Tiêu đề Antiderivatives And Its Application. Grade 11 Mathematics Additional Questions _ Solutions _ Khullakitab.pdf ✅

Grade 11 Mathematics Solution (h (http://www.khullakitab.com/) Back to Solutio (mathematics/publicat 11/12/solution  1. 2. 3. 4. 5. 6. 7. Chapters Sets,Real Number System and Logic. (setsreal-number- system-and- logic-/solution/grade- 11/mathematics/6/solutions) Relations,Functions and Graphs. (relationsfunctions- and-graphs-/solution/grade- 11/mathematics/42/solutions) Curve sketching. (curve- sketching-/solution/grade- 11/mathematics/1005/solutions) Trigonometric Equations and General Values. (trigonometric- equations-and-general- values-/solution/grade- 11/mathematics/47/solutions) Inverse Circular Functions. (inverse-circular- functions-/solution/grade- 11/mathematics/50/solutions) Properties of Triangle. (properties-of- triangle-/solution/grade- 11/mathematics/52/solutions) Solution of Triangle. (solution-of- triangle-/solution/grade- 11/mathematics/54/solutions) Antiderivatives and Its Application. Additional Questions Previous (antiderivatives-and-its-application-/solution/grade-11/mathematics/1069/solutions) Next (complex-number-/solution/grade-11/mathematics/1071/solutions) Exercise 19.1 (antiderivatives-and-its-application--exercise-19-1/solution/gr Exercise 19.2 (antiderivatives-and-its-application--exercise-19 Exercise 19.3 ade-11/mathematics/10 (antiderivati Exercise 19.4 (antiderivatives-and-its-application--exercise-19-4/solution/gr Exercise 19.5 (antiderivatives-and-its-application--exercise-19 Additional Questions ade-11/mathematics/10 (anti Share 10 Tweet Like 10 Share 13 1. Soln: Actually, antiderivatives is the reserve of derivative. If f(x) and F(x) are the functions of x such that (F(x)) then the antiderivative of f(x) with respect to x is th function F(x) and is written as: Or, .dx = F(x) If c is constant quantity and derivative of a constant quantity is zero, then, Or, (F(x) + c) = f(x) Hence, .dx = F(x) + c. Since, c is an unknown constant and it may be any real number, so there are a number of antiderivatives of function whoch are different from one another by a constant. Let us see the following example: If y = x be the given function then, Or, (x ) = 2x, then dx = x Similarly, (x + 1) = 2x, then + 1 Or, = 2x, then + . If a is constant, (x + a) = 2x, then = x + a and so on. Above examples show that there are a number of antiderivatives of 2x which are different from one another by constant. d dx ∫ f(x) d dx ∫ f(x) 2 d dx 2 ∫ 2x. 2 d dx 2 ∫ 2x.dx = x 2 ( + ) d dx x 2 3 5 ∫ 2x.dx = x 2 3 5 d dx 2 ∫ 2x.dx 2 Go Top  Home (http://www.khullakitab.com/) Grade 11 (grade-11/subjects/solutions) Mathematics (mathematics/publications/grade-11/12/solutions) Sukunda Pustak Bhawan Antiderivatives and Its Application.    
