of topics : Indefinite & Definite Integrals (Definite integrals is used to find the volume of revolution, in chapter 5) Basic Rules of Integration : i) Constant Rule ii) Power Rule (for n −1) iii) Logarithmic Rule (for n = −1) iv) Sum and Difference Rules v) Exponential Rule vi) Trigonometric Rule Integration with u-substitution Powers of sine and cosine (even and odd power) Mean value theorem for Integrals Second Fundamental Theorem of Calculus
MAT183/ MAT421 : Calculus I 178 Basic Rules of Integration Differentiation Integration i) Constant Rule Example : If y = 2 dx dy = 0 In general, Constant Rule Example 1 : Indefinite integral : 2 dx = 2x + C definite integral : 2 2 2(1) 2(0) 2 1 0 1 0 = = − = dx x Example 2 : Indefinite integral : 5 dx =5x + C definite integral : 5(2) 5(1) 5 5 5 2 1 2 1 = = − = x dx In general, a dx = ax + C (a = constant) ii) Power Rule (for all values of n) Example : If y = x 3 2 3x dx dy = In general, If y = axn (a = constant) dx dy = na x n−1 Power Rule (for n −1) Example : Indefinite integral : C x x dx = + 4 4 3 definite integral : ( ) ( ) 4 15 4 1 4 2 4 4 4 2 1 2 4 1 3 = − = = x x dx In general, C n x ax dx a x dx a n n n + + = = + 1 1 = , (ℎ = ) = 0
Chapter 4 : Integration 179 Differentiation Integration iii) Logarithmic Rule (for n = −1) Wrong method : C x x dx = + − 0 0 1 = undefined Correct method : dx ln x C x x dx = = + −1 1 Why?? Because ( ) dx x d lnx 1 = iv) Sum and Difference Rules (differentiate one by one) Example : If y = x x x x 2 3 2 − + − y = 2 1 2 1 3 2 2 − x − x + x − x 2 3 2 1 2 1 2 3 − − = x − + x + x dx dy 2 3 2 1 2 3 − = − + + x x x dx dy Sum and Difference Rules (integrate one by one) Example : Indefinite integral : definite integral : ∫ ( 2 − 3 + √ − 2 √ ) = ∫ ( 2 − 3 + 1 2 − 2 − 1 2) = 3 3 − 3 2 2 + 3 2 ( 3 2 ) − 2 1 2 ( 1 2 ) + = 1 3 3 − 3 2 2 + 2 3 3 2 − 2 ( 2 1 ) 1 2 + = 1 3 3 − 3 2 2 + 2 3 3 2 − 4 √ + ∫ ( 2 − 3 + √ − 2 √ ) 1 0 = ∫ ( 2 − 3 + 1 2 − 2 − 1 2) 1 0 = [ 3 3 − 3 2 2 + 3 2 ( 3 2 ) − 2 1 2 ( 1 2 ) ] 0 1 = [ 1 3 3 − 3 2 2 + 2 3 3 2 − 2 ( 2 1 ) 1 2] 0 1 = [ 1 3 3 − 3 2 2 + 2 3 3 2 − 4 √] 0 1 = ( 1 3 − 3 2 + 2 3 − 4 ) − 0 = − 9 2
MAT183/ MAT421 : Calculus I 180 Differentiation Integration v) Exponential Rule a) ( ) x x e dx d e = b) ( ) x x e dx d e 2 2 = 2 c) ( ) x x e dx d e 3 3 = 3 In general, ( ) ax ax ae dx d e = Exponential Rule a) e dx e C x x = + b) C e e dx x x = + 2 2 2 c) C e e dx x x = + 3 3 3 In general, C a e e dx ax ax = + vi) Trigonometric Rule • ( ) cos x dx d sin x = • ( ) sin x dx d cos x = − • ( ) sec x dx d tan x 2 = • ( ) sec x tan x dx d sec x = • ( ) cosecx cot x dx d cosecx = − • ( ) cosec x dx d cot x 2 = − Trigonometric Rule Note : never use product, quotient, generalized and chain rule • ∫ = + → ∫ = 1 + • ∫ = − + → ∫ = − 1 + • ∫ 2 = + → ∫ 2 = 1 + • ∫ = + → ∫ = 1 + • ∫ = − + → ∫ = − 1 + • ∫ 2 = − + → ∫ 2 = − 1 +
Chapter 4 : Integration 181 Examples for multiplication function (that possible to expand) a) ( x) dx x − + 2 1 2 3 2 1 7 = ( x x ) dx x − + + 2 1 2 9 12 4 1 7 (expand bracket square) = x dx x x x − + + − + 2 1 2 12 4 9 63 84 28 (expand) = dx x x x − + + 2 1 2 1 51 80 28 9 (simplify) = 2 1 2 3 9ln 3 28 2 80 51 − + + x x x x (integrate with respect to x) = − − + + − + + 9ln1 3 28 9ln 2 51 40 3 224 102 160 = 3 11 9ln 2 − = 2.5717 # b) (t )(t t ) dt − + − + 3 1 2 2 2 4 = (t t t t t ) dt − − + + − + 3 1 3 2 2 2 4 2 4 8 (expand) = (t ) dt − + 3 1 3 8 (simplify) = 3 1 4 8 4 − + t t (integrate with respect to t) = − − + 8 4 1 24 4 81 = 52 # Refer page 179 (chapter 4), dx ln x C x x dx = = + −1 1
