RD Sharma Class 12 Ex 19.2 Solutions Chapter 19 Indefinite Integrals

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter19
Exercise19.2
CategoryRD Sharma Solutions

RD Sharma Class 12 Ex 19.2 Solutions Chapter 19 Indefinite Integrals

Question 1. Evaluate (∫(3x√5 + 4√x + 5)dx

Solution:

We have, (∫(3x√5+4√x+5)dx

= ∫3x√5 dx + ∫4√x dx + ∫5dx 

= ∫3x3/2dx + 4∫x1/2 dx + 5∫dx

= x(3/2)+1/(3/2 + 1) + 4x(1/2)+1/(1/2 + 1) + 5x + c

= 6/5 x5/2 + 8/3 x3/2 + 5x + c

Question 2. Evaluate ∫(2+ 5/x – 1/x1/3)dx

Solution:

We have, ∫(2+ 5/x – 1/x1/3)dx

= ∫2xdx + 5∫1/x dx – ∫1/x1/3 dx

= 2x/(log⁡2) + 5log⁡x – 3/2 x2/3 + c

Question 3. Evaluate ∫{√x (ax+ bx + c)}dx

Solution:

We have, ∫{√x (ax+ bx + c)}dx

= ∫√x × ax2 dx + ∫√x × bx dx + ∫c√x dx

= ∫ax5/2 dx + ∫bx3/2dx + ∫cx1/2 dx

= (ax(5/2)+1)/(5/2 + 1) + (bx(3/2)+1)/(3/2 + 1) + (cx(1/2)+1)/(1/2 + 1) + d

= (2ax7/2)/7 + (2bx5/2)/5 + (2cx3/2)/3 + d

Question 4. Evaluate ∫(2 – 3x) (3 + 2x)(1 – 2x)dx

Solution:

We have, ∫(2 – 3x) (3 + 2x)(1 – 2x)dx

= ∫(6 + 4x – 9x – 6x2)(1 – 2x)dx

= ∫(-6x– 5x + 6)(1 – 2x)dx

= ∫(-6x+ 12x– 5x + 10x+ 6 – 12x)dx

= ∫(4x+ 12x– 17x + 6)dx

= ∫(12x+ 4x– 17x + 6)dx 

= 12/4 x+ 4/3 x– 17/2 x+ 6x + c 

= 3x+ 4/3 x– 17/2 x+ 6x + c

Question 5. Evaluate ∫(m/x + x/m + m+ x+ mx)dx

Solution:

We have, ∫(m/x + x/m + m+ x+ mx)dx

= m∫1/x dx + 1/m ∫xdx + ∫mxdx + ⌋xmdx + m∫xdx

= mlog⁡|x| + x2/2m + mx/(log⁡m) + xm+1/(m + 1) + (mx2)/2 + c

Question 6. Evaluate ∫ (√x – 1/√x)2 dx

Solution:

We have, ∫ (√x – 1/√x)2 dx

By using formula (x + y)2 = x2 + y2 +2xy 

We get,  ∫(x + 1/x – 2)dx

= ∫xdx + ∫1/x dx – 2∫1.dx

= x2/2 + log⁡|x| – 2x + C

Question 7. Evaluate ∫((1 + x)3)/√xdx 

Solution:

We have, ∫((1 + x)3)/√xdx 

By using formula (x + y)3 = x3 + y3 +3x2y + 3xy2

We get, ∫(1 + x+ 3x+ 3x)/√x dx

= 1/√x dx + ∫x3/√x dx + ∫(3x2)/√x dx + ∫3x/√x dx

= ∫x-1/2 dx + ∫x5/2 dx + 3∫x3/2 dx + 3∫x1/2 dx

= x(-1/2)+1/((-1)/2 + 1) + (x(5/2)+1)/(5/2 + 1) + (3x(3/2)+1)/(3/2 + 1) + 3 x(1/2)+1/(1/2 + 1) + c