8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. Sequence and Series and Mathematical Induction. (sequence-and-series-and- mathematical- induction-/solution/grade- 11/mathematics/55/solutions) Matrices and Determinants. (matrices-and- determinants-/solution/grade- 11/mathematics/60/solutions) System of Linear Equations. (system-of-linear- equations-/solution/grade- 11/mathematics/1060/solutions) Complex Number. (complex- number-/solution/grade- 11/mathematics/1061/solutions) Polynomial Equations. (polynomial- equations-/solution/grade- 11/mathematics/64/solutions) Straight Line. (straight- line-/solution/grade- 11/mathematics/68/solutions) Pair of Straight Lines. (pair- of-straight- lines-/solution/grade- 11/mathematics/1059/solutions) The circle. (the- circle-/solution/grade- 11/mathematics/71/solutions) Limits and Continuity. (limits-and- continuity-/solution/grade- 11/mathematics/73/solutions) The Derivatives. (the- derivatives-/solution/grade- 11/mathematics/79/solutions) Application of Derivatives. (application-of- derivatives-/solution/grade- 11/mathematics/1062/solutions) Antiderivatives and Its Application. 2. Soln: If f(x) and F(x) are two functions such that (F(x)) = f(x) then (F(x) + c) = f(x) as derivative of a constant ter is zero, so Or, .dx = F(x) + c Where, c is an arbitrary constant. Since this constant is the unknown quantity, so it is known as the indefini integral. The integral of a function independent of arbitrary constant is known as definite integral. The indefinite integral contains an arbitrary constant where definite integral does not. .dx is the definit integral of f(x) w.r.t. ‘x’ between the limits a and b. b is the upper limit ad a is the lower limit. This gives the are bounded by the curve y = f(x), x – axis and the two ordinates x = a and x = b. 3. (i) Soln: = .dx = .dx = .dx = + 3logx + + c. = x + x + 3logx – + c. (ii) Soln: = .dx = .dx = .dx = .dx = + c = (6x + 8) + c = (6x + 8) + c. = (36x + 55)(6x + 8) + c. (iii) Soln: = .dx = .dx = ..dx = .dx = + c = + c. = + c = + c = + c. (iv) d dx d dx ∫ f(x) ∫ f(x) a b ∫ (x + ) 1 x 3 ∫(x + 3 . + 3x. + ) 3 x 2 1 x 1 x 2 1 x 3 ∫(x + 3x + 3. + ) 3 1 x x−3 + x 4 4 3x 2 2 x−3+1 −2 1 4 4 3 2 2 1 2x 2 ∫(2x + 3)(6x + 8) 5 ∫(6x + 9) 1 3 (6x + 8) 5 ∫{(6x + 8) + 1} 1 3 (6x + 8) 5 ∫ {6x + 8 + } 1 3 ) 6 (6x + 8) 5 { + } 1 3 (6x+8) 7 7∗6 (6x+8) 6 6∗6 1 18 6{ + } 6x+8 7 1 6 1 18 6 36x+48+7 42 1 756 6 ∫ x+1 (5x+2) 3 ∫ 1 5 5x+5 (5x+2) 3 ∫ 1 5 (5x+2)+3 (5x+2) 3 ∫{ + } 1 5 5x+2 (5x+2) 3 3 (5x+2) 3 { + 3 } 1 5 (5x + 2)−2 (5x + 2)−3 1 5 { + 3. } (5x+2)−2+1 −1∗5 (5x+2)−3+1 −2∗5 − 1 25 { + } 1 5x+2 3 2(5x+2) 2 − { } 1 25 2(5x+2)+3 2(5x+2) 2 − 1 50 10x+7 (5x+2) 2 Go Top 