MAT183/ MAT421 : Calculus I 182 c) t (t ) dt − 3 2 1 1 = t (t t )dt − 2 +1 3 2 1 (expand bracket square) = t t t dt − + 3 1 3 4 3 7 2 (expand and simplify using the properties of indices) = ( ) ( ) ( ) C t t t − • + + 3 4 3 7 3 10 3 4 3 7 3 10 2 (integrate with respect to t) = t − • t + t 3 + C 4 3 7 3 10 4 3 7 3 2 10 3 = t − t + t 3 + C 4 3 7 3 10 4 3 7 6 10 3 # Examples for division function with single denominator (that possible to separate) a) ( ) dx x x − 3 2 5 = dx x x x − + 3 10 25 2 1 (expand bracket square) = dx x x x x x − + 3 3 3 10 25 2 1 (separate) = x x x dx − + 1−3 −3 −3 10 25 2 1 (simplify using the properties of indices) = x x x dx − + −2 − −3 10 25 2 5 = ( ) C x x x + − + • − − • − − − − 2 10 25 1 2 2 3 1 2 3 (integrate with respect to x) = C x x x − − • − • − + − 2 2 25 3 2 10 1 2 3 = C x x x − + − + 2 2 25 3 1 20 2 3 #
Chapter 4 : Integration 183 b) ( ) dy y y + 2 2 2 1 = dy y y y + + 2 4 2 2 1 (expand bracket square) = dy y y y y y + + 2 2 2 2 4 2 1 (separate) = (y y ) dy − + + 2 2 2 (simplify using the properties of indices) = C y y y + − + + − 1 2 3 3 1 (integrate with respect to y) = C y y + y − + 1 2 3 1 3 # c) ( ) dx e e x x + 3 2 3 2 = dx e e e x x x + + 3 3 6 2 2 2 (expand bracket square) = dx e e e e e x x x x x + + 3 6 3 3 3 2 2 2 (separate) = ( e e ) dx x x + + −3 3 2 2 2 (simplify using the properties of indices) = C e x e x x + + + − • − 3 2 2 3 2 3 3 (integrate with respect to x) = e x e C x x − + + + −3 3 3 1 2 2 3 2 # ∫ = +
MAT183/ MAT421 : Calculus I 184 Integration with u-substitution ☺ Power Rule with u-substitution, for n ≠ −1 (u in bracket) From C n x ax dx a x dx a n n n + + = = + 1 1 ( ) ( ) C n u a u du a u du a n n n + + = = + 1 1 Example 1 : evaluate i) ( x) dx 5 2 − ii) ( x) dx − 2 1 5 2 Solution : i) ( x) dx 5 2 − = u • −du 5 = u du 5 − = C u − + 6 6 = − ( − x) + C 6 2 6 1 # ii) ( x) dx − 2 1 5 2 = ( x) dx x x = = − 2 1 5 2 = u du u u • − = = 0 1 5 = u du u u = = − 0 1 5 = 0 1 6 6 − u = − − 6 1 6 0 = 6 1 # Let u = 2 − x = −1 dx = −du Let u = 2 − x = −1 dx = −du from u = 2 − x, when x = 2, u = 0 when x = 1, u = 1 For definite integral with u-substitution
Chapter 4 : Integration 185 Example 2 : evaluate i) + dx x x 3 1 2 ii) + 1 0 2 3 1 dx x x Solution : i) + dx x x 3 1 2 = ( ) − x x + dx 2 1 3 1 2 = • • − x du x u 6 2 1 = • − 6 2 1 du u = − u du 2 1 6 1 = ( ) c u + 2 1 2 1 6 1 = u + c • 2 1 1 2 6 1 = u + c 2 1 3 1 = 3x +1 + c 3 1 2 # ii) + 1 0 2 3 1 dx x x = ( ) − + 1 0 2 2 1 x 3x 1 dx = = = − • • 1 0 6 2 1 x x x du x u = = = − • 4 1 6 2 1 u u du u = = = − 4 1 2 1 6 1 u u u du = ( ) 4 1 2 1 2 1 6 1 u = 4 1 2 1 1 2 6 1 • u = 4 1 3 1 u = 4 1 3 1 − = 2 1 3 1 − = 3 1 # Let u = 3x 2 +1 = 6x dx = from u = 3x 2 +1, when x = 1, u = 4 when x = 0, u = 1 Let u = 3x 2 +1 = 6x dx = For definite integral with u-substitution
MAT183/ MAT421 : Calculus I 186 Example 3 : evaluate i) x x +1 dx ii) − + 0 1 x x 1 dx Solution : i) x x +1 dx = ( ) x x + 2 dx 1 1 = ( ) u − •u du 2 1 1 = u − u du 2 1 2 3 = ( ) ( ) c u u − + 2 3 2 5 2 3 2 5 = u − u + c 2 3 2 5 3 2 5 2 = (x + ) − (x + ) 2 + c 3 2 5 1 3 2 1 5 2 # ii) − + 0 1 x x 1 dx = ( ) = =− + 0 1 2 1 1 x x x x dx = ( ) = = − • 1 0 2 1 1 u u u u du = − 1 0 2 1 2 3 u u du = ( ) ( ) 1 0 2 3 2 5 2 3 2 5 − u u = 1 0 2 3 2 5 3 2 5 2 u − u = 0 3 2 5 2 − − = 15 4 − # Let u = x +1 = 1 dx = dx from u = x +1 x = u − 1 when x = 0, u = 1 when x = −1, u = 0 Let u = x +1 = 1 dx = dx For definite integral with u-substitution
Chapter 4 : Integration 187 ☺ Logarithmic Rule with u-substitution, for n = −1 (u = denominator) From dx ln x C x = + 1 du ln u C u = + 1 Example 4 Evaluate the following integrals. a) dx x x − 2 3 4 = x du u x 2 4 − • = du u − 1 2 = − 2lnu + C = − − x + C 2 2ln 3 # b) x dx cot = dx x x sin cos = x du u x cos cos • = du u 1 = lnu + C = ln sin x + C # .c) dx x x tan sec 2 = x du u x 2 2 sec sec • = du u 1 = lnu + C = ln tan x + C #
