RBSE Solutions for Class 12 Maths Chapter 11 Application of Integral:Quadrature Miscellaneous Exercise

RBSE Solutions for Class 12 Maths Chapter 11 Application of Integral:Quadrature Miscellaneous Exercise

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 = x−−√ and y = x is :
(a) 1 sq. unit
(d) 19 sq. unit
(c) 16 sq. unit
(d) 23 sq. unit
Solution:
Curve y = x−−√ is a parabola whose equation is y² = 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 y² = x and x² = y is :
(a) 1/3 sq. unit
(d) 1 sq. unit
(c) 1/2 sq. unit
(d) 2 sq. unit
Solution:
Sllving given equations
y² = x …..(i)
x² = y …..(ii)
We get integration (0,0) and (4,4)

Question 3.
Area of region enclosed by and its latus rectum parabola x² = 4y is :
(a) 5/3 sq. unit
(d) 2/3 sq. unit
(c) 4/3 sq. unit
(d) 8/3 sq. unit
Solution:
Graph of parabola x² = 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 x² = 4y is x = ± 2
Required area = Area AOCBA
= 2[Area OQCBO – Area OQCO)

Question 4.
Area of region enclosed by y = sin x, π/2 < x < 3π/2 and x-axis is :
(a) 1 sq. unit
(d) 2 sq. unit
(c) 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 y² = 2x and circle x² + y² = 8 is :

Solution:
y² = 2x is equation of parabola whose centre (0, 0) and circle whose centre is (0, 0) and radius is 2√2 units. Their graph of enclosed area is shown by shaded part.

Question 6.
Find the area of the region bounded by parabola y² = x and line x + y = 2.
Solution:
Area of region bounded by parabola y² = 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 y² = 2ax – x² and y² = ax
Solution:
Area enclosed by curve y² = 2ax – x² and y² = ax is shown by shaded part in the following graph.

Question 8.
Find the area of region bounded by parabola y = x² and y = | x |.
Solution:
Curve y = x² 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 = x² are O(0, 0) and A (1, 1).
Intersection points of y = – x and parabola y = x² and O(0, 0) and B (- 1, 1).
The region bounded by lines y = x and y = – x and parabola y = x² is shown in the following figure.

Question 9.
Find the area of the common region bounded by x² + y² = 16 and parabola y² = 6x.
Solution:
Centre of circle x² + y² = 16 is origin and radius is 4 unit. Vertex of parabola y² = 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 x² + y² = 16 …..(i)
y² = 6x …..(ii)

Question 10.
Find the area of the region bounded by curve x² + y² = 1 and x + y > 1.
Solution:
Area enclosed by circle x² + y² = 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 y² = 2x, y = 4x – 1 and y > 0.
Solution:

Question 14.
Find the area of the region bounded by curve y² = 4x, y-axis and line y = 3.
Solution:
Curve y² = 4x is a parabola whose vertex is origin and symmetric about x-axis. Area enclosed by curved y² = 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 x² + y² = 4 and (x – 2)² + y² = 4.
Solution:
Equations of given circles
x² + y² = 4 ….(i)
and (x – 2)² + y² = 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(1,3–√) and Q (1 – 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 + y² = 4)
+ Area CPAC (part of circle x² + y² = 4)
∴ Required Area
= 2 ∫ y dx (for circle (x – 2)² + y² = 4)
+ ∫ y dx (for circle x² + y² = 4)