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Textbook | NCERT |
Class | Class 12th |
Subject | Maths |
Chapter | 20 |
Exercise | 20.1 |
Category | RD Sharma Solutions |
RD Sharma Class 12 Ex 20.1 Solutions Chapter 20 Definite Integrals
Question 1: Ten cards numbered 1 through 10 are placed in a box, mixed
Evaluate the following definite integrals:
Question 1. 
Solution:
We have,
I =
I =
I =
I =
I = 2[√9 – √4 ]
I = 2 (3 − 2)
I = 2 (1)
I = 2
Therefore, the value of
is 2.
Question 2. 
Solution:
We have,
I =
I =
I = log (3 + 7) − log (−2 + 7)
I = log 10 − log 5
I =
I = log 2
Therefore, the value of
is log 2.
Question 3. 
Solution:
We have,
I =
Let x = sin t, so we have,
=> dx = cos t dt
Now, the lower limit is,
=> x = 0
=> sin t = 0
=> t = 0
Also, the upper limit is,
=> x = 1/2
=> sin t = 1/2
=> t = π/6
So, the equation becomes,
I =
I =
I =
I =
I =
I = π/6 – 0
I = π/6
Therefore, the value of
is π/6.
Question 4. 
Solution:
We have,
I =
I =
I =
I =
I = π/4
Therefore, the value of
is π/4.
Question 5. 
Solution:
We have,
I =
Let x2 + 1 = t, so we have,
=> 2x dx = dt
=> x dx = dt/2
Now, the lower limit is, x = 2
=> t = x2 + 1
=> t = (2)2 + 1
=> t = 4 + 1
=> t = 5
Also, the upper limit is, x = 3
=> t = x2 + 1
=> t = (3)2 + 1
=> t = 9 + 1
=> t = 10
So, the equation becomes,
I =
I =
I =
I = 1/2[log10 – log5]
I = 1/2[log10/5]
I = 1/2[log2]
I = log√2
Therefore, the value of
is log√2.
Question 6. 
Solution:
We have,
I =
I =
I =
I =
I =
I = 1/ab[tan-1∞ – tan-10]
I = 1/ab[π/2 – 0]
I = 1/ab[π/2]
I = π/2ab
Therefore, the value of
is π/2ab.
Question 7. 
Solution:
We have,
I =
I =
I = [tan-11 – tan-1(-1)]
I = [π/4 – (-π/4)]
I = [π/4 + π/4]
I = 2π/4
I = π/2
Therefore, the value of
is π/2.
Question 8. 
Solution:
We have,
I =
I =
I = -e–∞ – (-e0)
I = − 0 + 1
I = 1
Therefore, the value of
is 1.
Question 9. 
Solution:
We have,
I =
I =
I =
I =
I =
I = [1 − 0] − [log(1 + 1) − log(0 + 1)]
I = 1 − [log2 − log1]
I = 1 – log2/1
I = 1 − log 2
I = log e − log 2
I = loge/2
Therefore, the value of
is loge/2.
Question 10. 
Solution:
We have,
I =
I =
I =
I = [-cosπ/2 + cos0] + [sinπ/2 – sin0]
I = [−0 + 1] + 1
I = 1 + 1
I = 2
Therefore, the value of
is 2.
Question 11. 
Solution:
We have,
I =
I =
I = log(sinπ/2) – log(sinπ/4)
I = log1 – log1/√2
I =
I = log√2
Therefore, the value of
is log√2.
Question 12. 
Solution:
We have,
I =
I =
I = log(secπ/4 + tanπ/4 – log(sec0 + tan0)
I = log(√2 + 1) – log(1 + 0)
I =
I = log(√2 + 1)
Therefore, the value of
is log(√2 + 1).
Question 13. 
Solution:
We have,
I =
I =
I = [log|cosecπ/4 – cotπ/4|] – [log|cosecπ/6 – cotπ/6|]
I = [log|√2 – 1|] – [log|2 – √3|]
I =
Therefore, the value of
is
.
Question 14. 
