Dividing the interval (1, 2) into four equal parts and using Simp

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611.

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


A.

0.6932

We have, 12dxxHere, a = 1, b = 2, n = 4   nh = b - a  4h = 2 - 1 = 1    h = 14The value of f(x) = 1x are given below
x 1 54 64 74 2
f(x) y0 = 1 y145 y2 = 46 y347 y412

By Simpson's rule,

12dxx = h3y0 + y4 + 4y1 + y3 + 2y2= 1121 + 12 + 445 + 47 + 246= 11232 + 161235 + 43= 18 + 1635 + 19= 0.125 + 0.4571 + 0.1111= 0.6932


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612.

Taking four subintervals, the value of 01dx1 + x by usin trapezoidal rule will be

  • 0.6870

  • 0.6677

  • 0.6970

  • 0.5970


613.

2018x2017 + 2018x loge2018x2018 + 2018xdx =

  • log2018x + x2018 + c

  • 2018x + x2018-1 + c

  • 2018x + x2018 + c

  • 2018x + x2018 + c


614.

dxx2 + 4x +5 =

  • tan-1x + c

  • tan-1x + cx + 2

  • tan-1x + 2 + c

  • x + 2tan-1x + 2 + c


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615.

0π2cos2x2 - sin2x2 =

  • 0

  • 1

  • - 1

  • None of the above


616.

Evaluate - 21fxdx, where f(x) = 1 - 2x, x  01 + 2x, x  0

  • 0

  • 2

  • 4

  • 6


617.

dxxx +9 is equal to

  • 23tan-1x + C

  • 23tan-1x3 + C

  • tan-1x + C

  • tan-1x3 + C


618.

x + 12exdx is equal to

  • xex + C

  • x2xx + C

  • (x + 1)ex + C

  • (x2 + 1)ex + C


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619.

dxa2sin2x + b2cos2x is equal to

  • 1abtan-1atanxb + C

  • tan-1atanxb + C

  • 1abtan-1btanxa + C

  • tan-1btanxa + C


620.

0π2sin8xcos2xdx is equal to

  • π512

  • 3π512

  • 5π512

  • 7π512


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