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Application of Integrals

Multiple Choice Questions

141.

The area bounded by the curves y = x², y = - x² and y² = 4x - 3 is k, then the value of 6k is

142.

The area (in square units) of the region bounded by x² = 8y, x = 4 and X-axis, is

$\frac{2}{3}$
$\frac{4}{3}$
$\frac{8}{3}$
$\frac{10}{3}$

143.

The area bouned by y = x² + 2, x-axis, x = 1 and x = 2 is

$\frac{16}{3}$
$\frac{17}{3}$
$\frac{13}{3}$
$\frac{20}{3}$

144.

The area (in square units) bounded by the curves y² = 4x and x² = 4y in the plane is

$\frac{8}{3}$
$\frac{16}{3}$
$\frac{32}{3}$
$\frac{64}{3}$

145.

The area (in square unit) of the region enclosed by the curves y = x and y = x³ is

$\frac{1}{12}$
$\frac{1}{6}$
$\frac{1}{3}$
1

146.

The area (in sq unit) of the region bounded by the curves 2x = y² - 1 and x = 0 is

$\frac{1}{3}$
$\frac{2}{3}$
1
2

147.
The area (in square units) of the region enclosed by the two circles x² + y² = 1 and (x - 1)² + y² = 1 is

$\frac{2 π}{3} + \frac{\sqrt{3}}{2}$

$\frac{π}{3} + \frac{\sqrt{3}}{2}$

$\frac{π}{3} - \frac{\sqrt{3}}{2}$

$\frac{2 π}{3} - \frac{\sqrt{3}}{2}$

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$\frac{2 π}{3} - \frac{\sqrt{3}}{2}$

Intersection point of two circles

$x^{2} + y^{2} = 1 . . . (i) {(x - 1)}^{2} + y^{2} = 1 . . . (ii) is given by {(x - 1)}^{2} + {(1 - x)}^{2} = 1 \Rightarrow x^{2} + 1 - 2 x - x^{2} = 0 \Rightarrow x = \frac{1}{2} From eq . (i), \frac{1}{4} + y^{2} = 1 y^{2} = 1 - \frac{1}{4} \Rightarrow y = \pm \frac{\sqrt{3}}{2} Point A (\frac{1}{2}, \frac{\sqrt{3}}{2}) and C (\frac{1}{2}, \frac{- \sqrt{3}}{2}) So, Area of region OABDO = 2 \times Area of region OABDO . . . (iii) Area of OABDO = Area of OADO + Area of ABDA$

$= \int_{0}^{\frac{1}{2}} \sqrt{1 - {(x - 1)}^{2}} dx + \int_{\frac{1}{2}}^{1} \sqrt{1 - x^{2}} dx = {[\frac{1}{2} . (x - 1) \sqrt{1 - {(x - 1)}^{2}} + \frac{1}{2} \sin^{- 1} (\frac{x - 1}{1})]}^{\frac{1}{2}}_{0} + {[\frac{1}{2} x \sqrt{1 - x^{2}} + \frac{1}{2} \sin^{- 1} (\frac{x}{1})]}^{1}_{\frac{1}{2}} = [- \frac{1}{2} \frac{1}{2} \frac{\sqrt{3}}{2} + \frac{1}{2} \sin^{- 1} (\frac{- 1}{2}) - \frac{1}{2} 0 - \frac{1}{2} \sin^{- 1} (- 1)] + [\frac{1}{2} . 0 + \frac{1}{2} \sin^{- 1} (1) - \frac{1}{4} \frac{\sqrt{3}}{2} - \frac{1}{2} \sin^{- 1} (\frac{1}{2})]$

$= (- \frac{\sqrt{3}}{8}) - \frac{1}{2} \sin^{- 1} (\frac{1}{2}) + \frac{1}{2} \sin^{- 1} (1) + \frac{1}{2} \sin^{- 1} (1) - \frac{\sqrt{3}}{8} - \frac{1}{2} \sin^{- 1} (\frac{1}{2})$

$= \sin^{- 1} (1) - \sin^{- 1} (\frac{1}{2}) - \frac{\sqrt{3}}{4} = \frac{π}{2} - \frac{π}{6} - \frac{\sqrt{3}}{4} = (\frac{π}{3} - \frac{\sqrt{3}}{4}) From eq . (iii) Area of region OABCO = 2 \times (\frac{π}{3} - \frac{\sqrt{3}}{4}) = (\frac{2 π}{3} - \frac{\sqrt{3}}{2})$

148.

The values of a function f(x) at different values of x are as follows

x 0 1 2 3 4 5

f(x) 2 3 6 11 18 27

Then, the approximate area (in square units) bounded by the curve y = f(x) and x-axis between x = 0 and 5, using the Trapezoidal rule, is

50
75
52.5
62.5

149.

The area (in square units) of the region bounded by the curves x = y² and x = 3 - 2y² is

$\frac{3}{2}$
2
3
4

150.

The area (in sq units) bounded by the curves y² = 4x and x² = 4y is

$\frac{64}{3}$
$\frac{16}{3}$
$\frac{8}{3}$
$\frac{2}{3}$

Previous Year Papers

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Multiple Choice Questions

147. The area (in square units) of the region enclosed by the two circles x2 + y2 = 1 and (x - 1)2 + y2 = 1 is 2π3 + 32 π3 + 32 π3 - 32 2π3 - 32

147.
The area (in square units) of the region enclosed by the two circles x² + y² = 1 and (x - 1)² + y² = 1 is

$\frac{2 π}{3} + \frac{\sqrt{3}}{2}$

$\frac{π}{3} + \frac{\sqrt{3}}{2}$

$\frac{π}{3} - \frac{\sqrt{3}}{2}$

$\frac{2 π}{3} - \frac{\sqrt{3}}{2}$