Department of Mathematics, K.T.H.M. College, Nashik F.Y.B.Sc. Calculus Practical (Academic Year )

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1 F.Y.B.Sc. Calculus Practical (Academic Year 06-7) Practical : Graps of Elementary Functions. a) Grap of y = f(x) mirror image of Grap of y = f(x) about X axis b) Grap of y = f( x) mirror image of Grap of y = f(x) about Y axis c) Grap of y 3 = f(x a) Grap of y = f(x) displaced along X axis by amount a d) Grap of y 4 = f(x) + b Grap of y = f(x) displaced along Y axis by amount b. Grap of f (x) mirror image of Grap of f(x) about te line y = x F.Y.B.Sc. Calculus Practical, Page:

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6 F.Y.B.Sc. Calculus Practical (Academic Year 06-7). If 0 < a < b, sow tat a) a < ab < b b) b < a Ans: a) 0 < a < b = a.a < a.b and a.b < b.b = a < ab and ab < b = a < ab and ab < b = a < ab < b b) 0 < a < b = a < b ( ab > 0) ab ab = < b a Practical : Real Numbers. Find all x R satisfying te following inequalities. a) x > x + b) x + x + < Ans: a) x > x + x x + > 0... (I) End points are and. I Case: x In tis case x < 0 and x + 0. So from (I), (x ) ( (x + )) > 0 = > 0 Tis is true. So all x satisfy (I). Te solution set in I case is (, ] II Case: < x In tis case, x < 0 and x + > 0. So from (I), (x ) (x + ) > 0 = x > 0 = x < 0 Te solution set in II case is (, 0) III Case: x In tis case x > 0 and x + > > 0. So from (I), (x ) (x + ) > 0 = > 0. Tis is impossible. Te solution set in III case is φ Solution Set = (, ] (, 0) φ = (, 0) b) x + x + <... (II) End points are 0 and. I Case: x < F.Y.B.Sc. Calculus Practical, Page: 6

7 F.Y.B.Sc. Calculus Practical (Academic Year 06-7) In tis case x < 0 and x + 0. So from (II), x (x + ) < = x < = x < = x > Tis is impossible as x < Te solution set in I case is φ. II Case: x < 0 In tis case x < 0 and x + 0. So from (II), x + (x + ) < = < Tis is true. Te solution set in II case is [, 0). III Case: x 0 In tis case x 0 and x + > 0. So from (II), x + (x + ) < = x + < = x < Te solution set in III case is [0, ). Solution Set = φ [, 0) [0, ) = [, ) 3. Prove tat te following are not rational numbers. a) 3 b) Ans: a) Suppose on contrary tat 3 is rational. Take 3 = p, were p and q are integers q and p, q do not ave any common factor. Consider 3 = p q = 3 = p q = p = 3q = 3 divides p ( 3 is prime) = p = 3k (k is some integer) = q = 3k = 3 divides q ( 3 is prime) Tus 3 is a common factor of p and q. Tis contradicts te fact tat p, q do not ave any common factor. Hence 3 must be irrational. b) Consider is rational 3 5 is rational ( 3 5 = 3+ 5 ) = ( 3 + 5) + ( 3 5) is rational = 3 is rational = 3 is rational Tis is a contradiction. Hence is irrational. 4. Let K = {s + t : s, t Q}. Sow tat K satisfies te following conditions. F.Y.B.Sc. Calculus Practical, Page: 7

8 F.Y.B.Sc. Calculus Practical (Academic Year 06-7) a) If x, x K, ten x + x K b) If x 0 is in K, ten x K Ans: a) Take x = s + t, x = s + t in K were, s, s, t, t Q. Ten x + x = (s + s ) + (t + t ) = s + t, were s = s + s, t = t + t Q. Tis proves tat x + x K. b) Take x = s + t K, were s, t Q. Ten s+t (s t ) (s t = ) = x s+t = were s = s and t = t s t Tis proves tat K. x s s t s t are in Q. t s t = s + t 5. Find supremum and infimum of te following sets. a) { : n N} b) { : n N} c) {3 + ( n n 3 )n : n N} d) {( )n : n N} Ans: a) infimum of { : n N} = 0 and supremum of { : n N} = n n b) infimum of { : n N} = and supremum of { : n N} = 0 n n c) infimum of {3+( 3 )n : n N} = 3 and supremum of {3+( 3 )n : n N} = 3+ = 3 3 d)infimum of {( )n : n N} = 0 and supremum of {( )n : n N} = 6. If L R, L < M + ɛ for every ɛ > 0, prove L M. Ans: Suppose on contrary tat L > M. Take ɛ = L M > 0. By ypotesis, L < M + ɛ = L < M + (L M) = L < L Tis is a contradiction. Hence L M. 7. If L R, L M + ɛ for every ɛ > 0, prove L M. Ans: Suppose on contrary tat L > M. Take ɛ = L M > 0. By ypotesis, L < M + ɛ = L < M + L M = L M < L M = < Tis is a contradiction. Hence L M. F.Y.B.Sc. Calculus Practical, Page: 8

