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) ∫ || 2 −1 = ∫ − 0 −1 + ∫ 2 0 = 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 sin2 x 2sin2 x = 1 − cos2x sin2 x = 1 2 (1 − cos2) and use ∫ = + (never use u-substitution) Odd power use Phythagorean identity cos2 x + sin2 x = 1 sin2 x = 1 − cos2 x 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 = 2cos2 x − 1 2cos2 x = 1 + cos2x cos2 x = 1 2 (1 + cos2) and use ∫ = + (never use u-substitution) Odd power use Phythagorean identity cos2 x + sin2 x = 1 cos2 x = 1 − sin2 x 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 : ) ′() 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’
Chapter 5 : Applications of Definite Integrals 221 CHAPTER 5 APPLICATIONS OF DEFINITE INTEGRALS List of Topics : Area between Curves Volume by disk (solid object – axis of rotation is parallel to lower and upper limits of definite integration Volume by Washer (hollow object – axis of rotation is parallel to lower and upper limits of definite integration Volume by Shell (solid and hollow object – axis of rotation is perpendicular to lower and upper limits of definite integration 5.1 Area between Curves Area = ( ) = = − x b x a top bottom dx or Area = ( ) = = − y d y c right left dy Simple Example 1 The region R is bounded by the line y = 3, the line y = 1, the line x = 2, and the line x = 5 as shown in Figure 1 below. Find the area of the shaded region R. y x y = 3 Figure 1 R 0 y = 1 x = 2 x = 5
MAT183/ MAT421 : Calculus I 222 Solution : Area of rectangle = (width) • (length) = (3)(2) = 6 units2 # Area = ( ) = = − x b x a top bottom dx = ( ) = = − 5 2 3 1 x x dx = ( ) 5 2 2 dx = 5 2 2x = 2(5) − 2(2) = 6 units2 # Area = ( ) = = − y d y c right left dy = ( ) = = − 3 1 5 2 y y dy = ( ) 3 1 3 dx = 3 1 3y = 3(3) − 3(1) = 6 units2 # Top : line y = 3 bottom : line y = 1 x = 2 x = 5 left : line x = 2 Right : line x = 5 y = 1 y = 3 2 3
Chapter 5 : Applications of Definite Integrals 223 Simple Example 2 The region R is bounded by the line 2y + x = 4, the x-axis and the y-axis as shown in Figure 2 below. Find the area of the shaded region R. Solution : y x 2y + x = 4 Figure 2 R 0 y = x = 4 − 2y Area = ( ) = = − x b x a top bottom dx = ∫ ( 4− 2 − 0) =4 =0 = ∫ (2 − 1 2 ) 4 0 = [2 − 1 2 • 2 2 ] 0 4 = [2 − 2 4 ] 0 4 = [2(4) − (4) 2 4 ] − [2(0) − (0) 2 4 ] = 4 units2 # Area = ( ) = = − y d y c right left dy = ∫ (4 − 2) =2 =0 = [4 − 2 • 2 2 ] 0 2 = [4 − 2 ]0 2 = [4(2) − (2) 2 ] − [4(0) − (0) 2 ] = 4 units2 # Top : y = bottom : line y = 0 x = 0 x = 4 Right : x = 4 − 2y left : line x = 0 y = 0 y = 2
MAT183/ MAT421 : Calculus I 224 Area of triangle = 2 1 • base • height = 2 1 (4)(2) = 4 units2 # Examples from previous semester papers Example 1/ MAR 2013/ Q5b/ 4 marks The region R is bounded by the curve x = y 2 + 1 and the line y = 3 − x as shown in Figure 3 below. Find the area of the shaded region R. y x (5, −2) x = y 2 + 1 (2, 1) y = 3 − x Figure 3 0 R 2 4 Solution : = ∫ (ℎ − ) =1 =−2 = ∫ (3 − ) − ( 2 + 1) =1 =−2 = ∫ (3 − − 2 − 1) 1 −2 = ∫ (2 − − 2 ) 1 −2 = [2 − 2 2 − 3 3 ] −2 1 = (2 − 1 2 − 1 3 ) −(−4− 4 2 − −8 3 ) = (2 − 1 2 − 1 3 ) −(−4− 2 + 8 3 ) = 9 2 2 # = 3− = 3 −
