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Solved Examples
30.The value of ∫1000(√x)dx ( where {x} is the fractional part of x) is
(A) 50
(B) 1
(C) 100
(D) none of these
Solution: Given integral = ∫1000 (√x–[√x])dx ( by the def. of {x} )
Hence (D) is the correct answer.
31. The value of ∫10 (|sin 2 p x| dx is equal to
(A) 0 (B) 2/π
(C) 1/π (D) 2
Solution: Since |sin 2 π x | is periodic with period 1/2,
I = ∫10 |sin 2 π x| dx= 2 ∫10 sin 2 π x dx
= 2 [–cos2πx/2π]1/20 = 2/π
Hence (B) is the correct answer.
32. Let f : R —> R, f(x) = , where [.] denotes greatest integer function, then ∫4–2 f(z)dx is equal to
(A) 5/2 (B) 3/2
(C) 5 (D) 3
Solution: x – [x] = {x}
x – [x +1] ={x} – 1
∫4–2 f(x)dx = 6.1/2 (1.1) = 3
(A) 0 (B) 2
(C) e (D) none of these
Solution: I =
property ∫a–a f(x)dx = 0 (f (–x) = –f (x), odd function)
Hence I = 0
Hence (A) is the correct answer.
34. The value of ∫10–10 3x/3[x] dx is equal to (where [.] denotes greatest integer function) :
(A) 20 (B) 40 / In3
(C) 20 / In 3 (D) none of these
35. Values of ∫+1/2–1/2 cos x log 1+x/1–x dx is :
(A) 1/2 (B) – 1/2
(C) 0 (D) none of these
Solution: I = ∫+1/2–1/2 cos x log 1+x/1–x dx
f (x) = cos x ln 1+x/1–x
f (- x) = cox (- x) ln 1+x/1–x
= - cos (x) ln (1+x/1–x) = – f (x)
f (x) is an odd function
hence I = 0
Hence (C) is the correct answer.
36. f (x) = min (tan x, cot x), 0 < x < , then ∫π/20 f(x)dx is equal to :
(A) ln2 (B) ln √2
(C) 2 ln √2 (D) none of these
Solution: f (x) = min (tan x, cot x),
∈ [0, π/2]
f (x) = tan x, 0 < x < π/4
= cot x, π/4 < x < π/4
Hence
2 ln √2 = ln 2.
37. The value of is equal to :
(A) π/2 (B) 2π
(C) π (D) π/p
38. The value of is equal to :
(A) 2 – 1/e (B) 2 + 1/e
(C) e+1/e (D) none of these
Solution: I = = |x–e–x|10 (1 - e-1) - (0 - 1) = 2 - e-1
39. has the value is :
(A) 0 (B) 1/2
(C) 1 (D) 1/4
40. is :
(A) 0 (B) 1
(C) π/2 (D) π/4
41. The value of depends on :
(A) p (B) q
(C) r (D) p and q
= q (Since sin3x and sin5 x are odd functions)
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