9 F.Y.B.Sc. Calculus Practical (Academic Year 06-7) Practical 3: Limits and Continuity. a) Using ɛ δ definition of limit, sow tat lim(x + ) = 4. x b) Sow tat lim sin( ) does not exist. x 0 x Ans: a) Let ɛ > 0. Take δ = ɛ. Consider 0 < x < δ = x < ɛ = x < ɛ = x < ɛ = (x + ) 4 < ɛ = f(x) 4 < ɛ Tus 0 < x < δ = f(x) 4 < ɛ. Tis proves tat lim x (x + ) = 4. b) Take ɛ = and δ be any positive real number. Select a positive integer n suc tat < δ. Take y = and (nπ π ) (nπ π ) y = (nπ+ π ). Ten δ < y < y < δ. i.e. 0 < y 0 < δ, 0 < y 0 < δ Also f(y) f(y ) = ( ) = > = ɛ So 0 < x 0 < δ = f(x) f(0) < ɛ is not possible for any δ > 0. Tis proves tat lim sin( ) does not exist. x 0 x. If exists, evaluate te following limits. x+5 x +x a) lim b) lim c) lim x x+3 x 0 3x 5 x Ans: x+5 a) lim = +5 = 4 = 4 x x+3 ( )+3 sin( x 0 x x ) d) lim x x x F.Y.B.Sc. Calculus Practical, Page: 9

10 F.Y.B.Sc. Calculus Practical (Academic Year 06-7) b) lim x 0+ and lim x 0 So lim x 0+ x +x 3x 5 x x 0+ x+x 3x 5x x 0+ x +x x+x 3x 5 x x 0 3x+5x x 0+ c) Consider sin x x +x 3x 5 x lim x 0 3x = 3 x x 8x = 8 x +x 3x 5 x and lim x 0 x +x 3x 5 x does not exist. = x x sin x x if x 0 and x x sin x x if x 0 = lim x lim x sin lim x and lim x lim x sin lim x x 0+ x 0+ x x 0+ x 0 x 0 x x 0 = 0 lim x 0+ x sin x = lim x 0+ x sin x x 0 x sin x = 0 = lim x 0 x sin x = 0 = lim sin x x 0 x d) Consider lim x x x = 0 3. a) Sow tat lim x x+ x x x+ x e x x 0 e x + 0 and 0 lim x 0 x sin x 0 does not exist. x (x )( x+) x x+ = + = b) Discuss { te continuity of te function χ : R R defined as if x Q χ(x) = 0 if x Q Ans: e x a) lim x 0+ e x + x 0+ ( e x e x e x lim = 0 = x 0 e x + 0+ e x So lim lim e x x 0+ e x + x 0 e x + ( e x + e x ) ) x 0+ ( ( e x and lim x 0 e x + e x + e x ) ) 0 = x does not exist. b) Let c R. Take ɛ = > 0 and δ be any positive real number. Take a rational r and irrational s in te interval (c δ, c + δ) {c} (i.e. in deleted δ neigborood of c. Ten χ(r) χ(s) = 0 > = ɛ. So 0 < x c < δ = χ(x) χ(c) < ɛ is not possible for any δ > 0. F.Y.B.Sc. Calculus Practical, Page: 0

11 F.Y.B.Sc. Calculus Practical (Academic Year 06-7) Tis proves tat lim x c χ(x) does not exist. Terefore χ is discontinuous at c. Since c was arbitrary point of R, χ is discontinuous at all points of R. 4. Discuss te continuity of te following functions. x if x x if < x < a) f(x) = b) f(x) = 3x 4 if x < 4 x 3 if x 4 c) f(x) = for all x R Ans: x x+3 a) Possible points of discontinuity are, and 4. i) c = : lim f(x) (x ) = () = x x lim f(x) x + x + (x ) = So lim f(x) f(x) f(x) = = f(). x x + x f is continuous at. ii) c = : lim f(x) x x (x ) = = 4 lim f(x) (3x 4) = 3() 4 = x + x + So lim f(x) lim f(x) x x + lim f(x) does not exist. x f is discontinuous at. iii) c = 4 : lim f(x) (3x 4) = 3(4) 4 = 8 x 4 x 4 lim f(x) (x 3 ) = 4 3 = 8 x 4+ x 4+ So lim f(x) f(x) f(x) = 8 = f(4). x 4 x 4+ x 4 f is continuous at 4. f is continuous at all points of R except at c =. b) Possible point of discontinuity: 0. lim x 0 e x e x = 0 0 = 0 = f() e x e x if x 0 if x = 0 F.Y.B.Sc. Calculus Practical, Page:

12 F.Y.B.Sc. Calculus Practical (Academic Year 06-7) So f is discontinuous at c =0. f is continuous at all points of R except at c = 0. c) By te algebra of continuous functions, f(x) will be continuous at all points at wic it is defined. It is defined only wen x and x x+3 0. I Condition: x In tis case x 3 II Conditiion x x+3 0 In tis case, (x ) 0 and x OR x < 0 and x = x and x 3 OR x and x 3 = x OR x 3 Te function f(x) is continuous on (, 3) [, ) 5. Determine te set of points of discontinuity of te following functions. a) f(x) = x, b) f(x) = x +x+3 x + x+ Ans: a) f(x) = x is a rational function (ratio of two polynomials). It is discontinuous x + only were denominator vanises. So Points of discontinuity of f(x) = {x : x + = 0} = φ. Te function f(x) is continuous at all points of R. b) f(x) = x +x+3 x+ is a rational function (ratio of two polynomials). It is discontinuous only were denominator vanises. So Points of discontinuity of f(x) = {x : x + = 0} = { }. Te function f(x) is continuous at all points of R except at x =. 6. If function f(x) is continuous on [, ] were sin ax + if x < 0 x f(x) = 3x + 5 if 0 x x + 8 b if < x Sow tat a + b + = 0. Ans: f(x) is continuous at x = 0 = lim x 0 f(x) = f(0) = lim x 0 ( sin ax x + ) = 3(0) + 5 = lim x 0 ( sin ax ax a ) + = 5 = a + = 5 = a + = 5 F.Y.B.Sc. Calculus Practical, Page:

13 F.Y.B.Sc. Calculus Practical (Academic Year 06-7) = a = 3 Also f(x) is continuous at x = = lim x + f(x) = f() = + 8 b = 3() + 5 = 9 b = 8 = 3 b = 8 = b = 5 Tus a = 3 and b = 5. Terefore a + b + = 3 + ( 5) + = Sow tat te equation x x = 0 as at least one root in (0, ). Ans: Take f(x) = x x = 0 on [0, ]. Ten f(0) = 0 (0) = < 0 and f() = () = = > 0 Terefore by Bolzano s Teorem, tere is at least one c (0, ) suc tat f(c) = 0. Tis implies tat c c = 0 = c is root of x x = 0 in (0, ). 8. Sow tat te equation cos x = x as a solution in te interval [0, π ]. Ans: Take f(x) = cos x x on [0, π ]. Ten f(0) = cos 0 0 = 0 = > 0 and f( π ) = cos( π ) π = 0 π = π < 0 Terefore by Bolzano s Teorem, tere is at least one c (0, π ) suc tat f(c) = 0. Tis implies tat cos c c = 0 = c is solution of cos x = x in [0, π ]. F.Y.B.Sc. Calculus Practical, Page: 3

14 F.Y.B.Sc. Calculus Practical (Academic Year 06-7) Practical 4: Differentiation. Use te definition of derivative, to find derivative of te following functions. a) f(x) = x 3 b) f(x) = x Ans: a) f f(x+) f(x) (x) 0 (x+)3 x3 0 0 (x3 +3x +3x +3 ) x3 3x +3x (3x + 3x + ) 0 = 3x Ans: b) f (x) 0 x+ x x (x+) 0 x(x+) 0 0 = x x + x + f(x+) f(x) 0 f (x) = 3x f (x) = x. a) If f(x) = x, sow tat f (x) = wen x > 0 and f(x) = wen x < 0. Also sow tat te function f(x) is not differentiable at x = 0. b) If f(x) = x x ( x R), sow tat f (x) = wen x > 0 and f (x) = wen x < 0. Also sow { tat te function f(x) is differentiable at x = 0. x if x c) If f(x) = ( x) if x >, sow tat f +() = 0 and f () = Ans: a) f(x) = x = x if x > 0 implying tat f (x) = wen x > 0 f(x) = x = x if x < 0 implying tat f (x) = wen x < 0 f(0+) f(0) lim 0+ and f(0+) f(0) lim = = 0 f(0+) f(0) f(0+) f(0) f(0+) f(0) So lim lim and lim tat f(x) is not differentiable at x = 0. does not exist. Tis proves b) f(x) = x x = xx = x if x > 0 implying tat f (x) = x and f (x) = wen x > 0 f(x) = x x = x( x) = x if x > 0 implying tat f (x) = x and f (x) = wen x < 0 f(0+) f(0) lim 0+ f(0+) f(0) lim = 0 0+ = 0 f(0+) f(0) f(0+) f(0) f(0+) f(0) So lim = 0 and lim exists and equals to Tis proves tat f(x) is differentiable at x = 0 and f (0) = 0 c) f +() f(+) f() 0+ [ (+)] = 0 F.Y.B.Sc. Calculus Practical, Page: 4

15 F.Y.B.Sc. Calculus Practical (Academic Year 06-7) and f () f(+) f() 0 [ (+)] = 0 3. Test weter te conditions of Rolles Mean Value Teorem old for given functions on te interval and if so find te value of c. a) f(x) = log ( x +ab (a+b)x) on [a, b] b) f(x) = (x a) m (x b) n on [a, b] Ans: a) f(x) is continuous on [a, b] and f(x) is differentiable in (a, b). Also f(a) = log ( ) ( a +ab (a+b)a = log a (a +ab)x) = log = 0 and f(a) = log ( ) ( b +ab (a+b)b = log b +ab (a+b )x) = log = 0 Te ypotesis of Rolle s Mean Value teorem is satisfied. To find c, consider f (c) = 0 = (c +ab) (a+b)c (c) = 0 = c + ab = 0 = c = ab b) f(x) is continuous on [a, b] and f(x) is differentiable in (a, b). Also f(a) = (a a) m (a b) n = 0 and f(a) = (b a) m (b b) n = 0 Te ypotesis of Rolle s Mean Value teorem is satisfied. To find c, consider f (c) = 0 = m(c a) m (c b) n + n(c a) m (c b) n = 0 = (c a) m (c b) n (mc mb + nc na) = 0 = mc mb + nc na = 0 = c(m + n) (mb + na) = 0 F.Y.B.Sc. Calculus Practical, Page: 5

