The value of I = ∫01xx - 12dx from Mathematic

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 Multiple Choice QuestionsMultiple Choice Questions

121.

0π2sin2x . logtanxdx

  • 0

  • 2

  • 4

  • 7


122.

If x = 1cosx1 - cosx1 + sinxcosx1 + sinx + cosxsinxsinx1, then 0π4xdx is equal to

  • 14

  • 12

  • 0

  • - 14


123.

02πsinx + sinxdx is equal to

  • 0

  • 4

  • 8

  • 1


124.

The value of 02x2dx, where [.] is the greatest integer function, is

  • 2 - 2

  • 2 + 2

  • 2 - 1

  • 2 - 2


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125.

If l (m,n) = 01tm1 + tndt, then the expression for l (m, n) in terms of l (m + 1, n + 1) is

  • 2nm + 1 - nm + 1 . l m + 1, n - 1

  • nm + 1 . l m + 1, n - 1

  • 2nm + 1 + nm + 1 . l m + 1, n - 1

  • mn + 1 . l m + 1, n - 1


126.

1 + x - x- 1ex + x- 1dx is equal to

  • x + 1ex + x- 1 + C

  • x - 1ex + x- 1 + C

  • xex + x- 1 + C

  • xex + x- 1 x + C


127.

If f(x) = x - [x], for every real number x, where [x] is the integral part of x. Then - 11f(x)dx is equal to

  • 1

  • 2

  • 0

  • 12


128.

The value of integral

- 11x + 1x - 12 + x + 1x - 12 - 212dx is

  • log43

  • 4log34

  • 4log43

  • log34


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129.

dxsinx - cosx + 2 equals to

  • - 12tanx2 + π8 + C

  • 12tanx2 + π8 + C

  • 12cotx2 + π8 + C

  • - 12cotx2 + π8 + C


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130.

The value of I = 01xx - 12dx

  • 13

  • 14

  • 18

  • None of these


C.

18

Let I = 01xx - 12dx= - 012xx - 12dx + 121xx - 12dx= 012x2 - x2dx + 121x2 - x2dx= x24 - x33012 +x33 - x24121= 116 - 124 + 13 - 14 - 124 + 116= 6 - 496 + 32 - 24 - 4 + 696= 1296 = 18


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