Solution:
We have,
I =
Let x = cos 2t, so we have,
=> dx = –2 sin 2t dt
Now, the lower limit is,
=> x = 0
=> cos 2t = 0
=> 2t = π/2
=> t = π/4
Also, the upper limit is,
=> x = 1
=> cos 2t = 1
=> 2t = 0
=> t = 0
So, the equation becomes,
I =
I =
I =
I =
I =
Let cos t = z, so we have,
=> – sin t dt = dz
=> sin t dt = – dz
Now, the lower limit is,
=> t = 0
=> z = cos t
=> z = cos 0
=> z = 1
Also, the upper limit is,
=> t = π/4
=> z = cos t
=> z = cos π/4
=> z = 1/√2
So, the equation becomes,
I =
I =
I =
I =
I =
I =
I = -4[(log1/√2 – 1/2(2)) – (log1 – 1/2)]
I = -4[(log1/√2 – 1/4) – (0 – 1/2)]
I = -4[log1/√2 – 1/4 – 0 + 1/2]
I = -4[-log√2 + 1/4]
I = 4log√2 – 1
I = 4 × 1/2log2 – 1
I = 2log2 – 1
Therefore, the value of
is 2log2 – 1.
Question 15. 
Solution:
We have,
I =
I =
I =
I =
I =
I =
I =
I =
I =
I = [tan π – tan0] – [sec π – sec 0]
I = [0 – 0] – [–1 – 1]
I = 0 – (–2)
I = 2
Therefore, the value of
is 2.
Question 16. 
Solution:
We have,
I =
I =
I =
I =
I =
I =
I =
I =
I =
I =
I = [tan π/4 – tan(–π/4)] – [sec π/4 – sec (–π/4)]
I = [1 – (–1)] – [sec π/4 – sec (π/4)]
I = 2 – 0
I = 2
Therefore, the value of
is 2.
Question 17. 
Solution:
We have,
I =
I =
I =
I =
I =
I =
I = 1/2[π/2 – 0] + 1/4[sinπ – sin0]
I = 1/2[π/2] + 1/4[0 – 0]
I = π/4
Therefore, the value of
is π/4.
Question 18. 
Solution:
We have,
I =
I =
I =
I =
I =
I =
I = 1/12 [-1 – 0] + 3/4[1 – 0]
I = 3/4 – 1/12
I = (9 – 1)/12
I = 8/12
I = 2/3
Therefore, the value of
is 2/3.
Question 19. 
Solution:
We have,
I =
I =
I =
I =
I =
I =
I = 1/6[sinπ/2 – sin0] + 1/2[sinπ/6 – sin0]
I = 1/6[1 – 0] + 1/2[1/2 – 0]
I = 1/6 + 1/4
I = (4 + 6)/24
I = 10/24
I = 5/12
Therefore, the value of
is 5/12.
Question 20. 
Solution:
We have,
I =
I =
I =
I =
I =
I =
I = 1/2[sinπ/2 – sin0] – 1/6[sin3π/2 – sin0]
I = 1/2[1 – 0] – 1/6[-1 – 0]
I = 1/2 – 1/6(-1)
I = 1/2 + 1/6
I = (6 + 2)/12
I = 8/12
I = 2/3
Therefore, the value of
is 2/3.
Question 21. 
Solution:
We have,
I =
I =
I =
I =
I =
I =
I =
I =
I =
I =
I = 2[-cotπ/2 + cot2π/3]
I = 2[-1/√3 – 0]
I = -2/√3
Therefore, the value of
is -2/√3.
Question 22. 
Solution:
We have,
I =
I =
I =
I =
I =
I =
I =
I = 1/4[π/2 + π/4 + 0 + 0 – 0 – 0 – 0 – 0]
I = 1/4[3π/4]
I = 3π/16
Therefore, the value of
is 3π/16.
Evaluate the following definite integrals:
Question 23. 
Solution:
We have,
I =
I =
I =
I =
I =
I =
I =
I =
I =
I =
I = [(a2 + b2)/2][π/2]
I = π(a2 + b2)/4
Therefore, the value of
is π(a2 + b2)/4.
Question 24. 
Solution:
We have,
I =
I =
I =
I =
I =
I =
I =
I =
I = 2[sinπ/4 – cosπ/4 – 0 + 1]
I = 2[1/√2 – 1/√2 – 0 + 1]
I = 2 (1)
I = 2
Therefore, the value of
is 2.
Question 25. 
Solution:
We have,
I =
I =
I =
I =
I =
I =
I = 2√2[sinπ/4 – sin0]
I = 2√2[1/√2- sin0]
I = 2√2[1/√2]
I = 2
Therefore, the value of
is 2.