16 F.Y.B.Sc. Calculus Practical (Academic Year 06-7) = c = mb+na m+n 4. Test weter te conditions of Lagranges Mean Value Teorem old for given functions on te interval and if so find te value of c. a) f(x) = (x )(x )(x 3) on [0, 4] b) f(x) = tan x on [0, ] Ans: a) f(x) is continuous on [0, 4] and f(x) is differentiable in (0, 4). So te ypotesis of Lagrange s Mean Value Teorem is satisfied. To find value of c, consider f (c) = f(4) f(0) 4 0 = (c )(c 3) + (c )(c 3) + (c )(c ) = = 3c c + = 3 = 3c c + 8 = 0 = c = + 3 or c = = c = or c = b) f(x) is continuous on [0, ] and f(x) is differentiable in (0, ). So te ypotesis of Lagrange s Mean Value Teorem is satisfied. To find value of c, consider f (c) = f() f(0) 0 = = π +c 4 0 = +c = π 4 = + c = 4 π 4 = c = ± π = c = ( c (0, )) 4 π 5. a) Sow tat te value of c wen Caucy s Mean Value Teorem is applied to te functions f(x) = sin x and g(x) = cos x on [a, b] is te aritmetic mean of a and b. b) Sow tat te value of c wen Caucy s Mean Value Teorem is applied to te functions f(x) = x and g(x) = x on [a, b] is te geometric mean of a and b. Ans: a) Te functions f(x) = sin x and g(x) = cos x are continuous on [a, b] and differentiable in (a, b). So te ypotesis of Caucy s Mean Value Teorem is satisfied. To F.Y.B.Sc. Calculus Practical, Page: 6

17 F.Y.B.Sc. Calculus Practical (Academic Year 06-7) find c, consider f (c) = f(b) f(a) g (c) g(b) g(a) = cos c sin c = cot c = = sin b sin a cos b cos a b a b+a sin( ) cos( ) sin( b a = cot c = cot (b+a) = c = b+a ) sin( b+a ) = c is aritmetic mean of a and b. b) Te functions f(x) = x and g(x) = x are continuous on [a, b] and differentiable in (a, b). So te ypotesis of Caucy s Mean Value Teorem is satisfied. To find c, consider f (c) = f(b) f(a) g (c) g(b) g(a) = c c 3 b a = b = c = ba = c = ab a ( ) b a a b = c is geometric mean of a and b 6. Sow tat te function f(x) = x 3 3x + 3x 00 is increasing on R. Ans: f(x) = x 3 3x + 3x 00 = f (x) = 3x 6x + 3 = f (x) = 3(x ) = f (x) > 0 for all x R except for x =. = f(x) is increasing on R. 7. Find values of x for wic te function f(x) = x 3 9x + x + is decreasing. Ans: f(x) = x 3 9x + x + = f (x) = 6x 8x + F.Y.B.Sc. Calculus Practical, Page: 7

18 F.Y.B.Sc. Calculus Practical (Academic Year 06-7) = f (x) = 6(x )(x ) Terefore f (x) < 0 = 6(x )(x ) < 0 = x < 0 and x > 0 OR x > 0 and x < 0 = x < and x > OR x > and x < = x > and x < = < x < = x (, ) So f(x) is decreasing for x (, ). 8. Differentiate log( + x ) wit respect to tan x Ans: Take u = log( + x ) and v = tan x. Ten du = x = x +x +x and dv = +x So, du dv = du dv = x +x +x = x 9. Differentiate tan ( cos x+sin x cos x sin x) w.r.t x Ans: Take y = tan ( cos x+sin x cos x sin x). Ten y = tan ( ) +tan x tan x = y = tan ( tan( π 4 + x)) = y = π 4 + x = dy = F.Y.B.Sc. Calculus Practical, Page: 8

19 F.Y.B.Sc. Calculus Practical (Academic Year 06-7) Practical 5: Integration. Evaluate te following integrals. a) x + (x )(x+) b) x 3 +4x 7x+5 x+ Ans: a) x + (x )(x+) = x + (x+)(x )(x+) Partial fractions: x + (x )(x+) = A (x+) + B + (x ) C (x+) x + = A(x )(x + ) + B(x + )(x + ) + C(x )(x + )... (I)... (II) Put x = in (II), = A( )(( ) + ) = A = Put x = in (II), = B( + )(() + ) = B = 3 Put x = in (II), 5 4 = C( + )( ) = C = 5 3 Substituting tese values of A, B and C in (I), we get x + = + (x )(x+) (x+) 3 Terefore 5 (x ) 3 (x+) x + = log x + + log x 5 log x + + c (x )(x+) 3 6 b) x + x x + ) x 3 + 4x 7x + 5 x 3 x Terefore x x 7x 4x x + 5 x + x 3 + 4x 7x + 5 = (x + )(x + x ) + 7 = x3 +4x 7x+5 = (x + x ) + 7 x+ x+ x 3 +4x 7x+5 = x3 + x+ 3 x x + 7 log x + + c 7. Evaluate te following integrals. a) b) x x 3 x 4 +5x +6 F.Y.B.Sc. Calculus Practical, Page: 9