MAT183/ MAT421 : Calculus I 188 d) dx x x 1+ 2cos sin , u = 1 + 2cosx dx x x 1+ 2cos sin = x du u x 2sin sin − • = du u − 1 2 1 = − lnu + C 2 1 = − ln1+ 2cosx + C 2 1 # e) dt e e t t + ln 2 0 2 2 2 , u = e 2t + 2 dt e e t t + ln 2 0 2 2 2 = t t e du u e 2 ln 2 0 2 2 • = du u 6 3 1 2 1 = 6 3 ln 2 1 u = ln6 ln3 2 1 − = 3 6 ln 2 1 = ln2 2 1 = 2 1 ln2 = ln 2 # When , = 1 + 2 = 3 When , = 4 + 2 = 6
Chapter 4 : Integration 189 ☺ Exponential Rule with u-substitution (u = exponent/power) From e dx e C x x = + e du e C u u = + Example 5 Evaluate the integrals. a) dx e e x x − − 4 − = dx e e e x x x − − − − 4 (single denominator : so, we separate the fraction) = ( e )dx x 4 − 1 = e x C x 4 − + # b) x e dx x 4 3 = 3 3 4x du x e u • • = e du u 4 1 = e C u + 4 1 = e C x + 4 4 1 # c) dx e e e x x x + 2 2 = dx e e e e x x x x + 2 2 (single denominator : so, we separate the fraction) = ( e )dx x x + − 2 1 2 = ( e )dx x 2 + 1 = e x C x 2 + + #
MAT183/ MAT421 : Calculus I 190 ☺ Trigonometric Rule with u-substitution (u = angle, if the trigonometric function exist in table of integral) Properties of the Definite Integral From sin x dx = − cos x + c sinu du = − cosu + c From cos x dx = sin x + c cos u du = sinu + c From sec x dx = tan x + c 2 sec u du = tanu + c 2 From cos ec x dx = − cot x + c 2 cos ec u du = − cot u + c 2 From sec x tan x dx = sec x + c sec u tanu du = sec u + c From cos ec x cot x dx = − cos ec x + c cos ec u cot u du = − cos ec u + c Extra Formula Derivative Integration ( ) ax ax e ae dx d = c a e e dx ax ax = + ( ) a ax dx d ax cos sin = c a cos ax sinax dx + − = ( ) a ax dx d ax sin cos = − c a sinax cos ax dx = + ( ) a ax dx d ax 2 sec tan = c a tan ax sec ax dx = + 2 ( ) a ec ax dx d ax 2 cos cot = − c a cot ax cos ec ax dx + − = 2 ( ) a ax ax dx d ax sec tan sec = c a sec ax sec ax tan ax dx = + ( ) a ecx ax dx d ecax cos cot cos = − c a cos ec ax cos ec ax cot ax dx + − = Important : this formula only suitable for the angle with x the power of positive one
Chapter 4 : Integration 191 Properties of the Definite Integral Sometimes we add or subtract definite integrals, multiply their integrand by constant and compare them with other definite integrals. This can be done by the following rules. Properties of the Definite Integral Multiply integrand by constant cf x dx b a ( ) = c f x dx b a ( ) Definite integral at a point ( ) = 0 f x dx a a Opposite of the definite integral f x dx b a ( ) = − f x dx a b ( ) Subdivision rule f x dx b a ( ) = f x dx c a ( ) + f x dx b c ( ) where a c b Example Given ( ) 3 8 2 = f x dx . Find, a. f x dx 8 2 3 ( ) b. 2 8 f(x)dx c. f x dx − 2 8 ( ) 3 d. f x dx 3 2 ( ) + f x dx 8 3 ( ) e. If f x kdx + 8 2 ( ) = 20, find k.
MAT183/ MAT421 : Calculus I 192 Solution a. f x dx 8 2 3 ( ) = 3 8 2 f(x) dx = 3(3) = 9 b. 2 8 f(x)dx = ( ) 3 8 2 − = − f x dx c. f x dx − 2 8 ( ) 3 = 2 8 f(x)dx − 2 8 3dx = − 8 2 f(x)dx − 2 8 3dx = −3 − 2 8 3x = −3 − (6 − 24)= 15 d. f x dx 3 2 ( ) + f x dx 8 3 ( ) = 8 2 f(x)dx = 3 e. f x kdx + 8 2 ( ) = 20 f x dx 8 2 ( ) + kdx 8 2 = 20 3 + 8 2 k x = 20 3 + 8k − 2k = 20 6k = 17 k = 6 17 #
Chapter 4 : Integration 193 Examples (from previous semester papers) Example 1/ APR 2011/ MAT183/ Q4 a) Evaluate (x 2)(x 5x) dx 3 2 1 − + (3 marks) b) Solve the following integrals using suitable substitutions. i) + dx x x 1 3 2 (3 marks) ii) x( x ) dx + 4 sin cos 1 (3 marks) Solution : a) (x 2)(x 5x) dx 3 2 1 − + = (x x x x)dx + − − 2 1 4 2 3 5 2 10 = 2 1 5 3 4 2 2 10 4 2 3 5 5 + − − x x x x = 2 1 2 5 3 4 5 3 2 5 5 + − − x x x x = − + − − + − − 5 2 1 3 5 5 1 8 20 3 40 5 32 = 30 139 − # b. i) + dx x x 1 3 2 = • x du u x 2 3 = du u 1 2 3 = lnu + C 2 3 = ln x + 1 + C 2 3 2 # b. ii) x( x ) dx + 4 sin cos 1 = x du x u sin sin 4 − • • = u du − 4 = C u − + 5 5 = − ( x + ) + C 5 cos 1 5 1 #