Chapter 5 : Applications of Definite Integrals 225 Example 2/ OCT 2012/ Q5b/ 5 marks The shaded region R is bounded by the curve x = y 2 and the line x = y + 2 as shown in Figure 4. Find the area of the shaded region R. Solution : y x (1, −1) x = y 2 (4, 2) x = y + 2 Figure 4 = ∫ (ℎ − ) =2 =−1 = ∫ ( + 2)− ( 2 ) =2 =−1 = ∫ ( +2 − 2 ) 2 −1 = [ 2 2 + 2 − 3 3 ] −1 2 = ( 4 2 +4 − 8 3 ) −( 1 2 − 2 − −1 3 ) = (2 +4 − 8 3 ) −( 1 2 − 2 + 1 3 ) = 9 2 2 #
MAT183/ MAT421 : Calculus I 226 Example 3/ MAR 2012/ Q5b/ 5 marks Find the area of the shaded region R enclosed by the curve y = x 2 , the line y = 2 − x and y-axis as shown in Figure 5. Solution : y x R y = x 2 −2 y = 2 − x Figure 5 1 = ∫ ( −) =1 =−2 = ∫ (2 − )− ( 2 ) =1 =−2 = ∫ (2 − − 2 ) 1 −2 = [2 − 2 2 − 3 3 ] −2 1 = (2 − 1 2 − 1 3 ) −(−4 − 4 2 − −8 3 ) = (2 − 1 2 − 1 3 ) −(−4 −2 + 8 3 ) = 9 2 2 #
Chapter 5 : Applications of Definite Integrals 227 Example 4/ APR 2010/ Q5a/ 6½ marks Find the area bounded by the curve g(x) = x 2 − 1 and the line f(x) = x + 1 as shown in Figure 6. Solution : y R g(x) = x 2 − 1 −1 Figure 6 1 x f(x) = x + 1 = ∫ ( −) =1 =−2 = ∫ (2 − )− ( 2 ) =1 =−2 = ∫ (2 − − 2 ) 1 −2 = [2 − 2 2 − 3 3 ] −2 1 = (2 − 1 2 − 1 3 ) −(−4 − 4 2 − −8 3 ) = (2 − 1 2 − 1 3 ) −(−4 −2 + 8 3 ) = 9 2 2 # Solving simultaneous equation : = 2 − 1 −− − −① = + 1− − − −② ① ② ∶ 2 −1 = +1 2 − − 2 = 0 ( + 1)( − 2) = 0 = −1, = 2 y R −1 1 x = 2 − 1 2
MAT183/ MAT421 : Calculus I 228 Example 5/ JUN 2013/ MAT421/ Q5a/ 5 marks Find the area of the shaded region as shown in Figure 7 Solution : Figure 7 y x y = −1 y = x + 1 y = 2 y 2 = x − 1 = ∫ (ℎ − ) =2 =−1 = ∫ ( 2 + 1)− ( − 1) =2 =−1 = ∫ ( 2 + 1 − + 1) 2 −1 = ∫ ( 2 − + 2) 2 −1 = [ 3 3 − 2 2 + 2] −1 2 = ( 8 3 − 4 2 + 4) −(− 1 3 − 1 2 − 2) = ( 8 3 −2 + 4) −(− 1 3 − 1 2 − 2) = 15 2 2 # y x −1 = − 1 2 = 2 + 1
Chapter 5 : Applications of Definite Integrals 229 Example 6/ OCT 2004/ MAT183/ Q5a/ 8 marks Consider the graphs as shown in Figure 8 below. Calculate the total area of the shaded regions Solution : A1 = ( ) = = − 2 0 x x top bottom dx = (( ) ( )) = = + − + 2 0 2 2 1 1 x x x x dx = ( ) − 2 0 2 2x x dx = 2 0 2 3 2 3 2 − x x = 2 0 3 2 3 − x x = 0 3 2 2 3 2 − − = 0 3 8 4 − − = 3 4 units2 y Figure 8 x y = 2x +1 y = 10 y = x 2 + 1 (2, 5) (0, 1) 1 2 y = x + --------- y = 10 --------- into : 1 10 2 x + = 9 2 x = x = 3 x = 3 (on the right side) y = 2x +1 --------- 1 2 y = x + --------- into : 2x +1 = 10 2x = 9 2 = 9 x
MAT183/ MAT421 : Calculus I 230 A2 = A2a + A2b Where, A2b = ( ) = = − 2 9 3 x x top bottom dx = ( ( )) = = − + 2 9 3 10 2 1 x x x dx = ( ) − 2 9 3 9 2x dx = 2 9 3 2 2 2 9 − x x = 2 9 3 2 9x − x = ( ( ) ( ) ) ( ( ) ) 2 2 2 9 2 9 9 − − 9 3 − 3 = ( ) 18 4 81 2 81 − − = 4 9 units2 A2a = ( ) = = − 3 2 x x top bottom dx = (( ) ( )) = = + − + 3 2 2 1 2 1 x x x x dx = ( ) − 3 2 2 x 2x dx = 3 2 3 2 2 2 3 − x x = 3 2 2 3 3 − x x = − − − 2 3 2 3 2 3 2 3 3 3 = − − 3 4 0 = 3 4 units2 ℎ ℎ = 1 + 2 = 1 + 2 + 2 = 4 3 + 4 3 + 9 4 = 59 12 2 #