= x1/2/(1/2) + x7/2/(7/2) + (3x5/2)/(5/2) + 3 x3/2/(3/2) + c

= 2x1/2 + 2/7 x7/2 + 6/5 x5/2 + 6/3 x3/2 + c

= 2x1/2 + 2/7 x7/2 + 6/5 x5/2 + 2x3/2 + c

 Question 8. Evaluate ∫{x+ elogx + (e/2)x }dx

Solution:

We have, ∫{x+ elogx + (e/2)x }dx

= ∫x2 dx + ∫elogx dx + ∫(e/2)x dx

= x3/3 + ∫xdx + ∫(e/2)xdx

= x3/3 + x2/2 + 1/(log⁡(e/2)) × (e/2)+ c

Question 9. Evaluate ∫ (x+ e+ ee)dx

Solution:

We have, ∫ (x+ e+ ee)dx                             

= ∫xe dx + ∫exdx + ∫eedx

= xe+1/(e + 1) + e+ eex + c

Question 10. Evaluate ∫√x (x– 2/x)dx 

Solution:

We have, ∫√x (x– 2/x)dx 

= ∫ x7/2 dx – 2∫ x-1/2 dx

= x(7/2)+1/(7/2 + 1) – 2 x(-1/2)+1/((-1)/2 + 1) + c

= x9/2/(9/2) – (2x-1/2)/((-1)/2) + c

= 2/9 x9/2 – 4x-1/2 + c

Question 11. Evaluate ∫1/√x (1 + 1/x)dx

Solution:

We have, ∫1/√x (1 + 1/x)dx

= ∫ (1/√x + 1/(√x × x))dx

= ∫x-1/2 + ∫x-3/2 dx

= 2x1/2 – 2x-1/2 + c

Question 12. Evaluate ∫(x+ 1)/(x+ 1) dx

Solution:

We have, ∫(x+ 1)/(x+ 1) dx

= ∫((x2)+ (1)3)/(x+ 1) dx

= ∫(x+ 1)(x+ 1 – x2)/(x+ 1) dx

= ∫(x– x+ 1)dx

= ∫x4dx – ∫x2dx + ∫1dx

= x5/5 – x3/3 + x + c

Question 13. Evaluate ∫ (x-1/3 + √x + 2)/∛x dx

Solution:

We have, (x-1/3 + √x + 2)/∛x dx

= ∫(x-1/3 dx)/x1/3 + ∫x1/2/x1/3dx + ∫2/x1/3dx

= ∫x-2/3 dx + ∫x1/6 dx + 2∫ x-1/3dx

= 3x1/3 + 6/7 x7/6 + 3x2/3 + c

Question 14. Evaluate ∫((1 + √x)2)/√x dx

Solution:

We have, ∫((1 + √x)2)/√x dx

= ∫(1 + x + 2√x)/x1/2dx

= ∫x-1/2+∫ x1/2 dx + 2∫dx

= 2x1/2 + 2/3 x3/2 + 2x + c

= 2√x + 2x + 2/3 x3/2 + c

Question 15. Evaluate ∫√x(3 – 5x)dx

Solution:

We have, ∫√x(3 – 5x)dx

= 3∫√x dx – 5x3/2 dx

= 3x3/2/(3/2) – 5 x5/2/(5/2) + c

= 2x3/2 – 2x5/2 + c

Question 16. Evaluate ∫((x + 1)(x – 2))/√x dx

Solution:

We have, ∫((x + 1)(x – 2))/√x dx

= ∫(x– 2x + x – 2)/x1/2dx

= ∫(x– x – 2)/x1/2dx

= ∫x2/x1/2dx – ∫x1/2dx – 2∫x-1/2dx

= (2x5/2)/5 – (2x3/2)/3 – 4x1/2 + c

= 2/5 x5/2 – (2x3/2)/3 – 4√x + c

Question 17. Evaluate ∫(x+ x-2 + 2)/x2dx

Solution:

We have, ∫(x+ x-2 + 2)/x2dx

= ∫(x5/x2 +x-2/x2 +2/x2)dx

= ∫x3dx + ∫x-4 + 2∫x-2 dx

= x4/4 + x-3/(-3) + (2x-1)/(-1) + c

= x4/4 – x-3/3 – 2/x + c

Question 18. Evaluate ∫(3x + 4)2 dx

Solution:

We have, ∫(3x + 4)2 dx

By using formula (x + y)2 = x2 + y2 +2xy 

We get, ∫ (9x+ 16 + 24x)dx

= ∫9x2 dx + ∫16dx + ∫24xdx

= 9 x3/3 + 16x + 24 x2/2 + c

= 3x+ 16x + 12x+ c

Question 19. Evaluate ∫(2x+ 7x+ 6x2)/(x+ 2x) dx

Solution:

We have, ∫(2x+ 7x+ 6x2)/(x+ 2x) dx

= ∫x(2x+ 7x+ 6x)/(x(x + 2))dx

= ∫(2x+ 7x+ 6x)/(x + 2)dx

= ∫ (2x+ 4x+ 3x+ 6x)/((x + 2))dx

= ∫(2x2(x + 2) + 3x(x + 2))/(x + 2) dx

= ∫(x + 2)(2x+ 3x)/(x + 2) dx

= ∫(2x+ 3x)dx

= ∫2x2 dx + ∫3xdx

= 2/3 x+ 3/2 x+ c

Question 20. Evaluate ∫(5x+ 12x+ 7x2)/(x+ x) dx

Solution:

We have, ∫(5x+ 7x+ 5x+ 7x2)/(x+ x) dx

= ∫(5x+ 7x+ 5x+ 7x)/(x + 1) dx

= ∫5x2 (x + 1) + 7x(x + 1)/(x + 1) dx

= ∫(5x+ 7x)dx

= (5x3)/3 + (7x2)/2 + C

Question 21. Evaluate ∫(sin2x)/(1 + cos⁡x) dx

Solution:

We have, ∫(sin2x)/(1 + cos⁡x) dx

= ∫(1 – cos2⁡x)/(1 + cos⁡x) dx

= ∫((1 – cos⁡x)(1 + cos⁡x))/(1 + cos⁡x) dx

= ∫(1 – cos⁡x)dx

= x – sin⁡x + c

Question 22. Evaluate ∫(sec2x + cos⁡ec2 x)dx

Solution:

We have, ∫(sec2x + cos⁡ec2 x)dx

= ∫ sec2xdx + ∫ cosec2⁡xdx

= tan⁡x – cot⁡x + c

 = tan⁡x – cot⁡x + c

Question 23. Evaluate ∫(sin3⁡x – cos3x)/(sin2⁡xcos2x) dx

Solution:

We have, ∫(sin3⁡x – cos3x)/(sin2⁡xcos2x) dx

= ∫((sin3⁡x)/(sin⁡2x cos2⁡x) – (cos3x)/(sin⁡2x cos2x))dx 

= ∫ (sin⁡xsec2x – cos⁡xcos⁡ec2x)dx 

= ∫ (tan⁡xsec⁡x – cot⁡xcos⁡ecx)dx

= sec⁡x + cosec⁡x + c

Question 24. Evaluate ∫(5cos3⁡x + 6sin3x)/(2sin2xcos2x) dx

Solution:

We have, ∫(5cos3x + 6sin3x)/(2sin2⁡xcos2x) dx

= ∫(5cos3⁡x)/(2sin2xcos2x) dx + ∫(6sin3⁡x)/(2sin2⁡xcos2⁡x) dx

= 5/2 ∫(cos⁡x)/(sin2⁡x) dx + 3∫(sin⁡x)/(cos2⁡x) dx

= 5/2 ∫cot⁡xcosec⁡xdx + 3∫ sec⁡xtan⁡xdx

= (-5)/2 cosec⁡x + 3sec⁡x+c

= (-5)/2 cos⁡sec⁡x + 3sec⁡x+c

Question 25. Evaluate ∫(tan⁡x + cot⁡x)2 dx
Solution:

We have, ∫(tan⁡x + cot⁡x)2 dx
By using formula (x + y)2 = x2 + y2 + 2xy 
We get, ∫(tan2x + cot2⁡x + 2tan⁡x cot⁡x)dx
= ∫ (sec2⁡x – 1 + cosec2x – 1 + ((2 × 1)/cot⁡x) × cot⁡x)dx
= ∫ (sec2⁡x + cosec2⁡x)dx
= ∫sec2xdx + ∫cosec2⁡xdx
= tan⁡x – cot⁡x + c
Question 26. Evaluate ∫(1 – cos⁡2x)/(1 + cos⁡2x) dx
Solution:
We have, ∫(1 – cos⁡2x)/(1 + cos⁡2x) dx
= ∫(2sin2⁡x)/(2cos2⁡x) dx
= ∫tan2xdx
= ∫(sec2x – 1)dx
= ∫sec2⁡xdx – 1∫dx
= tan⁡x – x + c
Question 27. Evaluate ∫(cos⁡x)/(1 – cos⁡x) dx
Solution:
We have, ∫(cos⁡x)/(1 – cos⁡x) dx
= ∫(cos⁡x(1 + cos⁡x))/((1 – cos⁡x)(1 + cos⁡x)) dx
= ∫(cos⁡x + cos2⁡x)/(1 – cos2x) dx
= ∫(cos⁡x + cos2⁡x)/(sin2⁡x) dx
= ∫(cos⁡x)/(sin2⁡x) dx + ∫(cos2x)/(sin2⁡x) dx            [Since, cosx/sinx = cotx]
= ∫cot⁡x × cosec⁡xdx + ∫(cosec2⁡x – 1)dx                 [Since, cot2x = cosec2x – 1]
= -cosec⁡x – cot⁡x – x + c
Question 28. Evaluate ∫cos2x – sin2⁡x/√(1 + cos⁡4x) dx
Solution:
We have, ∫cos2x – sin2⁡x/√(1 + cos⁡4x) dx
= ∫(cos2⁡x – sin2x)/√(2cos2⁡2x) dx
= 1/√2 ∫(cos2x – sin2⁡x)/(cos⁡2x) dx
= 1/√2∣(cos2x – sin2⁡x)/(cos2⁡x – sin2⁡x) dx
= 1/√2∫1 × dx
= x/√2 + c
Question 29. Evaluate ∫ 1/(1 – cos⁡x) dx
Solution:
We have, ∫ 1/(1 – cos⁡x) dx
= ∫1/(1 – cos⁡x) × (1 + cos⁡x)/(1 + cos⁡x) × dx    
= ∫(1 + cos⁡x)/(1 – cos2x) × dx
= ∫(1 + cos⁡x)/(sin2x) × dx
= ∫1/(sin2x) dx + ∫(cos⁡x)/(sin22⁡x) dx
= ∫cosec2xdx + ∫cot⁡x × cosec⁡x dx
= -cot⁡x – cosec⁡x + c
Question 30. Evaluate ∫1/(1 – sin⁡x) dx
Solution:
We have, ∫1/(1 – sin⁡x) dx
= ∫1/(1 – sin⁡x) × (1 + sin⁡x)/(1 + sin⁡x) × dx
= ∫(1 + sin⁡x)/(1 – sin2⁡x) × dx
= ∫(1 + sin⁡x)/(cos2⁡x) × dx
= ∫(1/(cos2x) + (sin⁡x)/(cos2⁡x)) × dx
= ∫1/(cos2⁡x) dx + ∫(sin⁡x)/(cos2⁡x) × dx
= ∫sec2⁡xdx + ∫tan⁡x sec⁡x dx
= tan⁡x + sec⁡x + c
Question 31. Evaluate ∫(tan⁡x)/(sec⁡x + tan⁡x) dx
Solution:
We have, ∫(tan⁡x)/(sec⁡x + tan⁡x) dx
= ∫(tan⁡x)/(sec⁡x + tan⁡x) × (sec⁡x – tan⁡x)/(sec⁡x – tan⁡x) × dx
= ∫(tan⁡x(sec⁡x – tan⁡x))/(sec2⁡x – tan2⁡x) × dx
= ∫(tan⁡xsec⁡x – tan2⁡x)dx
= ∫sec⁡tan⁡xdx – ∫(sec2x – 1)dx
= ∫sec⁡xtan⁡xdx – ∫sec2⁡xdx + 1∫dx