20 F.Y.B.Sc. Calculus Practical (Academic Year 06-7) Ans: a) (x 3 ) = (x )(x +x+) Partial Fractions are: = (x )(x +x+) A + Bx+C (x ) x +x+... (I) = = A(x + x + ) + (Bx + c)(x ) = = A(x + x + ) + (Bx Bx + Cx C) = = (A + B)x + (A B + C)x + (A C) Equating coefficients of equal powers of x A + B = 0 A B + C = 0 A C = From first and last equation, B = A and C = A. Using tese in middle equation, A+A+A = 0. Terefore A =. So B = A = and C = A = = Substituting tese values of A, B and C in (I), we get = x (x )(x +x+) 3 (x ) 3 (x +x+) 3 (x +x+) = (x+) (x )(x +x+) 3 (x ) 6 (x +x+) 3 (x +x+) = = = = x+ (x )(x +x+) 3 (x ) 6 (x +x+) (x +x+) = x+ (x 3 ) 3 (x ) 6 (x +x+) (x+ ) +( 3 ) = = log x log x x + x + 3 tan ( x+ ) + c b) Put x = t. Ten x x 4 +5x +6 = Te partial fractions are t (t+)(t+3) 3 t = (t+)(t+3) A + (t+) B (t+3) t = A(t + 3) + B(t + )... (I)... (II) Put t = in (II). = A( + 3) = A = Put t = 3 in (II). 3 = B( 3 + ) = B = 3 Substituting tese values in (I) F.Y.B.Sc. Calculus Practical, Page: 0

21 F.Y.B.Sc. Calculus Practical (Academic Year 06-7) t = (t+)(t+3) = x x 4 +5x +6 = + 3 (t+) (t+3) + 3 x + x +3 = x = x 4 +5x +6 tan ( x ) tan ( x 3 ) + c = x x 4 +5x +6 = 3 tan ( x 3 ) tan ( x ) + c 3. Evaluate te following integrals. a) b) 4e x +6e x sin(x a) cos(x b) 9e x 4e x Ans: a) Put k =cos(a b). Consider sin(x a) cos(x b) = [ cos(a b) ] cos(a b) sin(x a) cos(x b) = k [ cos((x b) (x a)) sin(x a) cos(x b) ] = cos(x b) cos(x a)+sin(x b) sin(x a) [ ] k sin(x a) cos(x b) = (cot(x a) + tan(x b)) k Terefore sin(x a) cos(x b) = k (cot(x a) + tan(x b)) = [log sin(x a) + log sec(x b) ] + c k sin(x a) = log + c cos(a b) cos(x b) b) 4e x +6e x 9e x 4e x = 4(e x ) +6 9(e x ) 4 ex. Put t = e x. Ten e x = dt. Terefore 4e x +6e x 9e x 4e x = 4t +6 9t 4 dt... (I) 9t 4 ) 4t t t + 6 = 4 9 (9t 4) F.Y.B.Sc. Calculus Practical, Page:

22 F.Y.B.Sc. Calculus Practical (Academic Year 06-7) 4t +6 9t 4 = t 4 = 4t +6 t(9t 4) = 4 9t t(3t+)(3t )... (II) Te partial fractions are = A + t(3t+)(3t ) t B + (3t+) C (3t )... (III) = = A(3t + )(3t ) + Bt(3t ) + Ct(3t + )... (IV ) Taking t = 0 in (IV ), we get = A( ) = A = 4 Taking t = 3 in (IV ), we get = B( 3 ( )) = B = 3 8 Taking t = 3 in (IV ), we get = C( 3 ( + )) = C = 3 8 Substituting tese values in (III) = t(3t+)(3t ) 4t 8(3t+) 8(3t ) So from (II), = 4t +6 t(9t 4) = 4 9t ( 4t + 3 8(3t+) + 3 8(3t ) ) = 4t +6 t(9t 4) = 3 t + 35 (3t+) + 35 (3t ) Using tis in (I), 4e x +6e x 9e x 4e x = 3 = log t + log 3t + + log 3t + c log t + log (3t + )(3t ) + c 36 = 3 log ex log 9t 4 + c = 3 x log 9ex 4 + c 4. Evaluate te following integrals. a) x b) 3x x 4 + 4x + F.Y.B.Sc. Calculus Practical, Page:

23 F.Y.B.Sc. Calculus Practical (Academic Year 06-7) Ans: a) x x 4 + = ( x + x x x 4 + ) = ( + x x + = ( ( x ) Terefore x ) + ( x ) x + x (x x ) + ) + ( +( x ) ) (x+ x ) x x 4 + = ( x ) (x x ) + + +( x ) (x+ x ) Put u = x in First integral on R.H.S. and v = x + x x R.H.S. Ten... (I) in Second integral on ( ) = du and + ( ) = dv. So x x x = du + dv x 4 + u + v = x x 4 + = tan (u ) + 4 v log v+ + c = x = x 4 + tan ( x x ) + 4 log x x+ x + + c x+ b) 3x 4x + = 3 x 4x = 3 (x 3 ) = 3 (x 3 ) ( 3 ) We use te standard formula x a = x x a a log( x + x a ) + c log a 3x 4x + = 3 (x 3 ) ( 3 ) = 3 (x 3 ) (x 3 ) ( 3 ) 3 log( (x 3 ) + (x 3 ) ( 3 ) 8 ) + c 3 = (3x ) 6 3 (3x ) 3 = (3x ) log((3x ) + (3x ) ) + c 9x x log((3x ) + (3x ) ) + c F.Y.B.Sc. Calculus Practical, Page: 3

24 F.Y.B.Sc. Calculus Practical (Academic Year 06-7) = (3x ) 3x 4x + 3 log((3x ) + (3x ) 6 8 ) + c 5. Evaluate te following integrals. a) cos x +sin x+sin x b) ( x) 4x 5x + Ans: a) Put sin x = t. Ten cos x = dt. Terefore cos x +sin = x+sin x +t+t dt = t dt... (I) +t+ Consider t + t + = t + t + = (t + ) + 4 = (t + ) + ( 3 ) We use te standard formula x = log ( x + ) x +a +a + c dt t +t+ From (I), cos x +sin x+sin x = t +t+ dt = (t+ ) +( 3 dt ) (t+ ) +( 3 ) 3 = log ( (t+ )+ ) + c = log ( (t+) + (t+) +3) 3 +c = log ( t++ ) t +t+ 3 + c = log ( sin x++ a sin x+sin x+ 3 b) ( x) 4x 5x + ) + c = 4x 5x + x 4x 5x + F.Y.B.Sc. Calculus Practical, Page: 4

