∫ 1 + x - x- 1ex + x- 1dx is equal to˸
1 + xex + x- 1 + C
x - 1ex + x- 1 + C
- xex + x- 1 + C
xex + x- 1 + C
∫- 22xdx is equal to
1
2
3
4
∫01sin2tan-11 + x1 - xdx is equal to
π6
π4
π2
π
∫033x + 1x2 + 9dx is equal to :
log22 + π12
log22 + π2
log22 + π6
log22 + π3
If [2, 6] is divided into four intervals of equal length, then the approximate value of ∫261x2 - xdx using Simpson's rule, is
0.3222
0.2333
0.5222
0.2555
C.
Here, h = 6 - 24 = 1Let y = 1x2 - xAt x0 = 2, y0 = 122 - 2 = 14 - 2 = 12 x1 = 3, y1 = 132 - 3 = 19 - 3 = 16 x2 = 4, y2 = 142 - 4 = 116 - 4 = 112 x3 = 5, y3 = 152 - 5 = 125 - 5 = 120 x4 = 6, y4 = 162 - 6 = 136 - 6 = 130By Simpson's rule∫261x2 - xdx = h3y0 + y4 + 4y1 + y3 + 2y2 = 1312 + 130 + 416 + 120 + 2112 = 131630 + 426120 + 16 = 131630 + 2630 + 16 = 16 + 26 + 590 = 4790 = 0.5222∫261x2 - xdx = 0.5222
If fx = 1x2∫3x2t - 3f'tdt, then f't, then f'(3) is equal to
- 1 2
- 13
12
13
∫dxx + 100x + 99 = fx + c ⇒ fx
2(x + 100)1/2
3(x + 100)1/2
2tan-1x + 99
2tan-1x + 100
∫3 - x21 - 2x + x2exdx = exfx + c ⇒ fx
1 + x1 - x
1 - x1 + x
1 - xx - 1
x - 11 + x
∫cotxsinxcosxdx = - fx + c ⇒ fx
2tanx
- 2tanx
- 2cotx
2cotx
∫- π2π2log2 - sinθ2 + sinθdθ is equal to
0
- 1