= sec⁡x – tan⁡x + x + c
Question 32. Evaluate ∫(cosec⁡x)/(cosec⁡x – cot⁡x)dx
Solution:
We have, ∫(cosec⁡x)/(cosec⁡x – cot⁡x)dx
= ∫(cosec⁡x)/(cosec⁡x – cot⁡x) × (cosec⁡x + cot⁡x)/(cosec⁡x + cot⁡x) × dx
= ∫(cosec⁡x(cosec⁡x + cot⁡x))/(cosec2⁡x – cot2x) × dx
= ∫(cosec2⁡x + cosec⁡x cot⁡x)dx
= ∫cos⁡ec2xdx + ∫cosec⁡x cotx dx
= -cot⁡x – cosec⁡x + c
Question 33. Evaluate ∫1/(1 + cos⁡2x) dx
Solution:
We have, ∫1/(1 + cos⁡2x) dx
= ∫ 1/(2cos2⁡x) × dx
= 1/2 ∫sec2⁡x × dx
= 1/2 × tan⁡x + c
= (tan⁡x)/2 + c
Question 34. Evaluate∫1/(1 – cos⁡2x) dx
Solution:
We have, ∫1/(1 – cos⁡2x) dx
= ∫1/(2sin2⁡x)dx
= 1/2 ∫cosec2⁡x dx
= (-1)/2 × cot⁡x + c
= (-cot⁡x)/2 + c
Question 35. Evaluate ∫tan-1⁡[(sin⁡2x)/(1 + cos⁡2x)]dx
Solution:
We have, ∫tan-1⁡[(sin⁡2x)/(1 + cos⁡2x)]dx
= ∫tan-1[(2sin⁡xcos⁡x)/(2cos2⁡x)]dx
= ∫tan-1⁡[(sin⁡x)/(cos⁡x)]dx
= ∫tan-1(tan⁡x)dx
= ∫xdx
= x2/2 + c
Question 36. Evaluate ∫cos-1(sin⁡x)dx
Solution:
We have, ∫cos-1(sin⁡x)dx
= ∫cos-1⁡[cos⁡(π/2 – x)]dx
= ∫(π/2 – x)dx
= π/2 ∫dx – ∫xdx
= π/2 × x – x2/2 + c
Question 37. Evaluate ∫ cot-1⁡(sin⁡x)dx
Solution:
We have, ∫ cot-1⁡(sin⁡x)dx
= ∫cot-1⁡[(sin⁡2x)/(1 – cos⁡2x)]dx
= ∫cot-1⁡((cos⁡x)/(sin⁡x))dx
= ∫cot-1⁡(cotx)dx
= ∫xdx
= x2/2 + c 
Question 38. Evaluate ∫ sin-1⁡((2tan⁡x)/(1 + tan2⁡x))dx
Solution:
We have, ∫ sin-1⁡((2tan⁡x)/(1 + tan2⁡x))dx
= ∫ sin-1⁡(sin⁡2x)dx
= ∫2xdx
= 2∫xdx
= (2x2)/2 + c
= x+ c 
Question 39. Evaluate ∫((x+ 8)(x – 1))/(x– 2x + 4) dx
Solution:
We have, ∫((x+ 8)(x – 1))/(x– 2x + 4) dx
= ∫((x + 2)(x– 2x + 4)(x – 1))/(x– 2x + 4) dx
= ∫(x + 2)(x – 1)dx
= ∫(x– x+2x – 2)dx
= ∫(x+ x – 2)dx
= x3/3 + x2/2 – 2x + c
Question 40. Evaluate ∫(atan⁡x + bcot⁡x)2 dx
Solution:
We have, ∫(atan⁡x + bcot⁡x)2 dx
By using formula (x + y)2 = x2 + y2 + 2xy , we get
= ∫(a2 tan2⁡x + b2cot2x + 2ab tan⁡x cot⁡x)dx
= ∫[a2 (sec2⁡x – 1) + b2(cosec2x – 1) + 2ab]dx
= ∫[a2 sec2x – a+ b2cosec2⁡x – b+ 2ab]dx
= a2tan⁡x – a2x – b2 cot⁡x – b2x + 2abx + c
= a2tan⁡x – b2 cot⁡x – (a+ b– 2ab)x + c
Question 41. Evaluate ∫(x– 3x+ 5x – 7 + x2 ax)/(2x2) dx
