Dividing the interval (1, 2) into four equal parts and using Simpson's rule, the value of ∫12dxx will be
0.6932
0.6753
0.6692
0.7132
Taking four subintervals, the value of ∫01dx1 + x by usin trapezoidal rule will be
0.6870
0.6677
0.6970
0.5970
∫2018x2017 + 2018x loge2018x2018 + 2018xdx =
log2018x + x2018 + c
2018x + x2018-1 + c
2018x + x2018 + c
∫dxx2 + 4x + 5 =
tan-1x + c
tan-1x + cx + 2
tan-1x + 2 + c
x + 2tan-1x + 2 + c
C.
We have, I = ∫dxx2 + 4x + 5⇒ I = ∫dxx + 22 + 1⇒ I = tan-1x + 2 + c
∫0π2cos2x2 - sin2x2 =
0
1
- 1
None of the above
Evaluate ∫- 21fxdx, where f(x) = 1 - 2x, x ≤ 01 + 2x, x ≥ 0
2
4
6
∫dxxx + 9 is equal to
23tan-1x + C
23tan-1x3 + C
tan-1x + C
tan-1x3 + C
∫x + 12exdx is equal to
xex + C
x2xx + C
(x + 1)ex + C
(x2 + 1)ex + C
∫dxa2sin2x + b2cos2x is equal to
1abtan-1atanxb + C
tan-1atanxb + C
1abtan-1btanxa + C
tan-1btanxa + C
∫0π2sin8xcos2xdx is equal to
π512
3π512
5π512
7π512