MAT183/ MAT421 : Calculus I 194 Example 2/ SEP 2011/ MAT183/ Q4 a) Evaluate x (x x 7) dx 3 2 1 0 5 1 + − (4 marks) b) Solve the following integrals using suitable substitutions. i) + dx x x tan2 3 sec 2 2 ii) y(y ) dy − 5 4 (7 marks) Solution : a) x (x x 7) dx 3 2 1 0 5 1 + − = + − 1 0 5 1 5 11 5 16 x x 7x dx (expand) = ( ) ( ) ( ) 1 0 5 6 5 16 5 21 5 6 5 16 5 21 7 + − x x x = 1 0 5 6 5 16 5 21 7 6 5 16 5 21 5 x + x − • x = 0 6 35 16 5 21 5 − + − = 336 1775 − # b. i) + dx x x tan2 3 sec 2 2 = • x du u x 2sec 2 sec 2 2 2 = du u 1 2 1 = lnu + C 2 1 = ln tan2x + 3 + C 2 1 # b. ii) y(y ) dy − 5 4 = y u du • 5 = (u )u du + 5 4 (expand) = (u u )du + 6 5 4 = C u u + + 6 4 7 7 6 = ( ) ( ) C y y + − + − 6 4 4 7 4 7 6 − (y − ) + (y − ) + C 7 6 4 3 2 4 7 1 # y = u + 4
Chapter 4 : Integration 195 Example 3/ MAR 2012/ MAT183/ Q4 c) Evaluate ( ) x x + 3x dx 2 (3 marks) d) Solve the following integrals using suitable substitutions. i) ( ) + dx x x 4 3 5 (5 marks) ii) ( ) tan 2 d (4 marks) Solution : c) ( ) x x + 3x dx 2 = ( ) x x + 3x dx 2 2 1 (expand) = x + x dx 2 3 2 5 3 = ( ) ( ) C x x + + 2 5 2 7 2 5 2 7 3 = x + • x 2 + C 5 2 7 3 5 2 7 2 = x + x 2 + C 5 2 7 5 6 7 2 # d. i) ( ) + dx x x 4 3 5 = du u x 4 5 = ( ) − du u u 4 3 5 = − du u u 4 3 5 = − du u u u 4 4 3 5 = ( ) − − u − u du 3 4 5 3 = C u u + − − − − − 3 3 2 5 2 3 = C u u − + + 2 3 5 2 5 = ( ) ( ) C x x + + + + − 2 3 3 5 2 3 5 # x = u − 3
MAT183/ MAT421 : Calculus I 196 d. ii) ( ) tan 2 d = d sin2 cos2 = • 2cos2 cos2 du u = du u 1 2 1 = lnu + C 2 1 = ln sin2 + C 2 1 # Example 4/ JUN 2012/ MAT183/ Q4 a) Evaluate i) ( ) + dx x x 3 2 2 5 (4 marks) ii) ( ) y + y dy 10 5 3 2 (3 marks) Solution : a. i) ( ) + dx x x 3 2 2 5 = + + dx x x x 3 4 2 10 25 (expand numerator) = + + dx x x x x x 3 3 2 3 4 10 25 (separate) = + • + − x dx x x 3 25 1 10 (simplify) = C x x x + + + 3 2 25 10ln 2 # (integrate) a. ii) ( ) y + y dy 10 5 3 2 = • • 2 5 10 du y u = • • − • 2 2 3 5 10 du u u = ( ) u − u du 10 3 4 5 = ( ) u − u du 11 10 3 4 5 = C u u + − 11 3 4 12 5 12 11 = u − u + C 12 11 44 15 48 5 = ( + y ) − ( + y ) + C 12 11 3 2 44 15 3 2 48 5 # 2y = u − 3 2 − 3 = u y
Chapter 4 : Integration 197 Example 5/ OCT 2012/ MAT183/ Q4 a) Evaluate + − dx x x 5x 2 2 (3 marks) b) Solve the following integrals using suitable substitutions. i) ( ) 4 − sin cos d 3 (4 marks) ii) ( ) x x + dx 7 2 1 (5 marks) Solution : a) + − dx x x 5x 2 2 = + − dx x x x x x 2 1 2 1 5 2 2 (separate) = + − − x x dx 2 1 2 3 1 5 2 (simplify) = ( ) ( ) C x x x + − + 2 1 2 5 2 1 2 5 5 2 (integrate) = x + • x − • x 2 + C 1 2 5 2 1 2 5 5 2 = x + x − x 2 + C 1 2 5 2 4 # b. i) ( ) 4 − sin cos d 3 = ( ) − • • cos 4 cos 3 du u = ( ) − u du 3 4 = C u u − + 4 4 4 = − + C 4 sin 4 1 4sin # b. ii) ( ) x x + dx 7 2 1 = • 2 7 du x u = • − 2 2 1 7 du u u = ( ) u − u du 7 1 4 1 = ( ) u − u du 8 7 4 1 = C u u + − 4 9 8 1 9 8 = u − u + C 9 8 32 1 36 1 = ( x + ) − ( x + ) + C 9 8 2 1 32 1 2 1 36 1 # 2x = u − 1 2 −1 = u x
MAT183/ MAT421 : Calculus I 198 Example 6/ MAR 2013/ MAT183/ Q4 a) Evaluate dx x x x + 1 3 2 (3 marks) b) Solve the following integrals using suitable substitutions. i) ( ) x + x dx 3 2 4 sec 1 (3 ½ marks) ii) ( ) − dx x x 7 2 9 (4 ½ marks) Solution : a) dx x x x + 1 3 2 = (x ) dx + 3 2 1 (expand) = 3 2 2 2 + x x = 3 (2 2) 2 9 − + + = 2 7 # b. i) ( ) x + x dx 3 2 4 sec 1 = • 3 3 2 4 sec x du x u = u du 2 sec 4 1 = tanu + C 4 1 = # b. ii) ( ) − dx x x 7 2 9 = du u x 7 2 = ( ) + du u u 7 2 9 = + + du u u u 7 2 18 81 = + + du u u u u u 7 7 7 2 18 81 = ( ) − − − u + u + u du 2 7 1 7 7 18 81 = ( ) − − − u + u + u du 5 6 7 18 81 = C u u u + − + − + − − − − 6 81 5 18 4 4 5 6 = C u u u − − − + 4 5 6 2 27 5 18 4 1 = ( ) ( ) ( ) C x x x + − − − − − − 4 5 6 2 9 27 5 9 18 4 9 1 # ( + x ) + C 4 tan 1 4 1 x = u + 9