Solution:
We have, ∫(x– 3x+ 5x – 7 + x2 ax)/(2x2) dx
= 1/2 ∫x3/x2dx – 3/2∫x2/x2dx + 5/2∫x/x2dx – 7/2∫x-2dx + 1/2∫(x2ax)/x2dx
= 1/2 × x2/2 – 3/2x + 5/2 log⁡x – 7/2 x-1 + 1/2ax/(log⁡a) + c
= 1/2 [x2/2 – 3x + 5log⁡x + 7/x + ax/(log⁡a)] + c
Question 42. Evaluate ∫cos⁡x/(1 + cos⁡x) dx
Solution:
We have, ∫cos⁡x/(1 + cos⁡x) dx  …..(1)
Now solve
\frac{cos⁡x}{(1 + cos⁡x)} = \frac{cos^2\frac{⁡x}{2} - sin^2\frac{⁡x}{2}}{(2cos^2\frac{⁡x}{2})}  
Since, cos⁡x = cos2x/2 – sin2⁡x/2 and cos⁡x + 1 = 2cos2⁡x/2 
So, we get cos⁡x/(1 + cos⁡x) = 1/2[1 – tan2x/2] 
Now put this value in eq(1), we get
= 1/2 ∫(1 – tan2x/2)dx
= 1/2 ∫(1 – sec2⁡x/2 + 1)dx
= 1/2 ∫(2 – sec2⁡x/2)dx
= 1/2 [2x – (tan⁡x/2)/(1/2)] + c
= x – tan⁡x/2 + c
Question 43. Evaluate∫(1 – cos⁡x)/(1 + cos⁡x) dx
Solution:
We have, ∫(1 – cos⁡x)/(1 + cos⁡x) dx  ….(1)
Now solve
(1 – cos⁡x)/(1 + cos⁡x) = (2sin2⁡x)/(2cos2⁡x)
= tan2⁡x/2
= (sec2x/2 – 1)               [Since, 2sin2⁡x/2 = 1 – cos⁡x and 2cos2⁡x/2 = 1 + cos⁡x]
Now put this value in eq(1), we get
= ∫(sec2⁡x/2 – 1)dx
= tan(x/2)/(1/2) – x + c
= 2tan⁡x/2 – x + c
Question 44. Evaluate ∫{3sin⁡x – 4cos⁡x + 5/(cos2x) – 6/(sin2⁡x) + tan2⁡x – cot2⁡x}dx
Solution:
We have, ∫{3sin⁡x – 4cos⁡x + 5/(cos2x) – 6/(sin2⁡x) + tan2⁡x – cot2⁡x}dx
= 3∫sin⁡xdx – 4∫cos⁡xdx + 5∫sec2⁡dx – 6∫cosec2⁡x + ∫tan2⁡xdx – ∫cot2⁡xdx
= 3∫sin⁡xdx – 4∫cos⁡xdx + 5∫sec2⁡xdx – 6∫cosec2⁡x + ∫(sec2⁡x – 1)dx – ∫(cosec2⁡x – 1)dx
= 3∫sin⁡xdx – 4∫cos⁡xdx + 6∫sec2xdx – 7∫cosec2xdx
= -3cos⁡x – 4sin⁡x + 6tan⁡x + 7cot⁡x + c
Question 45. If f'(x) = x – 1/x2 and f(1) = 1/2, find f(x)?
Solution:
Given that ∫f'(x) = x – 1/x2
and f(1) = 1/2
We have to find f(x)
So, ∫f'(x) = ∫xdx – ∫1/x2dx
f(x) = x2/2 + x-1 + c
f(x) = x2/2 + 1/x + c
f(x) = x2/2 + 1/x + c     …..(i)
As we know that 
f(1) = 1/2
12/2 + 1/1 + c = 1/2
1/2 + 1 + c = 1/2
c = -1
On putting c = -1 in (i), we get
f(x) = x2/2 + 1/x – 1
Question 46. If f'(x) = x + b, f(1) = 5, f(2) = 13, find f(x)?
Solution:
 Given that f'(x) = x + b
 and f(1) = 5, f(2) = 13
