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Application of Integrals

Name: Zigya - For The Curious Learner
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Long Answer Type

91.

Using integration, find the area of the region enclosed between the two circles:
$straight x squared plus straight y squared space equals space 4 space and space left parenthesis straight x minus 2 right parenthesis squared plus straight y squared space equals space 4.$

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92.

Find the area bounded by the circle x² + y² = 16 and the line √3y=x in the first quadrant, using integration.

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93.

Using integration, find the area of region bounded by the triangle whose vertices are (–2, 1), (0, 4) and (2, 3).

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94.

Using integration, find the area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle x² + y² =32

95.
Using integration find the area of the region bounded by the parabola y² = 4x and the circle 4x² + 4y² = 9.

The respective equations for the parabola and the circle are:

$y^{2} = 4 x . . . . . . . . . . . . (1) 4 x^{2} + 4 y^{2} = 9 . . . . . . . . . . . . (2) or x^{2} + y^{2} = {(\frac{3}{2})}^{2}$

Equations (1) is a parabola with vertex ( 0, 0 ) which opens to the right and equation (2) is a circle with centre (0, 0 ) and radius $\frac{3}{2}$ .

From equations (1) and (2), we get:

$4 x^{2} + 4 (4 x) = 9 4 x^{2} + 16 x - 9 = 0 4 x^{2} + 18 x - 2 x - 9 = 0 2 x (2 x + 9) - 1 (2 x + 9) = 0 (2 x + 9) (2 x - 1) = 0 x = - \frac{9}{2}, \frac{1}{2} For x = - \frac{9}{2}, y^{2} = 4 (- \frac{9}{2}), which is not possible, hence x = \frac{1}{2} Therefore, the given curves intersect at x = \frac{1}{2} .$

Required area of the region bound by the two curves

$= 2 \int_{0}^{\frac{1}{2}} 2 \sqrt{x} dx + 2 \int_{\frac{1}{2}}^{\frac{3}{2}} \sqrt{\frac{9}{4} - x^{2}} dx = 4 {[\frac{2}{3} x^{\frac{3}{2}}]}_{0}^{\frac{1}{2}} + 2 {[\frac{x}{2} \sqrt{\frac{9}{4} - x^{2}} + \frac{9}{8} \sin^{- 1} (\frac{2 x}{3})]}_{\frac{1}{2}}^{\frac{3}{2}} = \frac{8}{3} {(\frac{1}{8})}^{\frac{1}{2}} + 2 [0 + \frac{9}{8} \sin^{- 1} 1 - \frac{1}{4} \sqrt{2} - \frac{9}{8} \sin^{- 1} (\frac{1}{3})] = \frac{8}{3} (\frac{1}{2 \sqrt{2}}) + \frac{9}{4} (\frac{π}{2}) - \frac{\sqrt{2}}{2} - \frac{9}{4} \sin^{- 1} (\frac{1}{3}) = \frac{2 \sqrt{2}}{3} + \frac{9 π}{8} - \frac{\sqrt{2}}{2} - \frac{9}{4} \sin^{- 1} (\frac{1}{3})$

96.

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

97.

Using integration, find the area of the following region:

$\{(x, y) : \frac{x^{2}}{9} + \frac{y^{2}}{4} \leq 1 \leq \frac{x}{3} + \frac{y}{2}\}$

98.

Using integration find the area of the triangular region whose sides have equations y=2x+1, y=3x+1 and x=4.

99.

Using the method of method of integration, find the area of the region bounded by the following lines:

3x – y – 3 = 0,

2x + y – 12 = 0,

x – 2y – 1 = 0

Multiple Choice Questions

100.

The area (in sq. units) enclosed between the curves y = x² and y = $|x|$ is

$\frac{2}{3}$
$\frac{1}{6}$
$\frac{1}{3}$
1