Chapter 4 : Integration 199 Example 7/ OCT 2013/ MAT183/ Q4 b) Evaluate ( ) − − dx x x x 3 2 5 2 3 (3 marks) c) Solve the following integrals using suitable substitutions. i) ( ) − dx x x 4 1 4 7 (5 marks) ii) ( ) + − dx x x 2 3 sin 3cos (4 marks) Solution : b) ( ) − − dx x x x 3 2 5 2 3 = − + − dx x x x x 3 2 25 20 4 3 (expand) = − + dx x x x 3 2 25 23 4 (simplify numerator) = − + dx x x x x x 3 2 3 3 25 23 4 (separate) = − + • − − dx x x x 1 25 23 4 3 2 (simplify) = x C x x + + − − − − − 4ln 1 23 2 25 2 1 (integrate) = x C x x − + + 4ln + 23 2 25 2 # c. i) ( ) − dx x x 4 1 4 7 = − • 4 7 4 du u x = − − x u du 4 4 7 = − − − u du u 4 4 1 4 7 = ( ) − − − u u du 4 1 16 7 = ( ) − − − u − u du 4 3 16 7 = C u u + − − − − − − 16 3 2 7 3 2 = C u u − + 3 2 32 7 48 7 = ( ) ( ) C x x + − − − 3 2 32 1 4 7 48 1 4 7 # u =1− 4x 4x = 1− u 4 1 u x − =
MAT183/ MAT421 : Calculus I 200 c. ii) ( ) + − dx x x 2 3 sin 3cos = • − x du u x cos 3cos 2 = − du u 2 1 3 = − − u du 2 3 = C u + − − − 1 3 1 = C u + 3 = C x + 3 + sin 3 Example 8/ MAR 2014/ MAT183/ Q4 a) Find i) ( x x) x dx 3 5 2 tan tan sec + (4 marks) ii) dx x x x − − 2 1 3 3 5 (4 marks) iii) x dx − 2 1 where |x| = − , 0 , 0 x x x x . (4 marks) b) evaluate x (x ) dx − 4 2 6 3 5 marks) Solution : a. i) ( x x) x dx 3 5 2 tan tan sec + = ( ) x du u u x 2 3 5 2 sec + sec • = (u u ) du + 3 5 = C u u + + 4 6 4 6 = C x x + + 6 tan 4 tan4 6 #
Chapter 4 : Integration 201 a. ii) = dx x x x x − − 2 1 3 3 5 (separate) = ( x ) dx − − 2 1 2 3 5 (simplify) = 5 1 3 5 3 3 − − x x (integrate) = 5 1 3 5 − x − x = (125 − 25) − (−1 + 5) = 96 # a. iii) − , 0 , 0 x x x x = x dx x dx − + − 2 0 0 1 = 2 0 2 0 1 2 2 2 + − − x x = 2 0 2 1 0 + − − − = 2 5 # b) x (x ) dx − 4 2 6 3 = x du x u 2 6 4 • • = u du 4 3 = C u + 5 3 5 = (x − ) + C 5 2 3 5 3 # dx x x x − − 2 1 3 3 5 where |x| =
MAT183/ MAT421 : Calculus I 202 Example 9/ SEP 2014/ MAT183/ Q4 a) Find i) − dx x x 2 3 cos 6cos 8 (4 marks) ii) (x 3)(x 2) dx 2 1 1 + − − (4 marks) iii) (e x )dx x − − − 2 1 3 1 2 (5 marks) b) Evaluate ( ) dx x x − 2 0 3 2 2 3 2 using the substitutions u = 2 − x 2 . (5 marks) Solution : a.i) − dx x x 2 3 cos 6cos 8 = − dx x x x 2 2 3 cos 8 cos 6cos (separate) = − • dx x x 2 cos 1 6cos 8 (simplify) = ( ) x − x dx 2 6cos 8sec = 6sin x − 8 tan x + C # (integrate) a.ii) (x 3)(x 2) dx 2 1 1 + − − = ( ) − − + − 1 1 3 2 x 2x 3x 6 dx (expand) = 1 1 4 2 3 6 3 3 2 2 4 − − + − x x x x (integrate) = 1 1 2 3 4 6 4 − − x + x − x x = − − − + − + − 1 1 6 4 1 1 1 6 4 1 = 10 #
Chapter 4 : Integration 203 a.iii) (e x )dx x − − − 2 1 3 1 2 = dx x e x − • − 2 1 3 1 2 = 2 1 3 2ln 3 − − − x e x (integrate) = 2 1 3 2ln 3 1 − − x e x = − − − − − 2ln1 3 1 2ln2 3 1 6 3 e e = 0 3 1 2ln2 3 1 6 3 − − + + e e = −1.3705 # b) ( ) dx x x − 2 0 3 2 2 3 2 = x( x ) dx − 2 0 3 2 2 2 3 = x du x u 2 2 3 2 0 3 − • • = u du − 2 0 3 4 3 = 2 0 4 4 4 3 = = − x x u (integrate) = 2 0 4 16 3 = − = x u x = ( ) 2 0 4 2 2 16 3 = = − − x x x = ( ) ( ) 4 4 2 4 2 0 16 3 − − − − = 16 16 16 3 − − = (0) 16 3 − = 0 #
MAT183/ MAT421 : Calculus I 204 Example 10/ MAR 2015/ MAT183/ Q4 a) Evaluate each of the following integrals using appropriate substitution. i x( x ) dx 3 1 0 2 + 1 (6 marks) ii) d 2 sin cos (4 marks) b) Given that ( ) 4 2 0 = f x dx and ( ) 3 2 0 = g x dx , find the value of k such that ( ) 2 ( ) 10 0 2 2 0 + + = f x dx g x kx dx (4 marks) Solution : a.i) x( x ) dx 3 1 0 2 +1 = 2 3 1 0 du x •u • = x u du 3 1 0 2 1 • = u du u 3 1 0 2 1 2 1 − = (u )u du 3 1 0 1 4 1 − = (u u )du − 1 0 4 3 4 1 = 1 0 5 4 4 5 4 1 = = − x x u u (integrate) = ( ) ( ) 1 0 5 4 4 2 1 5 2 1 4 1 = = + − + x x x x (replace u in terms of x) = − − − 4 1 5 1 4 3 5 3 4 1 5 4 5 4 (substitute the values of upper limit & lower limit) = − − + 4 1 5 1 4 81 5 243 4 1 = 5 142 4 1 = 10 71 # 2x = u − 1 2 −1 = u x
