## Rajasthan Board RBSE Class 12 Maths Chapter 11 Application of Integral:Quadrature Miscellaneous Exercise

Question 1.

Area of the region bounded by the curve y = \(\sqrt { x }\) and y = x is :

(a) 1 sq. unit

(d) \(\frac { 1 }{ 9 } \) sq. unit

(c) \(\frac { 1 }{ 6 } \) sq. unit

(d) \(\frac { 2 }{ 3 } \) sq. unit

Solution:

Curve y = \(\sqrt { x }\) is a parabola whose equation is y^{2} = x and centre is at origin y = x is a line which passes through origin which is shown in following figure.

Question 2.

Area of region enclosed by y^{2} = x and x^{2} = y is :

(a) \(\frac { 1 }{ 3 } \) sq. unit

(d) 1 sq. unit

(c) \(\frac { 1 }{ 2 } \) sq. unit

(d) 2 sq. unit

Solution:

Sllving given equations

y^{2} = x …..(i)

x^{2} = y …..(ii)

We get integration (0,0) and (4,4)

Question 3.

Area of region enclosed by and its latus rectum parabola x^{2} = 4y is :

(a) \(\frac { 5 }{ 3 } \) sq. unit

(d) \(\frac { 2 }{ 3 } \) sq. unit

(c) \(\frac { 4 }{ 3 } \) sq. unit

(d) \(\frac { 8 }{ 3 } \) sq. unit

Solution:

Graph of parabola x^{2} = 4y is given below :

Here, a=1

Thus, latus rectum intersects y axis at (0, 1).

Solving equation of latus rectum y = 1 and equation of parabola x^{2} = 4y is x = ± 2

Required area = Area AOCBA

= 2[Area OQCBO – Area OQCO)

Question 4.

Area of region enclosed by y = sin x, \(\frac { \pi }{ 2 } \) < x < \(\frac { 3\pi }{ 2 } \) and x-axis is :

(a) 1 sq. unit

(d) 2 sq. unit

(c) \(\frac { 1 }{ 2 } \) sq. unit

(d) 4 sq. unit

Solution:

The area enclosed by curve y = sin x and x = π/2 and x = 3rc/2 is shown by shaded part in figure for various values of x, table for values of y = sin x is given below :

Question 5.

Area of region enclosed by y^{2} = 2x and circle x^{2} + y^{2} = 8 is :

Solution:

y^{2} = 2x is equation of parabola whose centre (0, 0) and circle whose centre is (0, 0) and radius is 2\(\sqrt { 2 }\) units. Their graph of enclosed area is shown by shaded part.

Question 6.

Find the area of the region bounded by parabola y^{2} = x and line x + y = 2.

Solution:

Area of region bounded by parabola y^{2} = x and x + y = 2 is shown by shaded part in the following graph.

Question 7.

Find the area of the region bounded by curve y^{2} = 2ax – x^{2} and y^{2} = ax

Solution:

Area enclosed by curve y^{2} = 2ax – x^{2} and y^{2} = ax is shown by shaded part in the following graph.

Question 8.

Find the area of region bounded by parabola y = x^{2} and y = | x |.

Solution:

Curve y = x^{2} is a parabola whose vertex is (0, 0) and is symmetric about y-axis.

Equation y = | x | represents two lines

When x > 0, then y = x

When x < 0, then y = -x

Intersection points of y = x and parabola y = x^{2} are O(0, 0) and A (1, 1).

Intersection points of y = – x and parabola y = x^{2} and O(0, 0) and B (- 1, 1).

The region bounded by lines y = x and y = – x and parabola y = x^{2} is shown in the following figure.

Question 9.

Find the area of the common region bounded by x^{2} + y^{2} = 16 and parabola y^{2} = 6x.

Solution:

Centre of circle x^{2} + y^{2} = 16 is origin and radius is 4 unit. Vertex of parabola y^{2} = 6x is origin. Common part of these curves is shown in figure.

Two curves intersect at point P and Q by solving equations, coordinates of these can be obtained.

Equation of curves x^{2} + y^{2} = 16 …..(i)

y^{2} = 6x …..(ii)

Question 10.

Find the area of the region bounded by curve x^{2} + y^{2} = 1 and x + y > 1.

Solution:

Area enclosed by circle x^{2} + y^{2} = 1 and line x + y > 1 is shown by shaded part in figure.

Question 11.

By using intersection method, find the area of the triangle whose vertices are (- 1, 0), (1, 2) and (3, 2).

Solution:

Graph of triangle is given below. Required Area is shaded in the figure given below:

Now, putting values of area ∆ABP, area of trapezium BPAQ and area of ∆AQC in equation (i).

Required Area of ∆ABC

= 3 + 5 – 4 = 4 sq. unit

∴ Area of ∆ABC = 4 sq. unit

Question 12.

Find the area of region bounded by line y = 3 x + 2, x axis and ordinates x = – 1 and x = 1.

Solution:

Area enclosed by line y = 3x + 2, x axis and ordinates x = – 1 and x = 1 is shown below by shaded part.

Question 13.

Find the area of the region bounded by y^{2} = 2x, y = 4x – 1 and y > 0.

Solution:

Question 14.

Find the area of the region bounded by curve y^{2} = 4x, y-axis and line y = 3.

Solution:

Curve y^{2} = 4x is a parabola whose vertex is origin and symmetric about x-axis. Area enclosed by curved y^{2} = 4x and line y = 3 is shown by shaded part in figure given below.

Question 15.

Find the area of the region bounded by two circles x^{2} + y^{2} = 4 and (x – 2)^{2} + y^{2} = 4.

Solution:

Equations of given circles

x^{2} + y^{2} = 4 ….(i)

and (x – 2)^{2} + y^{2} = 4 ….(ii)

Centre of circle from equation –

(i) is at origin (0, 0) and radius is 2 unit. Centre of circle of equation

(ii) is (2, 0) and at x axis and radius is 2 unit. Required enclosed area is shown by shaded part in figure.

Solving equation (i) and (ii)

Obtained intersecting points of circles are p(1,\(\sqrt { 3 }\)) and Q (1 – \(\sqrt { 3 }\))

Out of two circles one symmetric about x axis

∴ Required area

= 2(Area OPACO) = 2 [Area OPCO + Area CPAC]

= 2[Area OPCO (part of circle (x – 2)^{2} + y^{2} = 4)

+ Area CPAC (part of circle x^{2} + y^{2} = 4)

∴ Required Area

= 2 ∫ y dx (for circle (x – 2)^{2} + y^{2} = 4)

+ ∫ y dx (for circle x^{2} + y^{2} = 4)

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