Question 26. 
Solution:
We have,
I =
By using integration by parts, we get,
I = x ∫sinxdx – ∫(∫sin x (1)dx)dx
I = -xcosx – ∫(∫sin xdx)dx
I = -xcosx + ∫cosxdx
I = -xcosx + sinx
So we get,
I =
I = [-π/2cosπ/2 + sinπ/2 + 0 – 0]
I = 0 + 1 + 0 – 0
I = 1
Therefore, the value of
is 1.
Question 27. 
Solution:
We have,
I =
By using integration by parts, we get,
I = x∫cosxdx – ∫(∫cos x (1)dx)dx
I = xsinx – ∫(∫cosxdx)dx
I = xsinx – ∫sinxdx
I = x sin x + cos x
So we get,
I =
I = [π/2sinπ/2 + cosπ/2 – 0 – cos0]
I = π/2 + 0 – 0 – 1
I = π/2 – 1
Therefore, the value of
is π/2 – 1.
Question 28. 
Solution:
We have,
I =
By using integration by parts, we get,
I = x2sinx – ∫(2x∫(cosx)dx)dx
I = x2sinx – ∫(2xsinx)dx
I = x2sinx – 2[-xcosx – ∫(1∫sinxdx)dx]
I = x2sinx – 2[-xcosx + ∫sinxdx]
I = x2sinx – 2[-xcosx + sinx]
I = x2sinx + 2xcosx – 2sinx
So we get,
I =
I = [(π/2)2sinπ/2 + 2(π/2)cosπ/2 – 2sinπ/2 – 0 – 0 + sin0]
I = [π2/4 + 0 – 2 – 0 – 0 + 0]
I = π2/4 – 2
Therefore, the value of
is π2/4 – 2.
Question 29. 
Solution:
We have,
I =
By using integration by parts, we get,
I = -x2cosx – ∫(2x∫sinxdx)dx
I = -x2cosx + ∫(2xcosx)dx
I = -x2cosx + 2[xsinx – ∫(∫cosxdx)dx]
I = -x2cosx + 2[xsinx – ∫sinxdx]
I = -x2cosx + 2[xsinx + cosx]
I = -x2cosx + 2xsinx + 2cosx
So we get,
I =
I = -(π/4)2cosπ/4 + 2π/4sinπ/4 + 2cosπ/4 + 0 – 0 – 2
I = –π2/16(1/ √2) + π/2(1/√2) + 2(1/√2) + 0 – 0 – 2
I = –π2/16√2 + π/2√2 + √2 – 2
Therefore, the value of
is -π2/16√2 + π/2√2 + √2 – 2.
Question 30. 
Solution:
We have,
I =
By using integration by parts, we get,
I = 1/2x2sin2x – ∫(2x∫cos2xdx)dx
I = 1/2x2sin2x – ∫(xsin2x)dx
I = 1/2x2sin2x – [-1/2xcos2x – ∫(∫sin2xdx)dx]
I = 1/2x2sin2x – [-1/2xcos2x + ∫1/2 cos2xdx]
I = 1/2x2sin2x – [-1/2xcos2x + 1/4sin2xdx]
I = 1/2x2sin2x + 1/2xcos2x – 1/4sin2xdx
So we get,
I =
I = [1/2(π2/4)sinπ + 1/2(π/2)cosπ – 0 – 0 – 0 + 0]
I = -π/4
Therefore, the value of
is -π/4.
Question 31. 
Solution:
We have,
I =
I =
I =
I =
By using integration by parts, we get,
I = 1/2[x3/3] + x2sin2x/2 – [x ∫sin2x – ∫(∫sin2xdx)dx]
I = 1/2[x3/3] + x2sin2x/2 + xcosx/2 – sin2x/4
So we get,
I =
I = [1/6[π3/8] + 0 + 0 – π/8]
I = π3/48 – π/8
Therefore, the value of
is π3/48 – π/8.
Question 32. 
Solution:
We have,
I =
By using integration by parts, we get,
I =
I = xlogx – ∫1dx
I = xlogx – x
So we get,
I =
I = 2log2 – 2 – log1 + 1
I = 2 log 2 – 1
Therefore, the value of
is 2 log 2 – 1.
Question 33. 