Chapter 4 : Integration 205 a.ii) d 2 sin cos = • cos cos 2 du u = du u 2 1 = − u du 2 = C u + − − 1 1 = C u − + 1 = − + C sin 1 = − cosec + C # b) ( ) 2 ( ) 10 2 0 2 0 − + = f x dx g x kx dx ( ) 2 ( ) 10 2 0 2 0 2 0 − − = f x dx g x dx kx dx ( ) 2 ( ) 10 2 0 2 0 2 0 − − = f x dx g x dx kx dx ( ) 10 2 4 2 3 2 0 2 = − − kx 4 − 6 − (2k − 0) = 10 − 2 − 2k = 10 − 2 − 10 = 2k −12 = 2k k = − 6 # given
MAT183/ MAT421 : Calculus I 206 Example 11/ SEP 2015/ MAT183/ Q4 a) Evaluate each of the following integrals using appropriate substitution. i) ( ) dx x x + 1 0 3 1 (5 marks) ii) x e dx sin2x cos2 (4 marks) b) Find ( ) + dx x x 2 4 2 . (4 marks) Solution : a.i) ( ) dx x x + 1 0 3 1 = du u x 1 0 3 = du u u − 1 0 3 1 = du u u u − 1 0 3 3 1 = (u u ) du − − − 1 0 2 3 = 1 0 1 2 1 2 = = − − − − − x x u u = 1 0 2 2 1 1 = = − + x u x u = ( ) ( ) 1 0 2 2 1 1 1 1 = = + + + − x x x x = − − + − + 2 1 1 8 1 2 1 = 8 1 # = 1
Chapter 4 : Integration 207 a.ii) x e dx sin2x cos2 = • • x du x e u 2cos2 cos2 = e du u 2 1 = e C u + 2 1 (integrate) = e C x + sin2 2 1 # (replace u in terms of x) b) ( ) + dx x x 2 4 2 = + + dx x x x 2 16 8 (expand numerator) = + + dx x x x x x 2 2 8 2 16 (separate) = + + − x dx x 2 1 4 8 1 2 1 (simplify) = • + + − x dx x 2 1 4 1 8 2 1 = ( ) x C x x + + + 2 4 1 8ln 2 1 2 1 (integrate) = C x x + x + + 2 8ln 8 #
MAT183/ MAT421 : Calculus I 208 Powers of sine and cosine (even and odd power) Example 1 Evaluate a) sin x dx b) x dx 2 sin (even power) c) x dx 3 sin (odd power) Solution : a) sin x dx = −cos x + c # (from table of integrals – trigonometric rule) b) x dx 2 sin = (1 cos2x) dx 2 1 − (even power) = ( x) dx 1− cos2 2 1 = c x x + − 2 sin2 2 1 = c x x − + 4 sin2 2 # c) x dx 3 sin = x x dx sin • sin 2 (odd power) = ( x) x dx 1− cos • sin 2 = ( ) x du u x sin 1 sin 2 − − • • = ( − u ) • (− du) 2 1 = (u ) du −1 2 = u c u − + 3 3 = cos x − cosx + c 3 1 3 # Even power use double-angle identity cos2x = 1 − 2 sin2x 2sin2x = 1 − cos2x sin2x = 1 2 (1 − cos2) and use ∫ = + (never use u-substitution) Odd power use Phythagorean identity cos2x + sin2x = 1 sin2x = 1 − cos2x and use u-substitution, u = cosx dx =
Chapter 4 : Integration 209 Example2 Evaluate a) cosx dx b) x dx 2 cos (even power) c) x dx 3 cos (odd power) Solution : a) cosx dx = sin x + c # (from table of integrals – trigonometric rule) b) x dx 2 cos = (1 cos2x) dx 2 1 + (even power) = ( x) dx 1+ cos2 2 1 = c x x + + 2 sin2 2 1 = c x x + + 4 sin2 2 # c) x dx 3 cos = x x dx cos • cos 2 (odd power) = ( x) x dx 1− sin • cos 2 = ( ) x du u x cos 1 cos 2 − • • = ( − u ) • du 2 1 = c u u − + 3 3 = x − x + c 3 sin 3 1 sin # Even power use double-angle identity cos2x = 2cos2x − 1 2cos2x = 1 + cos2x cos2x = 1 2 (1 + cos2) and use ∫ = + (never use u-substitution) Odd power use Phythagorean identity cos2x + sin2x = 1 cos2x = 1 − sin2x and use u-substitution, u = sin x dx =
MAT183/ MAT421 : Calculus I 210 Mean Value Theorem for Integrals Example 1/ JUN 2012/ MAT421/ Q4b/ 4 marks The Mean Value Theorem for Integrals states that if f(x) is continuous on [a, b], then there exist a number c in [a, b], such that f(x) dx f(c)(b a) b a = − If f(x) = 2x 2 − 3x is continuous on the interval [1, 7], determine the value of c. Solution : The Mean Value Theorem for Integrals states that if f(x) is continuous on [a, b], then there exist a number c in [a, b], such that Step 1 : identify the values of a and b = 1, = 7 Step 2 : from f(x), find f(c) () = 2 2 − 3 () = 2 2 − 3 Step 3 : find ∫ () ∫ () = ∫(2 2 − 3) 7 1 = [ 2 3 3 − 3 2 2 ] 1 7 = ( 2() 3 3 − 3() 2 2 ) − ( 2() 3 3 − 3() 2 2 ) = 156 Step 4 : use ∫ () = () ( − ) to find the value of c (note : the value of c should be between a and b) ∫ () = ()( − ) 156 = (2 2 − 3)(7 − 1) 156 = (2 2 − 3)(6) 156 6 = (2 2 − 3) 2 2 − 3 = 26 2 2 − 3 − 26 = 0 Solving quadratic equation (using calculator) = 4.43 = −2.93 = 4.43 [1, 7]