Soln: = .dx = .dx = .dx = 2 .dx = 2 .dx = 2(sinx + x.cosα) + c. (v) Soln: = .dx = .dx = .dx = .dx = xtanα + c. (vi) Soln: = .dx = .dx = .dx = .dx = .dx – .dx = e – x + c. (vii) Soln: = .dx = .dx = .dx = .dx = .dx (sec x – tan x = 1) = tanx – secx + c. (viii) Soln: = .dx = .dx = .dx = a.secx – b(cosecx) + c = a.secx – b.cosecx + c. 4. (i) Soln: = .dx = .dx = .dx ∫ cos2x−cos2α cosx−cosα ∫ 1 5 (2cos x−1)−(2 α−1) 2 cos2 (cosx−cosα) ∫ 2(cos x− α) 2 cos2 cosx−cosα ∫ (cosx+cosα)(cosx−cosα) cosx−cosα ∫(cosx + cosα) ∫ cos(x−α)−−−cos(x+α) sin(x+α)+sin(x−α) ∫ 2.sinx.sinα 2.sinx.cosα ∫ tanα ∫ 1 ∫ e +1 2x e +1 x ∫ 2(e x ) − 2 (1) 2 e +1 x ∫ (e −1)( +1) x e x e +1 x ∫(e − 1) x ∫ e x ∫ 1 x ∫ secx secx+tanx ∫ secx(secx−tanx) (secx+tanx)(secx−tanx) ∫ sec x−secx.tanx 2 sec x− x 2 tan2 ∫ sec x 2 ∫ secx.tanx 2 2 ∫ a.sin x+b. x 3 cos3 sin x. x 2 cos2 ∫( + ) asin x 3 sin x. x 2 cos2 b.cos x 3 sin x. x 2 cos2 ∫(a.secx.tanx + b. cosecx. cotx) ∫ 2x+3 √3x+1 ∫ 1 3 6x+9 √3x+1 ∫ 1 3 (6x+2)+7 (3x+1) 1 2 Go Top 
= .dx = .dx = + c = + c = (3x + 1) + c = (6x + 23)(3x + 1) + c. (ii) Soln: Or, .dx = .dx = .dx = .dx = .dx = .dx = (4x + 3) . + c = (42x + 29)(4x + 3) + c. (iii) Soln: Or, .dx = .dx = .dx = .dx = .dx = .dx = (3x + 2) . + c = .(9x + 5)(3x + 2) + c. (iv) Soln: Or, .dx = .dx = .dx = .dx = .dx = + c. = (5x + 1) . + c = (45x + 44)(5x + 1) + c. (v) ∫{ + } 1 3 2(3x+1) (3x+1) 1 2 7 (3x+1) 1 2 1 3 ∫{2(3x + 1) + 7 } 1 2 (3x + 1)− 1 2 [ + ] 1 3 2(3x+1) 3 2 ∗3 3 2 7(3x+1) 1 2 ∗3 1 2 [ + ] 1 3 4 9 (3x + 1) 3 2 14 3 (3x + 1) 1 2 2 9 1/2[ (3x + 1) + 7] 2 3 2 27 1/2 ∫(7x + 5). 4x + 3 −−−−− √ ∫(28x + 20) 1 4 4x + 3 −−−−− √ 1 4 ∫{(28x + 21) − 1} 4x + 3 −−−−− √ 1 4 ∫{7 (4x + 3) − 1} (4x + 3) 1 2 −−−−−−−− √ 1 4 ∫{7(4x + 3) − } 3 2 (4x + 3) 1 2 1 4 { − } 7(4x+3) 5 2 ∗4 5 2 (4x+3) 3 2 ∗4 3 2 1 8 3/2 { − } 7(4x+3) 5 1 3 1 60 3/2 ∫(5x + 3). 3x + 2 −−−−− √ ∫(15x + 9) 1 3 3x + 2 −−−−− √ 1 3 ∫{(15x + 10) − 1}(3x + 2) 1 2 1 2 ∫{5 (3x + 2) − 1}(3x + 2) 1 2 1 3 ∫{5(3x + 2) − } 3 2 (3x + 2) 1 2 1 3 { − } 5(3x+2) 5 2 ∗3 5 2 (3x+2) 3 2 ∗3 3 2 2 9 3/2 [(3x + 2) − ] 1 3 2 27 3/2 ∫(3x + 2). 5x + 1 −−−−− √ ∫(15x + 10) 1 5 5x + 1 −−−−− √ 1 5 ∫{(15x + 3) + 7} 5x + 1 −−−−− √ 1 5 ∫{3 (5x + 1) + 7}(5x + 1) 1 2 1 5 ∫{3(5x + 1) − 7 } 3 2 (5x + 1) 1 2 1 5 [ + ] 3(5x+1) 5 2 ∗5 5 2 7(5x+1) 3 2 ∗5 3 2 2 25 3/2 { + } 3(5x+1) 5 7 3 2 375 3/2 Go Top 