Solution:
We have,
I =
By using integration by parts, we get,
I =
I =
I =
I =
So we get,
I =
I = -log3/4 + log3 – log4 + log1/2 – log1 + log2
I = log3(1 – 1/4) – 2log2 + 0 – 0 + log2
I = 3/4log3 – log2
Therefore, the value of
is 3/4log3 – log2.
Question 34. 
Solution:
We have,
I =
I =
I =
By using integration by parts, we get,
I =
I = exlogx
So we get,
I =
I = eeloge – e1log1
I = ee (1) – 0
I = ee
Therefore, the value of
is ee.
Question 35. 
Solution:
We have,
I =
Let log x = t, so we have,
=> (1/x) dx = dt
Now, the lower limit is, x = 1
=> t = log x
=> t = log 1
=> t = 0
Also, the upper limit is, x = e
=> t = log x
=> t = log e
=> t = 1
So, the equation becomes,
I =
I =
I =
I = 1/2 – 0/2
I = 1/2
Therefore, the value of
is 1/2.
Question 36. 
Solution:
We have,
I =
I =
By using integration by parts, we get,
I =
I =
I =
I = x/logx
So we get,
I =
I =
I =
I = e2/2 – e
Therefore, the value of
is e2/2 – e.
Question 37. 
Solution:
We have,
I =
I =
I =
I =
I =
I =
I =
I = 1/2[3log2 – log4 + log3]
I = 1/2[3log2 – 2log2 + log3]
I = 1/2[log 2 – log 3]
I = 1/2[log6]
I = log6/2
Therefore, the value of
is log6/2.
Question 38. 
Solution:
We have,
I =
I =
I =
I =
I =
I =
I =
I =
I =
I =
I =
I = [1/5log6 + 3/√5tan-1(√5) – 1/5log1 – 3/√5tan-1(0)]
I = [1/5 log6 + 3√5 tan-1(√5) – 0 – 0]
I = 1/5 log6 + 3√5 tan-1(√5)
Therefore, the value of
is 1/5 log6 + 3√5 tan-1(√5).
Question 39. 
Solution:
We have,
I =
I =
I =
I =
I =
I =
Let x – 1/2 = t, so we have,
=> dx = dt
Now, the lower limit is, x = 0
=> t = x – 1/2
=> t = 0 – 1/2
=> t = 1/2
Also, the upper limit is, x = 2
=> t = x – 1/2
=> t = 2 – 1/2
=> t = 3/2
So, the equation becomes,
I =
I =
I =
I =
I =
I =
I =
I =
I =
I =
I =
I =
Therefore, the value of
is
.
Question 40. 
Solution:
We have,
I =
I =
I =
I =
I =
I =
I =
I =
I = 4/2√7[tan-1(5/√7) – tan-1(1/√7)]
I = 2/√7[tan-1(5/√7) – tan-1(1/√7)]
Therefore, the value of
is 2/√7[tan-1(5/√7) – tan-1(1/√7)].
Question 41. 
Solution:
We have,
I =
Let x = sin2 t, so we have,
=> dx = 2 sin t cos t dt
Now, the lower limit is, x = 0
=> sin2 t = 0
=> sin t = 0
=> t = 0
Also, the upper limit is, x = 1
=> sin2 t = 1
=> sin t = 1
=> t = π/2
So, the equation becomes,
I =
I =
I =
I =
I =
I =
I =
I =
I =
I =
I =
I = 1/4[π/2 – 0] – 1/16[sin2π – 0]
I = 1/4[π/2] – 1/16[0 – 0 ]
I = π/8
Therefore, the value of
is π/8.
Question 42. 
Solution:
We have,
I =
I =
I =
I =
I =
I = [sin-1(1/2) – sin-1(-1/2)]
I = π/6 -(-π/6)
I = π/6 + π/6
I = π/3
Therefore, the value of
is π/3.
Question 43.
Solution:
We have,
I =
I =
I =
I =
I =
I =
I = [sin-1(2/2) – sin-1(-2/2)]
I = sin-11 – sin-1(-1)
I = π/2 – (-π/2)
I = π/2 + π/2
I = π
Therefore, the value of
is π.
Question 44. 