Chapter 5 : Applications of Definite Integrals 231 You can also use the definite integral to find the volume of a solid that is obtained by revolving a plane region about a horizontal or vertical line that does not pass through the plane. This type of solid will be made up of one of three types of elements — disks, washers, or cylindrical shells — each of which requires a different approach in setting up the definite integral to determine its volume. 5.2 Volume by Disk (solid object – axis of rotation is parallel to lower and upper limits of definite integration VDisk = ( ) = = x b x a radius dx 2 (axis of rotation : horizontal line) VDisk = ( ) = = y d y c radius dy 2 (axis of rotation : vertical line) Where radius = curve/ line − axis of rotation 5.3 Volume by Washer (hollow object – axis of rotation is parallel to lower and upper limits of definite integration Vwasher = ( ) ( ) = = − x b x a outer radius inner radius dx 2 2 (axis of rotation : horizontal line) Vwasher = ( ) ( ) = = − y d y c outer radius inner radius dy 2 2 (axis of rotation : vertical line) Where outer radius = outer curve/ line − axis of rotation inner radius = inner curve/ line − axis of rotation x-axis or line y = k y-axis or line x = k x-axis or line y = k y-axis or line x = k
MAT183/ MAT421 : Calculus I 232 5.4 Volume by Shell (solid & hollow object– axis of rotation is perpendicular to lower and upper limits of definite integration) VShell = ( )( ) = = x b x a 2 radius height dx (axis of rotation : vertical line) VShell = ( )( ) = = y d y c 2 radius height dy (axis of rotation : horizontal line) Where radius = equation of axis of rotation Height = Top − Bottom (dx) Height = Right − Left (dy) Finding Radius (using Shell Method) Axis of rotation : vertical line axis of rotation n : line x = 2 (on the right side, compare to the shaded region) radius = 2 − x axis of rotation : line x = −1 (on the left side, compare to the shaded region) radius = x +1 Axis of rotation : horizontal line axis of rotation : line y = 3 (on the top, compare to the shaded region) radius = 3 − y axis of rotation : line y = −2 (on the bottom, compare to the shaded region) radius = y + 2 x-axis or line y = k y-axis or line x = k Note : • Equation of x-axis : line y = 0 • Equation of y-axis : line x= 0 x = 2 x − 2 = 0 0 = 2 − x Bring all the terms to left side Bring all the terms to right side Bring all the terms to left side Bring all the terms to right side x = −1 x + 1 = 0 0 = −1 − x y = 3 y − 3 = 0 0 = 3 − y Bring all the terms to left side Bring all the terms to right side Bring all the terms to left side Bring all the terms to right side y = −2 y + 2 = 0 0 = −2 − y Question : How to find the radius using Shell Method? Answer : radius = equation of axis of rotation
Chapter 5 : Applications of Definite Integrals 233 Simple Example 1 The region R is bounded by the line y = 3, the line y = 1, the line x = 2, and the line x = 5 as shown in Figure 9 below. Find the volume of the solid generated when the shaded region R is revolved about the a) line y = 1 b) line x = 2 c) x-axis d) y-axis Solution : a) y x y = 3 Figure 9 R 0 y = 1 x = 2 x = 5 x = 2 x = 5 line y = 3 y = 1 (axis of rotation) line y = 3 y = 1 (axis of rotation)