25 F.Y.B.Sc. Calculus Practical (Academic Year 06-7) = 4x 5x + 8 [(8x 5) 4x 5x + ] 5 8 4x 5x + = 8 [(8x 5) 4x 5x + ] x 5x + So 3x 4x + = (4x 5x + ) x 5x (I) Consider 4x 5x + = x 5x = (x 5 8 ) = = (x 5 8 ) 9 64 (x 5 8 ) ( 3 8 ) We use te standard formula x a = x x a a log( x + x a ) + c log a 4x 5x + = (x 5 8 ) ( 3 8 ) = ( (x 5 8 ) (x 5 8 ) ( 3 8 ) 9 log( (x 5 8 ) + (x 5 8 ) ( 3 ) ) 8 ) c 8 Using tis in (I), we get 3x 4x + = (4x 5x + ) ( (x 5 8 ) 6. Evaluate te following integrals. a) b) +x x (x )(x 3) Ans: a) + x x (x 5 8 ) ( 3 8 ) 9 log( (x 5 8 ) + (x 5 8 ) ( 3 ) ) 8 ) c 8 = x x = 5 4 (x x + 4 ) = ( 5 ) (x ) F.Y.B.Sc. Calculus Practical, Page: 5

26 F.Y.B.Sc. Calculus Practical (Academic Year 06-7) Terefore +x x = ( 5 = ) (x ) 5 5 log +(x ) 5 (x ) + c 5 log 5+ x + c = 5 +x b) (x )(x 3) = x 5x + 6 = x 5x = (x 5 ) ( ) Terefore, (x )(x 3) = (x 5 ) ( ) = log (x (x 5) + 5 ) ( ) + c = log (x 5) + x 5x c 7. Evaluate x 3 +4x +7x+ x +3x+ Ans: x + Terefore x + 3x + ) x 3 + 4x + 7x + x 3 3x x x + 5x + x 3x x x 3 + 4x + 7x + = (x + 3x + )(x + ) + x = x3 +4x +7x+ x +3x+ = x + + x x +3x+ = x + + x+3 x +3x+ 3 x +3x+ F.Y.B.Sc. Calculus Practical, Page: 6

27 F.Y.B.Sc. Calculus Practical (Academic Year 06-7) = x + + x+3 x +3x+ 3 x +3x = x + + x+3 x +3x+ 3 (x+ 3 ) ( ) Tus x 3 +4x +7x+ x +3x+ = x + + x+3 x +3x+ 3 (x+ 3 ) ( ) Tis implies tat x 3 +4x +7x+ x +3x+ = x + x + log x + 3x + 3 = x log + x + log x + 3x + 3 log x+ x+ + c x+ 3 x c F.Y.B.Sc. Calculus Practical, Page: 7

28 F.Y.B.Sc. Calculus Practical (Academic Year 06-7) Practical 6: Differential Equations Solve te following differential equations.. dy = 4x 3y 3x y Ans: Put y = vx. Ten dy = v + x dy = v + x dy = 4x 3vx 3x vx = v + x dv = 4 3v 3 v = x dv = 4 3v 3 v v = x dv = 4 6v+v 3 v Using variable seperable form 3 v dv = 4 6v+v x d (4 6v + dv v ) = 6 + 4v = (3 v) = (3 v) (4 6v+v ) dv = x = ( 6+4v dv = ) 4 6v+v x = = 6+4v dv = v 6v+4 x log(v 6v + 4) = log x log c = log(v 6v + 4) = log x + log c = log(v 6v + 4) = log x c = v 6v + 4 = x c Replacing v by y x ( y x ) 6 y x + 4 = c x Multiplying by x on bot sides, we get y 6xy + 4x = c. ( y sin( y x ) + x) x sin( y x ) dy = 0 F.Y.B.Sc. Calculus Practical, Page: 8

29 F.Y.B.Sc. Calculus Practical (Academic Year 06-7) Ans: y sin( y x ) + x = x sin( y x ) dy y sin( y x )+x x sin( y dy )= x y x + sin( y x ) = dy Put y = vx. dy = v + x dy v + sin v = v + x dv = x dv sin v Using variable seperable form = sin vdv x = x = sin vdv = log x = cos v + c Replacing v by y x, log x = cos( y x ) + c 3. x dy y = x y Ans: x dy y = x y = x dy = y + x y = dy = y+ Put y = vx dy = v + x dv x y x = v + x dv = vx+ x v x x = v + x dv = v + v = x dv = v Using variable seperable form, v dv = x F.Y.B.Sc. Calculus Practical, Page: 9