Solution:
We have,
I =
I =
I =
Let x + 1 = t, so we have,
=> dx = dt
Now, the lower limit is, x = –1
=> t = x + 1
=> t = – 1 + 1
=> t = 0
Also, the upper limit is, x = 1
=> t = x + 1
=> t = 1 + 1
=> t = 2
So, the equation becomes,
I =
I =
I = 1/2tan-12/2 – 1/2tan-10/2
I = 1/2tan-11 – 1/2tan-10
I = 1/2(π/4) – 0
I = π/8
Therefore, the value of
is π/8.
Evaluate the following definite integrals:
Question 45. 
Solution:
We have,
I =
Let 2x + 1 = t2, so we have,
=> 2 dx = 2t dt
=> dx = t dt
Now, the lower limit is, x = 1
=> t2 = 2x + 1
=> t2 = 2(1) + 1
=> t2 = 3
=> t = √3
Also, the upper limit is, x = 4
=> t2 = 2x + 1
=> t2 = 2(4) + 1
=> t2 = 9
=> t = 3
So, the equation becomes,
I =
I =
I =
I =
I =
I = 1/4 [35/5 – 3 – (√3)5/5 + √3]
I = 1/4[243/5 – 3 – 9√3/5 + √3]
I = 1/4((243 – 15 – 9√3 + 5√3)/5)
I = 1/4[(228 – 4√3)/5]
I = 1/4[4(57 – √3)/5]
I = (57 – √3)/5
Therefore, the value of
is (57 – √3)/5.
Question 46. 
Solution:
We have,
I =
By using binomial theorem in the expansion of (1 – x)5, we get,
I =
I =
I =
I =
I = 1/2 – 5/3 + 10/4 – 10/5 + 5/6 – 1/7
I = 1/2 – 5/3 + 5/3 – 2 + 5/6 – 1/7
I = 1/2 – 2 + 5/6 – 1/7
I = 1/42
Therefore, the value of
is 1/42.
Question 47. 
Solution:
We have,
I =
I =
I =
I =
By using integration by parts, we get,
I =
I =
I = ex/x
So we get,
I =
I = e2/2 – e1/1
I = e2/2 – e
Therefore, the value of
is e2/2 – e.
Question 48. 
Solution:
We have,
I =
By using integration by parts in first integral, we get,
I =
I = xe2x/2 – (1/2)(e2x/2) + 2/π[1 – 0]
I = xe2x/2 – e2x/4 + 2/π
So we get,
I =
I = [e2/2 + e2/4 – 0 + 1/4] + 2/π
I = e2/4 + 1/4 + 2/π
Therefore, the value of
is e2/4 + 1/4 + 2/π.
Question 49. 
Solution:
We have,
I =
By using integration by parts in first integral, we get,
I =
I =
I =
I =
So we get,
I =
I =
I = [e1(1 – 1) – e0(0 – 1)] + 2√2/π
I = [0 – (-1)] + 2√2/π
I = 1 + 2√2/π
Therefore, the value of
is 1 + 2√2/π.
Question 50. 
Solution:
We have,
I =
I =
I =
I =
I = -eπ cotπ/2 + eπ/2 cotπ/4
I = 0 + eπ/2(1)
I = eπ/2
Therefore, the value of
is eπ/2.
Question 51. 
Solution:
We have,
I =
I =
I =
I =
By using integration by parts in first integral, we get,
I =
I =
I =
I = 1/√2[sinπ(2eπ) – 0]
I = 1/√2[0 – 0]
I = 0
Therefore, the value of
is 0.
Question 52. 
Solution:
We have,
I =
By using integration by parts, we get,
I = excos(x/2 + π/4) + 1/2∫exsin(x/2 + π/4)
I = ex cos(x/2 + π/4) + 1/2[ exsin(x/2 + π/4) – 1/2 ∫excos(x/2 + π/4)dx]
I = excos(x/2 + π/4) + 1/2exsin(x/2 + π/4) – 1/4I
5I/4 = -3/ 2√2(e2π + 1)
I = -3√2/5(e2π + 1)
Therefore, the value of is -3√2/5(e2π + 1).
Question 53. 
Solution:
We have,
I =
I =
I =
I =
I =
I =
I = 2/3[23/2 – 1] + 2/3[1 – 0]
I =
I = 25/2/3
Therefore, the value of
is 25/2/3.
Question 54. 