MAT183/ MAT421 : Calculus I 234 Radius = curve/line − axis of rotation Radius = 3 − 1 = 2 VDisk = ( ) = = x b x a radius dx 2 VDisk = ( ) = = 5 2 2 2 x x dx = = = 5 2 4 x x dx = 5 4 2 x = 4(5) − 4(2) = 12 units3 # Radius = equation of axis of rotation Radius = y − 1 Height = Right − left = 5 − 2 = 3 VShell = ( )( ) = = y d y c 2 radius height dy VShell = ( )( ) = = − 3 1 2 1 3 y y y dy = ( ) − 3 1 6 y 1 dy = 3 1 2 2 6 − y y = ( ) ( ) ( ) ( ) − − − 1 2 1 3 2 3 6 2 2 = 6 (2) = 12 units3 # r = y −1 h = R − L x = 2 (L) x = 5 (R) line y = 3 y = 1 (axis of rotation) radius h = 3 r = 2 Volume of cylinder = = = 12 units3 #
Chapter 5 : Applications of Definite Integrals 235 b) line x = 5 x = 2 (axis of rotation) y = 3 y = 1 line x = 5 x = 2 (axis of rotation) Volume of cylinder = r h 2 = (3) 2 2 • = 18 units3 # r = 3 h = 2
MAT183/ MAT421 : Calculus I 236 Radius = curve/line − axis of rotation Radius = 5 − 2 = 3 VDisk = ( ) = = y d y c radius dy 2 VDisk = ( ) = = 3 1 2 3 y y dy = = = 3 1 9 y y dy = 3 1 9y = 9 (3) − 9 (1) = 18 units3 # Radius = equation of axis of rotation Radius = x − 2 Height = top − bottom = 3 − 1 = 2 VShell = ( )( ) = = x b x a 2 radius height dx VShell = ( )( ) = = − 5 2 2 2 2 x x x dx = ( ) − 5 2 4 x 2 dx = 5 2 2 2 2 4 − x x = ( ) ( ) ( ) ( ) − − − 2 2 2 2 2 5 2 5 4 2 2 = 2 9 4 = 18 units3 # line x = 5 x = 2 (axis of rotation) h = T − B x = 2 (axis of rotation) r = x − 2 y = 3 (T) y = 1 (B)
Chapter 5 : Applications of Definite Integrals 237 c) = (r ) h 2 1 − (r ) h 2 2 = (3) (3) 2 − (1) (3) 2 = 27 − 3 = 24 units3 # x = 2 x = 5 line y = 3 Line y = 0 (x-axis) (axis of rotation) line y = 1 = − r1 = 3 h = 3 h = 3 r2 = 1 line y = 3 Line y = 0 (x-axis) (axis of rotation) line y = 1
MAT183/ MAT421 : Calculus I 238 Outer radius = outer curve/line − axis of rotation Outer radius = 3 − 0 = 3 inner radius = inner curve/line − axis of rotation innter radius = 1 − 0 = 1 Vwasher = ( ) ( ) = = − x b x a outer radius inner radius dx 2 2 Vwasher = ( ) ( ) = = − 5 2 2 2 3 1 x x dx = 5 2 8 dx = 5 2 8x = 8 (5) − 8 (2) = 24 units3 # Radius = equation of axis of rotation Radius = y Height = Right − left = 5 − 2 = 3 VShell = ( )( ) = = y d y c 2 radius height dy VShell = ( )( ) = = 3 1 2 3 y y y dy = ( ) 3 1 6 y dy = 3 1 2 2 6 y = ( ) ( ) − 2 1 2 3 6 2 2 = 6 (4) = 24 units3 # line y = 3 (outer curve/line) Line y = 0 (x-axis) (axis of rotation) line y = 1 (inner curve/line) h = R − L Line y = 0 (x-axis) (axis of rotation) r = y x = 2 (L) x = 5 (R)
Chapter 5 : Applications of Definite Integrals 239 d) = (r ) h 2 1 − (r ) h 2 2 = (5) (2) 2 − (2) (2) 2 = 50 − 8 = 42 units3 # line x = 2 line x = 5 y = 3 line x = 0 (y-axis) (axis of rotation) y = 1 = − r2 = 2 h = 2 r1 = 5 h = 2 line x = 0 (y-axis) (axis of rotation) line x = 5 line x = 2