30 F.Y.B.Sc. Calculus Practical (Academic Year 06-7) By taking integration of bot sides v dv = x sin v = log x + c By replacing v = y x sin y x = log x + c 4. Sow tat te differential equations is (x 3 xy y + 3) (x y + x)dy = 0 exact and solve it. Ans: M = x 3 xy y + 3 and N = x y x = M y N = xy and = xy x = M y = N x Terefore te given differential equation is exact. Te solution is given by y const = M + (terms in N free from x) dy = c y const (x 3 xy y + 3)+ 0dy = c = x4 4 x y xy + 3x = c = x4 x y xy + 3x = c = x 4 x y 4xy + 6x = c 5. Find order and degree of te following differential equations. a) dy + xy = x3 b) Ans: a) dy + xy = x3 d y + ( dy ) + 5y = 0 c) ( + ( dy )) 3 = d y Te order is and degree is. b) d y + ( dy ) + 5y = 0 Te order is and degree is. c) ( + ( dy )) 3 = d y Te order is and degree is. F.Y.B.Sc. Calculus Practical, Page: 30

31 F.Y.B.Sc. Calculus Practical (Academic Year 06-7) 6. Wic of te following functions are omogeneous? a) f(x, y) = x +y b) g(x, y) = x + y c) (x, y) = x +xy Ans: a) f(tx, ty) = t x + t y = t (x + y ) = t f(x, y). Terefore f is omogeneous function of degree. b) g(tx, ty) = t x + t y = t (x + y ) = t x + y = tg(x, y). Terefore g is omogeneous function of degree. c) (tx, ty) = t x + txt y = t (x + txy ) t (x, y). Terfore tis is not omogeneous differential equation. 7. Solve te differential equation dy = x+y x+y 3 Ans: Compairing given differential equation wit dy, a =, b =, c = 0, c = but a a = b b = dy = = and dy dy = dy dx Terefore given differential becomes dy dx = dy dx = X+Y +(+k) X+Y +(+k 3) = a x+b y+c a x+b y+c We get a =, b = Terefore put x = X +, andy = Y + k = dx = = X++Y +k X++(Y +k) 3 Put + k = 0 and + k 3 = 0. We obtain by solving = 3 and k = 3. Terefore dy = X+Y dx X+Y Put Y = V X. Ten dy dx = V + X dv dx V + X dv dx = X+V X X+V X =!+V +V X dv dx = +V +V = V = +V V V +V = X dv = V dx +V = +V V dv = X dx = dv + V dv = dx V V X = ( V ) dv dv = dx ( V ) V X F.Y.B.Sc. Calculus Practical, Page: 3

32 F.Y.B.Sc. Calculus Practical (Academic Year 06-7) ( 4V ) = dv dv = dx ( V ) V X By taking integration of bot sides, we get dv ( 4V ) dv = ( V ) V dx X ] [log( V V + ) log( V ) = log X Multiplying by 4 on bot sides, we get ] [log( V V + ) log( V ) = 4 log X ] = [log( V V + ) = log X 4 + log C ( V ) = V V + ( V ) = X 4 C = v v+ ( V ) (+ V ) = X 4 C = v v+ ( V ) (+ V ) = X 4 C = ( V +)( V )(+ V ) = X 4 C = (V )(+ V ) = X 4 C Put V = Y X ( Y X )(+ Y X ) = X4 C On Simplifying, we get (Y X )(X + Y ) = C Replace X by x + 3 and Y by y 3 ((y 3) (x + 3) )((x + 3) + (y 3)) = C = d On Simplifying we get, (y y x 6x + 9)(x + y + 3( )) = d 8. Solve ( + x ) dy + xy = 0 Ans: ( + x ) dy + xy = 0 = dy + x y = +x +x Compairing given differential equation wit dy + P (x)y = Q(x) F.Y.B.Sc. Calculus Practical, Page: 3

33 F.Y.B.Sc. Calculus Practical (Academic Year 06-7) We get P (x) = x and Q(x) = +x +x Terefore integrating factor is I.F. = e P (x) x = e +x = e log(+x) = ( + x ) Terefore solution is y.i.f. = (I.F.)Q(x) = y.( + x ) = ( + x ) +x = y.( + x ) = = y.( + x ) = x + c = y = x+c +x 9. Solve ( + y ) = (tan y x)dy Ans: ( + y ) = (tan y x) dy ( + y ) dy = (tan y x) + x = tan y dy +y +y Comparing given differential equation wit + P (y)x = Q(y) dy We get P (y) = and Q(y) = tan y +y +y Terefore integrating factor is I.F. = e P (y)dy = e +y dy = e tan y Terefore solution is x.i.f. = (I.F.)Q(y)dy x.i.f. = e tan y ( tan y +y ) dy Put tan = y = sec udu = dy Terefore xe tan y = e ( ) tan tan u tan tan u +tan u sec udu = xe tan y = e u udu Using product rule for integration we get = xe tan y = u e u du u u ( e u du)du F.Y.B.Sc. Calculus Practical, Page: 33

34 F.Y.B.Sc. Calculus Practical (Academic Year 06-7) = xe tan y = u e u du ( e u du)du = xe tan y = ue u e u + C By repacing u by tan y we get = xe tan y = tan ye tan y e tan y + C = x = tan y + C e tan y = x = tan y + Ce tan y 0. Solve (x + y 3 ) dy = y Ans: (x + y 3 ) dy = y = y dy = x + y3 = dy x y = y... (I) Te above equation is of te form dy of y. + P.x = Q were P and Q are functions Terefore I.F. = e P dy = e y dy=e log y = y Multiplying equation (I) by I.F., ( x) = y dy y y y = d dy (x y ) = y Integrating w.r.t. y x y = ydy + c = x y = y + c = x = y(c + y ) Tis is te general solution of given differential equation.. Find ortogonal trajectories of te family x + y = a Ans: Differentiate te family x + y = a wit respect to x = x + y dy = 0 F.Y.B.Sc. Calculus Practical, Page: 34

