∫0xftdt = x2 + ex (x > 0), then f(1) is equal to
1 + e
2 + e
3 + e
e
B.
We have,∫0xftdt = x2 + exUsing Leibnitz Rule, fx = 2x + ex∴ f1 = 2 + e
∫x + 1x12dx =
- x3/2 + x1/2 + C
x1/2
x3/2 + 2x1/2 + C
2x323 + 2x12 + C
∫dxex + e- x + 2 is equal to
1ex + 1 + C
- 1ex + 1 + C
11 + e- x + C
1e- x - 1 + C
∫- 10dxx2 + x + 2 is equal to
π4
π2
π
47tan-117
∫- 221 - x2dx is equal to :
4
2
- 2
0
∫tanxsinxcosxdx is equal to :
2tanx + c
cotx + c
tan2x + c
∫dxx2 + 4x + 13 is equal to :
logx2 + 4x + 130 + c
13tan-1x + 23 + c
log2x + 4 + c
1x2 + 4x + 13 + c
∫01ddxsin-12x1 + x2dx is equal to :
∫0π2sinxsinx + cosxdx is equal to
π3
∫sinxsinx - αdx is equal to
x - αcosα + sinαlogsinx - α + c
sinx - α + sinx + c
x - αcosα + logsinx - α + c
cosx - α + cosx + c