MAT183/ MAT421 : Calculus I 240 line x = 0 (y-axis) (axis of rotation) Outer radius = outer curve/line − axis of rotation Outer radius = 5 − 0 = 5 inner radius = inner curve/line − axis of rotation innter radius = 2 − 0 = 2 Vwasher = ( ) ( ) = = − y d y c outer radius inner radius dy 2 2 Vwasher = ( ) ( ) = = − 3 1 2 2 5 2 y y dy = 3 1 21 dy = 3 21 1 x = 21(3) − 21(1) = 42 units3 # Radius = equation of axis of rotation Radius = x Height = top − bottom = 3 − 1 = 2 VShell = ( )( ) = = x b x a 2 radius height dx VShell = ( )( ) = = 5 2 2 2 x x x dx = ( ) 5 2 4 x dx = 5 2 2 2 4 x = ( ) ( ) − 2 2 2 5 4 2 2 = 2 21 4 = 42 units3 # line x = 0 (y-axis) (axis of rotation) line x = 5 (outer curve/line) line x = 2 (inner curve/line) line y = 1 (B) line y= 3 (T) r = x h = T − B
Chapter 5 : Applications of Definite Integrals 241 Simple Example 2 The region R is bounded by the line 2y + x = 4, the x-axis and the y-axis as shown in Figure 10 below. Find the volume of the solid generated when the shaded region R is revolved through 3600 about the a) x-axis b) y-axis c) line y = 2 d) line x = 4 Solution : a) y x 2y + x = 4 Figure 10 R 0 (0, 0) x = 4 line y = y = 0 (axis of rotation) y = 2 line y = y = 0 (axis of rotation)
MAT183/ MAT421 : Calculus I 242 Radius = curve/line − axis of rotation Radius = 2 4 − x − 0 = 2 4 − x VDisk = ( ) = = x b x a radius dx 2 VDisk = = = − 4 0 2 2 4 x x dx x = − + 4 0 2 4 16 8 dx x x = ( ) − + 4 0 2 16 8 4 x x dx = 4 0 2 3 2 3 8 16 4 − + x x x = 4 0 3 2 3 16 4 4 − + x x x = − − + 0 3 64 48 48 4 = 3 16 units3 # Radius = equation of axis of rotation Radius = y Height = Right − left = (4 − 2y) − 0 = 4 − 2y VShell = ( )( ) = = y d y c 2 radius height dy VShell = ( )( ) = = − 2 0 2 4 2 y y y y dy = ( ) − 2 0 2 2 4y 2y dy = 2 0 2 3 3 2 2 4 2 − y y = 2 0 3 2 3 2 2 2 − y y = − − 0 3 16 2 8 = 3 16 units3 # Volume of cone = r h 2 3 = (2) 4 3 2 • = 3 16 units3 # h = 4 r = 2 line y = y = 0 (axis of rotation) radius x = 0 x = 4 h = R − L radius = y line x = 4 − 2y (R) y = 0 (axis of rotation) x = 0 (L)
Chapter 5 : Applications of Definite Integrals 243 b) r = 4 h = 2 Volume of cone = = = units3 # (0, 0) x = 4 line x = 4 − 2y x = 0 (axis of rotation) y = 2 x = 0 (axis of rotation) line x = 4 − 2y
MAT183/ MAT421 : Calculus I 244 Radius = curve/line − axis of rotation Radius = (4 − 2y) − (0) = 4 − 2y VDisk = ( ) = = y d y c radius dy 2 VDisk = ( ) = = − 2 0 2 4 2 y y y dy = ( ) − + 2 0 2 16 16y 4y dy = 2 0 2 3 3 4 2 16 16 − + y y y = 2 0 3 2 3 4 16 8 − + y y y = − − + 0 3 32 32 32 = 3 32 units3 # Radius = equation of axis of rotation Radius = x Height = top − bottom = 2 4 − x − 0 = 2 4 − x VShell = ( )( ) = = x b x a 2 radius height dx VShell = ( ) = = − 4 0 2 4 2 x x dx x x = ( ) − 4 0 x 4 x dx = ( ) − 4 0 2 4x x dx = 4 0 2 3 2 3 4 − x x = 4 0 3 2 3 2 − x x = − − 0 3 64 32 = 3 32 units3 # x = 0 (axis of rotation) line x = 4 − 2y h = T − B r = x x = 0 (axis of rotation) line y = (T) y = 0 (B)
Chapter 5 : Applications of Definite Integrals 245 c) = (r ) h 2 − (r ) h 2 3 = (2) (4) 2 − (2) (4) 3 2 = 16 − 3 16 = 3 32 units3 # = − r = 2 h = 4 r = 2 h = 4 (0, 0) x = 4 y = 2 (axis of rotation) line y = y = 2 (axis of rotation) line y =