Chapter 4 : Integration 211 Example 2/ APR 2010/ MAT183/ Q4b/ 4 marks The Mean Value Theorem for Integrals states that if f(x) is continuous on [a, b], then there exist a number c in [a, b], such that f(x) dx f(c)(b a) b a = − Find the number of c, if ( ) 2 1 x f x = is continuous on the interval [1, 5]. Solution : Step 1 : identify the values of a and b = 1, = 5 Step 2 : from f(x), find f(c) () = 1 2 () = 1 2 Step 3 : find ∫ () ∫ () = ∫ ( 1 2 ) 5 1 = ∫ −2 5 1 = [ −1 −1 ] 1 5 = [− 1 ] 1 5 = (− 1 ) − (− 1 ) = − 1 5 + 1 = 4 5 Step 4 : use ∫ () = () ( − ) to find the value of c (note : the value of c should be between a and b) ∫ () = ()( − ) 4 5 = ( 1 2 ) (5 − 1) 4 5 = 4 2 2 = 5 = ±√5 = √5 = 2.236 [1,5]
MAT183/ MAT421 : Calculus I 212 Example 3/ NOV 2005/ MAT183/ Q4b/ 6 marks The Mean Value Theorem for Integrals states that if f(x) is continuous on [a, b], then there exist a number c in [a, b], such that f(x) dx f(c)(b a) b a = − Find the number of c, if ( ) 3 2 2 f x = x + x + is continuous on the interval [1, 4]. Solution : Step 1 : identify the values of a and b = 1, = 4 Step 2 : from f(x), find f(c) () = 2 + 3 + 2 () = 2 + 3 + 2 Step 3 : find ∫ () ∫ () = ∫( 2 + 3 + 2) 4 1 = [ 3 3 + 3 2 2 + 2] 1 4 = ( () 3 3 + 3() 2 2 + 2()) − ( () 3 3 + 3() 2 2 + 2()) = 99 2 Step 4 : use ∫ () = () ( − ) to find the value of c (note : the value of c should be between a and b) ∫ () = ()( − ) 99 2 = ( 2 + 3 + 2)(4 − 1) 99 2 = ( 2 + 3 + 2)(3) 99 = ( 2 + 3 + 2)(6) 99 = 6 2 + 18 + 12 0 = 6 2 + 18 − 87 6 2 + 18 − 87 = 0 Solving quadratic equation (using calculator) = 2.59 = −5.59 = 2.59 [1, 4]
Chapter 4 : Integration 213 Example 4/ MAR 2005/ MAT183/ Q3b/ 6 marks The Mean Value Theorem for Integrals states that if f(x) is continuous on [a, b], then there exist a number c in [a, b], such that f(x) dx f(c)(b a) b a = − Find the number of c, if ( ) x f x 1 = is continuous on the interval [2, 4]. Solution : Step 1 : identify the values of a and b = 2, = 4 Step 2 : from f(x), find f(c) () = 1 √ () = 1 √ Step 3 : find ∫ () ∫ () = ∫ ( 1 √ ) 4 2 = ∫ − 1 2 4 2 = [ 1 2 ( 1 2 ) ] 2 4 = [2√] 2 4 = 2√ − 2√ = 4 − 2√2 = 1.17 Step 4 : use ∫ () = () ( − ) to find the value of c (note : the value of c should be between a and b) ∫ () = ()( − ) 4 − 2√2 = ( 1 √ ) (4 − 2) 4 − 2√2 = 2 √ √(4 − 2√2) = 2 √ = 2 4 − 2√2 = ( 2 4 − 2√2 ) 2 = 2.914 [2, 4]
MAT183/ MAT421 : Calculus I 214 Example 5/ OCT 2004/ MAT183/ Q4c/ 6 marks The Mean Value Theorem for Integrals states that if f(x) is continuous on [a, b], then there exist a number c in [a, b], such that f(x) dx f(c)(b a) b a = − Find the number of c, if ()= 2 is continuous on the interval [ 3 , 4 ]. (Leave your answer in three (3) decimal places) Solution : Step 1 : identify the values of a and b = 4 , = 3 Step 2 : from f(x), find f(c) () = 2 () = 2 Step 3 : find ∫ () ∫ () = ∫ 2 3 4 = [ 1 2 2] 4 3 = ( 1 2 2 ( )) − ( 1 2 2 ( )) = 1 2 ( ) − 1 2 ( ) = 1 2 (°) − 1 2 (°) = 1 2 ( √3 2 ) − 1 2 (1) = √3 4 − 2 4 = √3 − 2 4 Step 4 : use ∫ () = () ( − ) to find the value of c (note : the value of c should be between a and b) ∫ () = ()( − ) √3 − 2 4 = ( 2) ( 3 − 4 ) √3 − 2 4 = ( 2) ( 4 − 3 12 ) √3 − 2 4 = ( 2) ( 12) 2 = ( √3 − 2 4 ) ( 12 ) 2 = 3(√3 − 2) 2 = −1 ( 3(√3 − 2) ) = 1 2 −1 ( 3√3 − 6 ) = 0.915 [ 4 , 3 ] [0.785,1.047 ] ∫ = + Note : Set your calculator in mode ‘Radian’
