The area bounded by the curve y = x |x|, x-axis and the ordinate

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 Multiple Choice QuestionsMultiple Choice Questions

81.

Area lying between the curves y2 = 4x and y = 2x  is

  • 2 over 3
  • 1 third
  • 1 fourth
  • 1 fourth
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82. Area bounded by the curve y = x3  the x-axis and the ordinates x - 2 and a = 1 is
  • -9

  • fraction numerator negative 15 over denominator 4 end fraction
  • 15 over 4
  • 15 over 4
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83. The area bounded by the curve y = x |x|, x-axis and the ordinates x = - 1 and x = 1 is given by
  • 0

  • 1 third
  • 2 over 3
  • 2 over 3


C.

2 over 3

We want to find the area bounded by the curve straight y equals straight x space open vertical bar straight x close vertical bar comma x-axis and the ordinates x = -1 and x = 1
Required area = integral subscript negative 1 end subscript superscript 1 space straight y space dx space equals space integral subscript negative 1 end subscript superscript 1 space straight x space open vertical bar straight x close vertical bar space dx
              equals space integral subscript negative 1 end subscript superscript 0 open vertical bar straight x open vertical bar straight x close vertical bar close vertical bar space dx space plus space integral subscript 0 superscript 1 space straight x space open vertical bar straight x close vertical bar space dx
                                      open square brackets because space space space open vertical bar straight x space open vertical bar straight x close vertical bar close vertical bar space equals space open vertical bar straight x close vertical bar space open vertical bar straight x close vertical bar space equals straight x squared close square brackets
                 equals space integral subscript negative 1 end subscript superscript 0 left parenthesis straight x squared right parenthesis dx plus integral subscript 0 superscript 1 straight x squared dx space equals space open square brackets straight x cubed over 3 close square brackets subscript negative 1 end subscript superscript 0 plus open square brackets straight x cubed over 3 close square brackets subscript 0 superscript 1
equals space open square brackets 0 minus fraction numerator left parenthesis negative 1 right parenthesis cubed over denominator 3 end fraction close square brackets plus open square brackets 1 third minus 0 close square brackets space equals space 1 third plus 1 third equals 2 over 3


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84. The area of the circle x +y =16 exterior to the parabola y2 = 6x is
  • 4 over 3 left parenthesis 4 straight pi minus square root of 3 right parenthesis
  • 4 over 3 left parenthesis 4 straight pi plus square root of 3 right parenthesis
  • 4 over 3 left parenthesis 8 straight pi minus square root of 3 right parenthesis
  • 4 over 3 left parenthesis 8 straight pi minus square root of 3 right parenthesis
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85.

The area bounded by the x-axis, y = cosx and y = sin x when 0 less or equal than straight x less or equal than straight pi over 2

  • 2 left parenthesis square root of 2 minus 1 right parenthesis
  • square root of 2 minus 1
  • square root of 2 plus 1
  • square root of 2 plus 1
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 Multiple Choice QuestionsLong Answer Type

86. Prove that the curves y2 = 4x and x2 = 4y divide the area of the square bounded by x = 0, x = 4, y = 4 and y = 0 into three equal parts.
640 Views

87.

Using the method of integration, find the area of the triangular region whose vertices are (2, -2), (4, 3) and (1, 2).

1133 Views

88.

Sketch the region bounded by the curves straight y equals square root of 5 minus straight x squared end root space and space straight y space equals open vertical bar straight x minus 1 close vertical bar and find its area using integration. 

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89.

Using integration, find the area of the region bounded by the triangle whose vertices are (-1, 2), (1, 5) and (3, 4).

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90.

Using integration, find the area bounded by the curve x2 = 4y and the line x = 4y – 2.

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