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Application of Integrals

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Multiple 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
$fraction numerator negative 15 over denominator 4 end fraction$
$15 over 4$
$15 over 4$

110 Views

83. The area bounded by the curve y = x |x|, x-axis and the ordinates x = - 1 and x = 1 is given by

0
$1 third$
$2 over 3$
$2 over 3$

122 Views

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

4 over 3 left parenthesis 8 straight pi minus square root of 3 right parenthesis

The equation of circle is
$straight x squared plus straight y squared space equals space 16$ ...(1)
The equation of parabola is
   $straight y squared space equals space 6 straight x$ ...(2)
From (1) and (2),
   $straight x squared plus 6 straight x space equals space 16$
or $straight x squared plus 6 straight x minus 16 space equals space 0$
$therefore space space left parenthesis straight x plus 8 right parenthesis space left parenthesis straight x minus 2 right parenthesis space equals space 0$
$rightwards double arrow$ x = -8, 2
Rejecting x = -8 as parabola lies in 1st or 4th quadrant, we get x = 2
When x = 2,    $straight y equals square root of 12 space equals space 2 square root of 3$
$therefore$ (1) and (2) intersect in $straight P left parenthesis 2 comma space 2 square root of 3 right parenthesis$
Required Area = Area of the circle - area of circle interior to the parabola
   $equals space 16 straight pi minus 2 space integral subscript 0 superscript 2 straight y space dx space minus space 2 integral subscript 2 superscript 4 straight y space dx equals space 16 straight pi minus 2 integral subscript 0 superscript 2 square root of 6 space straight x to the power of 1 half end exponent dx space minus space 2 integral subscript 0 superscript 4 square root of 16 minus straight x squared end root space dx space equals space 16 straight pi space minus space 2 square root of 6 open square brackets 2 over 3 straight x to the power of 3 over 2 end exponent close square brackets subscript 0 superscript 2 space minus space 2 open square brackets straight x over 2 square root of 16 minus straight x squared end root plus 16 over 2 sin to the power of negative 1 end exponent straight x over 4 close square brackets subscript 2 superscript 4$

$equals 16 straight pi minus 4 over 3 square root of 6. space 2 square root of 2 minus 2 open square brackets 4 over 2 square root of 16 minus 6 end root plus 16 over 2 sin to the power of negative 1 end exponent 1 minus 2 over 2 square root of 16 minus 4 end root minus 16 over 2 sin to the power of negative 1 end exponent 2 over 4 close square brackets equals space 16 straight pi minus fraction numerator 16 over denominator square root of 3 end fraction minus 16. space straight pi over 2 plus 4 square root of 3 plus fraction numerator 8 straight pi over denominator 3 end fraction equals negative fraction numerator 4 over denominator square root of 3 end fraction plus fraction numerator 32 straight pi over denominator 3 end fraction equals space 4 over 3 left parenthesis 8 straight pi minus square root of 3 right parenthesis$

196 Views

85.

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

$2 left parenthesis square root of 2 minus 1 right parenthesis$
$square root of 2 minus 1$
$square root of 2 plus 1$
$square root of 2 plus 1$

162 Views

Long 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.