We have to find f(x)
So, ∫f'(x) = ∫(x + b)dx
 f(x) = x2/2 + bx + c   …….(i)
As we know that 
f(1) = 5
12/2 + b × 1 + c = 5
1/2 + b + c = 5
 b + c = 9/2    …….(ii)
Also, f(2) = 13
22/2 + b × 2 + c = 13
2 + 2b + c = 13
2b + c = 11     …….(iii)
Now, subtract eq(ii) from eq(iii), we get
b = 11 – 9/2
b = 13/2
Now, put b = 13/2 in eq(ii), we get
13/2 + c = 9/2
 c = 9/2 – 13/2
 c = (9 – 13)/2 
= (-4)/2 
= -2
Now, on putting b = 13/2 and c = -2 in equation (i), we get
f(x) = x2/x + 13/2x – 2
f(x) = x2/2 + 13/2x – 2
Question 47. If f'(x) = 8x– 2x, f(2) = 8, find f(x)?
Solution:
Given that f'(x) = 8x– 2x
and f(2) = 8
We have to find f(x)
So, ∫f'(x)dx = ∫(8x– 2x)dx
f(x) = ∫(8x– 2x)dx
= ∫8x3dx – ∫2xdx
= (8x4)/4 – (2x2)/2 + c
= 2x– x+ c
f(x) = 2x– x+ c  ……….(i)
As we know that f(2) = 8
So, f(2) = 2(2)– (2)+ c = 8
32 – 4 + c = 8
28 + c = 8
c = -20
Now, Put c = -20 in eq(i), we get
f(x) = 2x– x– 20
Question 48. If f'(x) = asin⁡x + bcos⁡x and f'(0) = 4, f(0) = 3, f(π/2) = 5, find f(x)?
Solution:
Given that, f'(x) = asin⁡x + bcos⁡x 
and f'(0) = 4, f(0) = 3, f(π/2) = 5
We have to find f(x)
So, 
∫f'(x) = ∫(asin⁡x + bcos⁡x)dx
f(x) = -acos⁡x + bsin⁡x + c
f(x) = -acos⁡x + bsin⁡x + c ………(i)
As we know that f'(0) = 4
So, f'(0) = asin⁡0 + bcos⁡0 = 4
a × 0 + b × 1 = 4
b = 4
Also, f(0) = 3
f(0) = -acos⁡0 + bsin⁡0 + c = 3
-a + 0 + c = 3
 c – a = 3                  ……..(ii)
Also, f(π/2) = 5
f(π/2) = -acos⁡(π/2) + bsin⁡(π/2) + c = 5
-a × 0 + b × 1 + c = 5
b + c = 5
4 + c = 5                    [Since, b = 4]
c = 5 – 4
c = 1
Now, put c = 1 in eq(ii), we get 1 – a = 3
-a = 3 – 1
-a = 2
 a = -2
Now, put a = -2, b = 4, and c = 1 in eq(i), we get
f(x) = -(-2)cos⁡x + 4sin⁡x + 1
f(x) = 2cos⁡x + 4sin⁡x + 1
Question 49. Write the primitive or anti-derivative of f(x) = √x + 1/√x.
Solution:
We have, f(x) = √x + 1/√x
∫f(x) = ∫(√x + 1/√x)dx
= ∫x1/2dx + ∫ x-1/2 dx
= 2/3 x3/2 + 2x1/2 + c
Hence, the primitive or anti-derivative of f(x) is 2/3 x3/2 + 2x1/2 + c.

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