The value of ∫12 dx / [x (1 + x4)] is 1) (1 / 4) log (17 / 32) 2) (1 / 4) log (32 / 17) 3) log (17 / 2) 4) (1 / 4) log (17 / 2) Solution: (2) (1 / 4) log (32 / 17)... View Article
The value of ∫01 [√(1 – x) / (1 + x)] dx is 1) (π / 2) + 1 2) (π / 2) - 1 3) -1 4) 1 Solution: (2) (Ï€ / 2) - 1... View Article
The value of ∫23 [x + 1] / [x2 (x – 1)] dx is 1) [log (16 / 9) + (1 / 6)] 2) [log (16 / 9) - (1 / 6)] 3) 2 log 2 - (1 / 6) 4) log (4 / 3) - (1 / 6) Solution: (2) [log (16 / 9) -... View Article
∫0∞ dx / [(x + √x2 + 1)3] = 1) 3/8 2) 1/8 3) -3/8 4) None of these Solution: (1) ⅜... View Article
The value of ∫-π/4π/4 sin-4 x dx is 1) - (8/3) 2) 3/2 3) 8/3 4) None of these Solution: (1) - (8/3)... View Article
If k ∫01 x . f (3x) dx = ∫03 t . f (t) dt, then the value of k is 1) 9 2) 3 3) 1/9 4) 1/3 Solution: (1) 9... View Article
The value of ∫12 dx / [(x + 1) (√x2 – 1)] is 1) 1 2) √3 3) 2/√3 4) -2/√3 Solution: (2) √3... View Article
The value of ∫π/4π/2 ex (log sin x + cot x) dx = 1) eπ/4 log 2 2) - eπ/4 log 2 3) (1 / 2) eπ/4 log 2 4) (- 1 / 2) eπ/4 log 2 Solution: (3) (1 / 2) eπ/4 log 2... View Article
The value of ∫0a [√(a – x) / x] dx is 1) a/2 2) a/4 3) Ï€a/2 4) Ï€a/4 Solution: (3) Ï€a/2... View Article
The value of ∫0∞ dx / [(x2 + 4) (x2 + 9)] is 1) π/60 2) π/20 3) π/40 4) π/80 Solution: (1) π/60... View Article
∫510 1 / [(x – 1) (x – 2)] dx = 1) log (27 / 32) 2) log (32 / 27) 3) log (8 / 9) 4) log (3 / 4) Solution: (2) log (32 / 27)... View Article
∫ab √(x – a) (b – x) dx (b > a) = 1) [Ï€ (b - a)2 / 8] 2) [Ï€ (b + a)2] / 8 3) (b - a)2 4) (b + a)2 Solution: (1) [Ï€ (b - a)2 / 8]... View Article
Find the value of ∫02 f’ (x2) 2x dx 1) f(4) - f(0) 2) f(4) + f(0) 3) f(0) - f(4) 4) None of these Solution: (1) f(4) - f(0)... View Article
∫01 x tan-1 x dx = 1) π/4 2) [(π / 4) + (1 / 2)] 3) [(π /4) - (1 / 2)] 4) 1/2 Solution: (3) [(π /4) - (1 / 2)]... View Article
If ∫0π/3 [cos x / (3 + 4 sin x)] dx = k log [(3 + 2√3) / 3], then k = 1) 1/2 2) 1/3 3) 1/4 4) 1/8 Solution: (3) ¼... View Article
If p(x) be a polynomial of degree three that has a local maximum value 8 at x = 1 and a local minimum value 4 at x = 2; then p(0) is equal to a) 12 b) -12 c) -24 d) 6 Solution: p’(1) = 0 and p’(2) = 0 p’(x) = a(x-1)(x-2) p(x) = a[(x3/3)-(3x2/2)+2x]+b p(1) = 8... View Article