35 F.Y.B.Sc. Calculus Practical (Academic Year 06-7) = x + y dy = 0 = dy = x y By replacing dy = dy = x y = dy = x y by dy we get By using variable separable form = y dy = x Taking integration on bot sides y dy = x log y = log x + log c = log y = log cx = y = cx Terefore ortogonal trajectories of te family x + y = a is te family of straigt line pasiing troug origine i.e. y = cx were c is a slope of te line F.Y.B.Sc. Calculus Practical, Page: 35

36 F.Y.B.Sc. Calculus Practical (Academic Year 06-7) Practical 7: Miscellaneous. By knowing te grap of f(x) = e x, draw te graps of a) f(x) = e x b) f(x) = e x Ans: a) Grap of f(x) mirror image of Grap of f(x) about X axis = Grap of e x mirror image of Grap of e x about X axis b) Grap of f( x) mirror image of Grap of f(x) about Y axis = Grap of e x mirror image of Grap of e x about Y axis F.Y.B.Sc. Calculus Practical, Page: 36

37 F.Y.B.Sc. Calculus Practical (Academic Year 06-7). By knowing te grap of f(x) = sin x, draw te graps of a) f(x) = sin(x) + b) f(x) = sin(x) c) f(x) = sin(x ) d) f(x) = sin(x + ) Ans: a) Grap of f(x) + b Grap of f(x) displaced along Y axis by amount b = Grap of sin(x) + Grap of sin x displaced along Y axis by amount b) Grap of f(x) + b Grap of f(x) displaced along Y axis by amount b = Grap of sin(x) Grap of sin x displaced along Y axis by amount c) Grap of f(x a) Grap of f(x) displaced along X axis by amount a F.Y.B.Sc. Calculus Practical, Page: 37

38 F.Y.B.Sc. Calculus Practical (Academic Year 06-7) = Grap of sin(x ) Grap of sin x displaced along X axis by amount d) Grap of f(x a) Grap of f(x) displaced along X axis by amount a = Grap of sin(x + ) Grap of sin x displaced along X axis by amount 3. Evaluate te integral x +3 (x )(x +4) Ans: Partial fractions are x +3 = (x )(x +4) A + Bx+c x x (I) = x + 3 = A(x + 4) + (Bx + C)(x ) = x + 3 = (A + B)x + (C B)x + (4A C) F.Y.B.Sc. Calculus Practical, Page: 38

39 F.Y.B.Sc. Calculus Practical (Academic Year 06-7) = A + B =, C B = 0, 4A C = 3 = B = C, A + C =, 4A C = 3 = 5A = 4, C = 5 = B = A = 4 5, B = 5, C = 5 So from (I), x +3 (x )(x +4) = ( (x+) ) x x +4 = x 5 = 4 5 = 4 5 = 4 5 log x + 0 x+ x 4 ( x x +4 + x +4 ) log x + x + 0 x +4 5 x +4 log x + 0 log x tan x + c 4. If y = (sin x), sow tat ( x )y n+ (n + )xy n+ n y n = 0 Ans: y = (sin x) = y = sin x x = ( x )y 4 sin x = 0 = ( x )y 4y = 0 = ( x )y y xy 4y = 0 = y [( x )y xy ] = 0 = ( x )y xy = 0 Differentiating n times by using Leibnitz s Teorem, we get [ n C 0 ( x )y n+ + n C ( x)y n+ + n C ( )y n ] [ n C 0 xy n+ + n C y n ] = 0 = ( x )y n+ nxy n+ n(n )y n xy n+ ny n = 0 = ( x )y n+ (n + )xy n+ n y n = 0 5. If y = tan x, sow tat ( + x )y n+ + [(n + )x ]y n+ + n(n + )y n = 0 F.Y.B.Sc. Calculus Practical, Page: 39

40 F.Y.B.Sc. Calculus Practical (Academic Year 06-7) Ans: y = tan x = y = e tan x +x = ( + x )y = y Differentiating (n + ) times by using Leibnitz s Teorem, we get [( + x )y ] n+ = y n+ = (n+) C 0 ( + x )y n+ + (n+) C y n+ (x) + (n+) C y n () = y n+ = ( + x )y n+ + (n + )xy n+ + (n+)n. y n () = y n+ = ( + x )y n+ + [(n + )x ]y n+ + n(n + )y n = 0 6. Solve (x y + 3) (x y + )dy = 0. Given tat y = wen x = 0. Ans: (x y + 3) (x y + )dy = 0 = (x y + )dy = (x y + 3) = dy = x y+3 x y+ were v = x y Now, v = x y = dv = dy = dv = dy = (x y)+3 (x y)+ = v+3 v+... (I) = dv = v+3 v+ from (I) = v+3 v+ = dv = dv = v+ v 3 = v+ v+ v+ = v+ dv = v+ = v+ dv = + c v+ = ( ) dv = x + c v+ = v log v + = x + c F.Y.B.Sc. Calculus Practical, Page: 40

41 F.Y.B.Sc. Calculus Practical (Academic Year 06-7) = x y log x y + = x + c = x y log x y + = c... (II) Tis is te general solution. To find particular solution, y = wen x = 0. Terefore (0) log 0 + = c = = c = c = = x y log x y + = from (II) = x y log x y + + = 0 F.Y.B.Sc. Calculus Practical, Page: 4