RD Sharma Class 12 Ex 19.2 Solutions Chapter 19 Indefinite Integrals

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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 

= ∫3x^3/2dx + 4∫x^1/2 dx + 5∫dx

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

= 6/5 x^5/2 + 8/3 x^3/2 + 5x + c

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

Solution:

We have, ∫(2^x + 5/x – 1/x^1/3)dx

= ∫2^xdx + 5∫1/x dx – ∫1/x^1/3 dx

= 2^x/(log⁡2) + 5log⁡x – 3/2 x^2/3 + c

Question 3. Evaluate ∫{√x (ax² + bx + c)}dx

Solution:

We have, ∫{√x (ax² + bx + c)}dx

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

= ∫ax^5/2 dx + ∫bx^3/2dx + ∫cx^1/2 dx

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

= (2ax^7/2)/7 + (2bx^5/2)/5 + (2cx^3/2)/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

= ∫(-6x² – 5x + 6)(1 – 2x)dx

= ∫(-6x² + 12x³ – 5x + 10x² + 6 – 12x)dx

= ∫(4x² + 12x³ – 17x + 6)dx

= ∫(12x³ + 4x² – 17x + 6)dx 

= 12/4 x⁴ + 4/3 x³ – 17/2 x² + 6x + c 

= 3x⁴ + 4/3 x³ – 17/2 x² + 6x + c

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

Solution:

We have, ∫(m/x + x/m + m^x + x^m + mx)dx

= m∫1/x dx + 1/m ∫xdx + ∫m^xdx + ⌋x^mdx + m∫xdx

= mlog⁡|x| + x²/2m + m^x/(log⁡m) + x^m+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 + x³ + 3x² + 3x)/√x dx

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

= ∫x^-1/2 dx + ∫x^5/2 dx + 3∫x^3/2 dx + 3∫x^1/2 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

= x^1/2/(1/2) + x^7/2/(7/2) + (3x^5/2)/(5/2) + 3 x^3/2/(3/2) + c

= 2x^1/2 + 2/7 x^7/2 + 6/5 x^5/2 + 6/3 x^3/2 + c

= 2x^1/2 + 2/7 x^7/2 + 6/5 x^5/2 + 2x^3/2 + c

 Question 8. Evaluate ∫{x² + e^logx + (e/2)^x }dx

Solution:

We have, ∫{x² + e^logx + (e/2)^x }dx

= ∫x² dx + ∫e^logx dx + ∫(e/2)^x dx

= x³/3 + ∫xdx + ∫(e/2)^xdx

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

Question 9. Evaluate ∫ (x^e + e^x + e^e)dx

Solution:

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

= ∫x^e dx + ∫e^xdx + ∫e^edx

= x^e+1/(e + 1) + e^x + e^ex + c

Question 10. Evaluate ∫√x (x³ – 2/x)dx 

Solution:

We have, ∫√x (x³ – 2/x)dx 

= ∫ x^7/2 dx – 2∫ x^-1/2 dx

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

= x^9/2/(9/2) – (2x^-1/2)/((-1)/2) + c

= 2/9 x^9/2 – 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

= 2x^1/2 – 2x^-1/2 + c

Question 12. Evaluate ∫(x⁶ + 1)/(x² + 1) dx

Solution:

We have, ∫(x⁶ + 1)/(x² + 1) dx

= ∫((x²)³ + (1)³)/(x² + 1) dx

= ∫(x² + 1)(x⁴ + 1 – x²)/(x² + 1) dx

= ∫(x⁴ – x² + 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)/x^1/3 + ∫x^1/2/x^1/3dx + ∫2/x^1/3dx

= ∫x^-2/3 dx + ∫x^1/6 dx + 2∫ x^-1/3dx

= 3x^1/3 + 6/7 x^7/6 + 3x^2/3 + c

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

Solution:

We have, ∫((1 + √x)²)/√x dx

= ∫(1 + x + 2√x)/x^1/2dx

= ∫x^-1/2+∫ x^1/2 dx + 2∫dx

= 2x^1/2 + 2/3 x^3/2 + 2x + c

= 2√x + 2x + 2/3 x^3/2 + c

Question 15. Evaluate ∫√x(3 – 5x)dx

Solution:

We have, ∫√x(3 – 5x)dx

= 3∫√x dx – 5x^3/2 dx

= 3x^3/2/(3/2) – 5 x^5/2/(5/2) + c

= 2x^3/2 – 2x^5/2 + c

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

Solution:

We have, ∫((x + 1)(x – 2))/√x dx

= ∫(x² – 2x + x – 2)/x^1/2dx

= ∫(x² – x – 2)/x^1/2dx

= ∫x²/x^1/2dx – ∫x^1/2dx – 2∫x^-1/2dx

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

= 2/5 x^5/2 – (2x^3/2)/3 – 4√x + c

Question 17. Evaluate ∫(x⁵ + x^-2 + 2)/x²dx

Solution:

We have, ∫(x⁵ + 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, ∫ (9x² + 16 + 24x)dx

= ∫9x² dx + ∫16dx + ∫24xdx

= 9 x³/3 + 16x + 24 x²/2 + c

= 3x³ + 16x + 12x² + c

Question 19. Evaluate ∫(2x⁴ + 7x³ + 6x²)/(x² + 2x) dx

Solution:

We have, ∫(2x⁴ + 7x³ + 6x²)/(x² + 2x) dx

= ∫x(2x³ + 7x² + 6x)/(x(x + 2))dx

= ∫(2x³ + 7x² + 6x)/(x + 2)dx

= ∫ (2x³ + 4x² + 3x² + 6x)/((x + 2))dx

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

= ∫(x + 2)(2x² + 3x)/(x + 2) dx

= ∫(2x² + 3x)dx

= ∫2x² dx + ∫3xdx

= 2/3 x³ + 3/2 x² + c

Question 20. Evaluate ∫(5x⁴ + 12x³ + 7x²)/(x² + x) dx

Solution:

We have, ∫(5x⁴ + 7x³ + 5x³ + 7x²)/(x² + x) dx

= ∫(5x³ + 7x² + 5x² + 7x)/(x + 1) dx

= ∫5x² (x + 1) + 7x(x + 1)/(x + 1) dx

= ∫(5x² + 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 + tan^2⁡x))dx
Solution:
We have, ∫ sin^-1⁡((2tan⁡x)/(1 + tan^2⁡x))dx
= ∫ sin^-1⁡(sin⁡2x)dx
= ∫2xdx
= 2∫xdx
= (2x²)/2 + c
= x² + c 
Question 39. Evaluate ∫((x³ + 8)(x – 1))/(x² – 2x + 4) dx
Solution:
We have, ∫((x³ + 8)(x – 1))/(x² – 2x + 4) dx
= ∫((x + 2)(x² – 2x + 4)(x – 1))/(x² – 2x + 4) dx
= ∫(x + 2)(x – 1)dx
= ∫(x² – x+2x – 2)dx
= ∫(x² + 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 – a² + b²cosec²⁡x – b² + 2ab]dx
= a²tan⁡x – a²x – b² cot⁡x – b²x + 2abx + c
= a²tan⁡x – b² cot⁡x – (a² + b² – 2ab)x + c
Question 41. Evaluate ∫(x³ – 3x² + 5x – 7 + x² a^x)/(2x²) dx
Solution:
We have, ∫(x³ – 3x² + 5x – 7 + x² a^x)/(2x²) dx
= 1/2 ∫x³/x²dx – 3/2∫x²/x²dx + 5/2∫x/x²dx – 7/2∫x^-2dx + 1/2∫(x²a^x)/x²dx
= 1/2 × x²/2 – 3/2x + 5/2 log⁡x – 7/2 x^-1 + 1/2a^x/(log⁡a) + c
= 1/2 [x²/2 – 3x + 5log⁡x + 7/x + a^x/(log⁡a)] + c
Question 42. Evaluate ∫cos⁡x/(1 + cos⁡x) dx
Solution:
We have, ∫cos⁡x/(1 + cos⁡x) dx  …..(1)
Now solve
 
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) = 8x³ – 2x, f(2) = 8, find f(x)?
Solution:
Given that f'(x) = 8x³ – 2x
and f(2) = 8
We have to find f(x)
So, ∫f'(x)dx = ∫(8x³ – 2x)dx
f(x) = ∫(8x³ – 2x)dx
= ∫8x³dx – ∫2xdx
= (8x⁴)/4 – (2x²)/2 + c
= 2x⁴ – x² + c
f(x) = 2x⁴ – x² + c  ……….(i)
As we know that f(2) = 8
So, f(2) = 2(2)⁴ – (2)² + c = 8
32 – 4 + c = 8
28 + c = 8
c = -20
Now, Put c = -20 in eq(i), we get
f(x) = 2x⁴ – x² – 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
= ∫x^1/2dx + ∫ x^-1/2 dx
= 2/3 x^3/2 + 2x^1/2 + c
Hence, the primitive or anti-derivative of f(x) is 2/3 x^3/2 + 2x^1/2 + c.

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