The value of ∫π4π2exlogsinx + cotxdx is
eπ4log2
- eπ4log2
12eπ4log2
- 12eπ4log2
Considering four sub-intervals, the value of ∫0111 + xdx by Trapezoidal rule, is
0.6870
0.6677
0.6977
0.5970
By Simpson's rule, the value of ∫12dxx dividing the interval (1, 2) into four equal parts, is
0.6932
0.6753
0.6692
7.1324
The value of ∫xsinxsec3xdx is
12sec2x - tanx + c
12xsec2x - tanx + c
12xsec2x + tanx + c
12sec2x + tanx + c
The value of ∫0xxsin3xdx is
4π3
2π3
0
None of these
Which of the following is true ?
∫01exdx = e
∫012xdx = log2
∫01xdx = 23
∫01xdx = 13
∫sinlogx + coslogxdx is equal to
xcoslogx + c
coslogx + c
xsinlogx + c
sinlogx + c
∫exx - 1x2dx is equal to
exx2 + c
- exx2 + c
exx + c
- exx + c
∫5101x - 1x - 2dx
log2732
log3227
log89
log34
B.
Let I = ∫5101x - 1x - 2dx = ∫510- 1x - 1 + 1x - 2dx = - logx - 1 + logx - 2510 = - log9 + log8 + log4 - log3 = - 2log3 + 3log2 + 2log2 - log3 = - 3log3 + 5log2 = - log27 + log32 = log3227
∫xlogxdx is equal to
x242logx - 1 + c
x222logx - 1
x242logx + 1 + c
x222logx + 1