Solution:
We have,
I =
I =
I =
I =
I = -log3 + log2 + 2[log4 – log3]
I = -log3 + log2 + 2[2log2 – log3]
I = -log3 + log2 + 4log2 – 2log3
I = 5log2 – 3log3
I = log25 – log33
I = log32 – log27
I = log32/27
Therefore, the value of
is log32/27.
Question 55. 
Solution:
We have,
I =
I =
I =
Let cos x = t, so we have,
=> – sin x dx = dt
Now, the lower limit is, x = 0
=> t = cos x
=> t = cos 0
=> t = 1
Also, the upper limit is, x = π/2
=> t = cos x
=> t = cos π/2
=> t = 0
So, the equation becomes,
I =
I =
I =
I = [0 – 1/3] – [0 – 1]
I = [-1/3] – [-1]
I = -1/3 + 1
I = 2/3
Therefore, the value of
is 2/3.
Question 56. 
Solution:
We have,
I =
I =
I =
I =
I = -sinπ + sin0
I = 0
Therefore, the value of
is 0.
Question 57. 
Solution:
We have,
I =
Let 2x = t, so we have,
=> 2x dx = dt
Now, the lower limit is, x = 1
=> t = 2x
=> t = 2(1)
=> t = 2
Also, the upper limit is, x = 2
=> t = 2x
=> t = 2(2)
=> t = 4
So, the equation becomes,
I =
I =
I =
By using integration by parts in first integral, we get,
I =
I =
I =
I = e4/4 – e2/2
Therefore, the value of
is e4/4 – e2/2.
Question 58. 
Solution:
We have,
I =
I =
I =
I =
I =
I =
I = [sin-1(1) – sin-1(-1)]
I = π/2 – (-π/2)
I = π/2 + π/2
I = π
Therefore, the value of
is π.
Question 59. If
, find the value of k.
Solution:
We have,
=>
=>
=>
=>
=> tan-12k/4 – tan-10 = π/16
=> tan-12k/4 – 0 = π/16
=> tan-12k/4 = π/16
=> tan-12k = π/4
=> 2k = tanπ/4
=> 2k = 1
=> k = 1/2
Therefore, the value of k is 1/2.
Question 60. If
, find the value of k.
Solution:
We have,
=>
=>
=>
=>
=> a3 – 0 = 8
=> a3 = 8
=> a = 2
Therefore, the value of a is 2.
Question 61.
Solution:
We have,
I =
I =
I =
I =
I = -[√2cos3π/2 – √2cosπ]
I = -(-√2 – 0)
I = √2
Therefore, the value of
is √2.
Question 62. 
Solution:
We have,
I =
I =
I =
I =
I =
I =
I =
I = [-4cosπ/2 + 4cos0] + [4sinπ/2 – 4sin0]
I = 0 + 4 + 4 – 0
I = 8
Therefore, the value of
is 8.
Question 63. 
Solution:
We have,
I =
I =
I =
I =
I =
I =
I =
I =
I =
I =
I =
I = (π/8 – 1/4) – (3/4(π/8 – 1/4) – 1/16)
I = π/8 – 1/4 – (3π/32 – 3/16 – 1/16)
I = π/8 – 1/4 – (3π/32 – 1/4)
I = π/8 – 1/4 – 3π/32 + 1/4
I = π/8 – 3π/32
I = (4π – 3π)/32
I = π/32
Therefore, the value of
is π/32.
Question 64. 
Solution:
We have,
I =
By using integration by parts we get,
I =
I =
I =
I =
I =
So we get,
I =
I = log3/2 – 1/8log3
I = 3/8log3
Therefore, the value of
is 3/8log3.
Question 65. 
Solution:
We have,
I =
I =
I =
I =
I =
I =
I =
I = [tanπ/3 – tanπ/6] + [-cotπ/3 + cotπ/6]
I = [√3 – 1/√3] + [- 1/√3 – √3]
I = 2[√3 – 1/√3]
I = 4/√3
Therefore, the value of
is 4/√3.
Question 66. 
Solution:
We have,
I =
I =
I =
I =
I =
I =
I =
I =
I =
I =
I =
Therefore, the value of
is
.
Question 67. 
Solution:
We have,
I =
I =
I =
I =
I = -log2/4 + log2/2 – 1/4 + 1/2
I = log2/4 + 1/4
Therefore, the value of
is log2/4 + 1/4.
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