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
A.
h = 2 - 14 = 14Now, x0 = 1, x1 = 1 + 14, x2 = 1 + 2 × 14, x3 = 1 + 3 × 14, x4 = 1 + 4 × 14i.e., x0 = 1, x1 = 1.25, x2 = 1.5, x3 = 1.75, x4 = 2⇒ y0 = 1, y1 = 0.8, y2 = 0.667, y3 = 0.571, y4 = 0.5∴ Using Simpson's 13rd rule∫12dxx = 1121 + 0.5 + 40.8 + 0.571 + 20.667 = 1121.5 + 5.484 + 1.334 = 1128.318 = 0.6932
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
∫xlogxdx is equal to
x242logx - 1 + c
x222logx - 1
x242logx + 1 + c
x222logx + 1