Chapter 4 : Integration 215 Second Fundamental Theorem of Calculus Example 1/ SEP 2011/ MAT183/ Q4c/ 4 marks Use the Second Fundamental Theorem of Calculus to find F ‘(x) if ( ) dt t t F x x + − = 3 2 4 5 3 4 . ( ) ( ) ( ) ( ) ( ) = = f t dt f g x g x dx d H F x g x a int : ' ' Solution : Theorem : Using the Second Fundamental Theorem of Calculus, Step 1 : identify () and find ′() (note : a = lower limit, g(x) = upper limit) () = 3 , ′() = 3 2 Step 2 : use () to find (()) () = 3 − 4 4 + 5 (()) = ( 3) = 3( 3) − 4 ( 3) 4 + 5 = 3 3 − 4 12 + 5 Step 3 : use ′() = (()) ∙ ′() ′() = (()) ∙ ′() = 3 3 − 4 12 + 5 ∙ 3 2 = 3 2(3 3 − 4) 12 + 5
MAT183/ MAT421 : Calculus I 216 Example 2/ APR 2011/ MAT183/ Q4d/ 4 marks Let ( ) dt t t F x x + + = 2 1 0 2 2 2 3 . Use the Second Fundamental Theorem of Calculus to find i) F’(x) ii) F’(1) ( ) ( ) ( ) ( ) ( ) = = f t dt f g x g x dx d H F x g x a int : ' ' Solution : ) ′() Step 1 : identify () and find ′() (note : a = lower limit, g(x) = upper limit) () = 2 + 1, ′() = 2 Step 2 : use () to find (()) () = 2 2 + 3 2 (()) = (2 + 1) = (2 + 1) 2 2 + 3(2 + 1) 2 Step 3 : use ′() = (()) ∙ ′() ′() = (()) ∙ ′() = (2 + 1) 2 2 + 3(2 + 1) 2 ∙ 2 = 2 (2 + 1) 2 2 + 3(2 + 1) 2 ) ′(1) = 2 (2 + 1) 2 2 + 3(2 + 1) 2 = 2 (9) 2 + 3(9) = 18 29
Chapter 4 : Integration 217 Example 3/ OCT 2010/ MAT183/ Q4c/ 6 marks Let ( ) dt t t F x x − + = 2 0 2 2 3 . Use the Second Fundamental Theorem of Calculus to find i) F’(x) ii) F’(1) ( ) ( ) ( ) ( ) ( ) = = f t dt f g x g x dx d H F x g x a int : ' ' Solution : ) ′() Step 1 : identify () and find ′() (note : a = lower limit, g(x) = upper limit) () = √2 − = (2 − ) 1 2 ′() = 1 2 (2 − ) − 1 2(−1) = − 1 2√2 − Step 2 : use () to find (()) () = √3 + 2 2 (()) = (√2 − ) = √3 + (√2 − ) 2 2√2 − = √3 + (2 − ) 2√2 − = √5 − 2√2 − Step 3 : use ′() = (()) ∙ ′() ′() = (()) ∙ ′() = √5 − 2√2 − ∙ − 1 2√2 − = − √5 − (2√2 − ) 2 = − √5 − 4(2 − ) ) ′(1) = − √5 − 1 4(2 − 1) = − √4 4(1) = 2 4 = 1 2
MAT183/ MAT421 : Calculus I 218 Example 4/ APR 2010/ MAT183/ Q4c/ 4 marks Let ( ) dt t F x x = 3 0 1 . Use the Second Fundamental Theorem of Calculus to find i) F(0) ii) F’(1) iii) F’’(x) ( ) ( ) ( ) ( ) ( ) = = f t dt f g x g x dx d H F x g x a int : ' ' Solution : ) () = ∫ 1 3 0 () = ∫ 1 0 0 = 0 ) ′(1) Step 1 : identify () and find ′() (note : a = lower limit, g(x) = upper limit) () = 3 ′() = 3 2 Step 2 : use () to find (()) () = 1 (()) = ( 3) = 1 3 Step 3 : use ′() = (()) ∙ ′() ′() = (()) ∙ ′() = 1 3 ∙ 3 2 = 3 ′(1) = 3 1 = 3 ) ′() = 3 = 3 −1 ′′() = −3 −2 = 3 2 Note : If the upper and lower limits are the same then there is no work to do, the integral is zero.
Chapter 4 : Integration 219 Example 5/ OCT 2010/ MAT421/ Q4b/ 5 marks Let () = ∫ 3−2 3 2−1 2+2 2 . Use the Second Fundamental Theorem of Calculus to find i) F’(x) ii) F’(1) ( ) ( ) ( ) ( ) ( ) = = f t dt f g x g x dx d H F x g x a int : ' ' Solution : ) ′() Step 1 : identify () and find ′() (note : a = lower limit, g(x) = upper limit) () = 2 + 2 ′() = 2 Step 2 : use () to find (()) () = 3 − 2 3 2 − 1 (()) = ( 2 + 2) = 3( 2 + 2) − 2 3( 2 + 2) 2 − 1 = 3 2 + 6 − 2 3( 2 + 2) 2 − 1 = 3 2 + 4 3( 2 + 2) 2 − 1 Step 3 : use ′() = (()) ∙ ′() ′() = (()) ∙ ′() = 3 2 + 4 3( 2 + 2) 2 − 1 ∙ 2 = 2(3 2 + 4) 3( 2 + 2) 2 − 1 ) ′(1) = 2(3 + 4) 3(1 + 2) 2 − 1 = 2(7) 3(9) − 1 = 14 26 = 7 13
MAT183/ MAT421 : Calculus I 220 END OF CHAPTER 4 Example 6/ OCT 2008/ MAT183/ Q4b/ 4 marks Let ( ) dt t t F x x = 0 cos . Use the Second Fundamental Theorem of Calculus to find i) F’(x) ii) F’(1) ( ) ( ) ( ) ( ) ( ) = = f t dt f g x g x dx d H F x g x a int : ' ' Solution : END OF CHAPTER 4 ) ′() Step 1 : identify () and find ′() (note : a = lower limit, g(x) = upper limit) () = √ = 1 2 ′() = 1 2 − 1 2 = 1 2√ Step 2 : use () to find (()) () = (()) = (√) = √ √ Step 3 : use ′() = (()) ∙ ′() ′() = (()) ∙ ′() = √ √ ∙ 1 2√ = √ 2 ) ′(1) = (1) 2 = 0.27 Note : Set your calculator in mode ‘Radian’