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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).

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.

600 Views

89.

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

363 Views

90.
Using integration, find the area bounded by the curve x² = 4y and the line x = 4y – 2.

The shaded area OBAO represents the area bounded by the curve x² = 4y and the line x = 4y – 2.

Let A and B be the points of intersection of the line and parabola.
Co-ordinates of point A are $open parentheses negative 1 comma space 1 fourth close parentheses. space$ Co-ordinates of point B are (2, 1).
Area OBAO = Area OBCO + Area OACO ...(1)
Area OBCO =
$equals integral subscript 0 superscript 2 fraction numerator straight x plus 2 over denominator 4 end fraction dx minus integral subscript 0 superscript 2 straight x squared over 4 dx equals 1 fourth open square brackets straight x squared over 2 plus 2 straight x close square brackets subscript 0 superscript 2 minus 1 fourth open square brackets straight x cubed over 3 close square brackets subscript 0 superscript 2 equals 1 fourth open square brackets 2 plus 4 close square brackets minus 1 fourth open square brackets 8 over 3 close square brackets equals 3 over 2 minus 2 over 3 equals 5 over 6$
Area OACO =
$equals integral subscript negative 1 end subscript superscript 0 fraction numerator straight x plus 2 over denominator 4 end fraction dx minus integral subscript negative 1 end subscript superscript 0 straight x squared over 4 dx equals 1 fourth open square brackets straight x squared over 2 plus 2 straight x close square brackets subscript negative 1 end subscript superscript 0 space minus space 1 fourth open square brackets straight x cubed over 3 close square brackets subscript negative 1 end subscript superscript 0 equals 1 fourth open square brackets negative fraction numerator left parenthesis negative 1 right parenthesis squared over denominator 2 end fraction minus 2 left parenthesis negative 1 right parenthesis close square brackets minus 1 fourth open square brackets negative open parentheses negative 1 close parentheses cubed over 3 close square brackets equals 1 fourth open square brackets negative 1 half plus 2 close square brackets minus 1 fourth open square brackets 1 third close square brackets equals 3 over 8 minus 1 over 12 equals 7 over 24$
Therefore, required area = $open parentheses 5 over 6 plus 7 over 24 close parentheses space equals 9 over 8 sq. units$