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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

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

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.

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87.

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

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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).

Consider the vertices, A(-1, 2), B(1, 5) and C(3, 4).
Let us find the equation of the sides of the triangle $increment ABC.$
Thus, the equation of AB is:

$fraction numerator straight y minus 5 over denominator 5 minus 2 end fraction space equals space fraction numerator straight x minus 1 over denominator 1 plus 1 end fraction rightwards double arrow 3 straight x minus 2 straight y plus 7 space equals space 0 Similarly comma space the space equation space of space BC space is colon fraction numerator straight y minus 4 over denominator 4 minus 5 end fraction equals fraction numerator straight x minus 3 over denominator 3 minus 1 end fraction rightwards double arrow straight x plus 2 straight y minus 11 space equals 0 Also comma space the space equation space of space CA space is colon fraction numerator straight y minus 4 over denominator 4 minus 2 end fraction space equals fraction numerator straight x minus 3 over denominator 3 plus 1 end fraction rightwards double arrow straight x minus 2 straight y plus 5 space equals 0$

Now the area of $increment ABC$ = Area of $increment ADB$ + Area of $increment BDC$

$therefore space Area space of space increment ADB space equals space integral subscript negative 1 end subscript superscript 1 open square brackets fraction numerator 3 straight x plus 7 over denominator 2 end fraction minus fraction numerator straight x plus 5 over denominator 2 end fraction close square brackets dx$

$Similarly comma space Area space of space increment BDC equals space integral subscript 1 superscript 3 open square brackets fraction numerator 11 minus straight x over denominator 2 end fraction minus fraction numerator straight x plus 5 over denominator 2 end fraction close square brackets dx Thus comma space Area space of space increment ADB space plus space Area space of space increment BDC equals space integral subscript negative 1 end subscript superscript 1 open square brackets fraction numerator 3 straight x plus 7 over denominator 2 end fraction minus fraction numerator straight x plus 5 over denominator 2 end fraction close square brackets dx plus integral subscript 1 superscript 3 open square brackets fraction numerator 11 minus straight x over denominator 2 end fraction minus fraction numerator straight x plus 5 over denominator 2 end fraction close square brackets dx equals integral subscript negative 1 end subscript superscript 1 open square brackets fraction numerator 2 straight x plus 2 over denominator 2 end fraction close square brackets dx plus integral subscript 1 superscript 3 open square brackets fraction numerator 6 minus 2 straight x over denominator 2 end fraction close square brackets dx equals integral subscript negative 1 end subscript superscript 1 open square brackets straight x plus 1 close square brackets dx plus integral subscript 1 superscript 3 open square brackets 3 minus straight x close square brackets dx equals open square brackets straight x squared over 2 plus straight x close square brackets subscript negative 1 end subscript superscript 1 space plus space open square brackets 3 straight x minus straight x squared over 2 close square brackets subscript 1 superscript 3 equals 0 plus 2 plus 9 minus 9 over 2 minus 3 plus 1 half equals 2 plus 9 over 2 minus 5 over 2 equals 4 space square space units$