The value of ∫0lnπ2cosex2xex2dx is
1
1 + sin(1)
1 - sin(1)
(sin(1) - 1
If f(x) = ∫2xdt1 + t4 and g is the inverse of f. Then, the value of g'(0) is
17
None of the above
∫0100ex - xdx is equal to
50(e - 1)
75(e - 1)
90(e - 1)
100(e - 1)
By Simpson's 13rd rule, the approximate value of the integral ∫12e- x2dx using four intervals, is
0.377
0.487
0.477
0.387
For n = 4, using trapezoidal rule, the value of ∫02dx1 + x will be
1.116625
1.1176
1.1180
None of these
The value of ∫05dx1 + x2 by chosing six sub-intervals and by using Simpson's rule will be
1.3562
1.3662
1.3456
1.2662
B.
Here, n = 6∵ ∆x = 6 - 06 = 1Now, y0 = f0 = 11 + 0 = 1y1 = f0 + ∆x = 11 + 1 = 0.5y2 = f0 + 2∆x = 11 + 4 = 15 = 0.2y3 = f0 + 3∆x = 11 + 9 = 110 = 0.1y4 = f0 + 4∆x = 1 1 + 16 = 117 = 0.058y5 = f0 + 5∆x = 11 + 25 = 126 = 0.038y6 = f0 + 6∆x = 11 + 36 = 137 = 0.027∴ ∫06dx1 + x2 = 131 + 40.5 + 0.1 + 0.038 + 20.2 + 0.058 + 0.027= 131 + 2.552 + 0.516 + 0.027= 13 × 4.095 = 1.365 = 1.3662
If ∫cos4x + 1cotx - tanxdx = Acos4x + B, then the value of A is
12
18
- 18
14
If f(x) = fx = x, gx = ex - 1 and ∫fogxdx = Afogx + Btan-1fogx + C, then the value of A + B is
2
3
The value of ∫01tan-12x - 11 + x - x2dx is
0
- 1
∫0π2sin2xtan-1sinxdx =
π2
π2 - 1