NCERT Solutions for 12th Class Maths: Chapter 8-Application of Integrals
NCERT Solutions for 12th Class Maths: Chapter 8-Application of Integrals

Class 12: Maths Chapter 8 solutions. Complete Class 12Maths Chapter 8 Notes.

NCERT Solutions for 12th Class Maths: Chapter 8-Application of Integrals

Class 12:Maths Chapter 8 solutions. Complete Class 12Maths Chapter 8 Notes.

Ex 8.1 Class 12 Maths Question 1.
Find the area of the region bounded by the curve y² = x and the lines x = 1, x = 4, and the x-axis.
Solution:
The curve y² = x is a parabola with vertex at origin.Axis of x is the line of symmetry, which is the axis of parabola. The area of the region bounded by the curve, x = 1, x=4 and the x-axis. Area LMQP

Ex 8.1 Class 12 Maths Question 2.
Find the area of the region bounded by y² = 9x, x = 2, x = 4 and x-axis in the first quadrant
Solution:
The given curve is y² = 9x, which is a parabola with vertex at (0, 0) and axis along x-axis. It is symmetrical about x-axis, as it contains only even powers of y. x = 2 and x = 4 are straight lines parallel toy-axis at a positive distance of 2 and 4 units from it respectively.
∴ Required area = Area ABCD

Ex 8.1 Class 12 Maths Question 3.
Find the area of the region bounded by x² = 4y, y = 2, y = 4 and the y-axis in the first quadrant.
Solution:
The given curve x² = 4y is a parabola with vertex at (0,0). Also since it contains only even powers of x,it is symmetrical about y-axis.y = 2 and y = 4 are straight lines parallel to x-axis at a positive distance of 2 and 4 from it respectively.

Ex 8.1 Class 12 Maths Question 4.
Find the area of the region bounded by the ellipse 


Solution:
The equation of the ellipse is 


The given ellipse is symmetrical about both axis as it contains only even powers of y and x.

Ex 8.1 Class 12 Maths Question 5.
Find the area of the region bounded by the ellipse 


Solution:

 It is an ellipse with centre (0,0) and length of major axis = 2a = 2×3 = 6 and length of minor axis = 2b = 2 × 2 = 4.

Ex 8.1 Class 12 Maths Question 6.
Find the area of the region in the first quadrant enclosed by x-axis, line x = √3y and the circle x² + y² = 4. .
Solution:
Consider the two equations x² + y² = 4 … (i)
and x = √3y i.e. 

 …(ii)
(1) x² + y² = 4 is a circle with centre O (0,0) and radius = 2.

Ex 8.1 Class 12 Maths Question 7.
Find the area of the smaller part of the circle x² + y² = a² cut off by the line 


Solution:
The equation of the given curve are
x² + y² = a² …(i) and 

 …(ii)
Clearly, (i) represent a circle and (ii) is the equation of a straight line parallel to y-axis at a

Ex 8.1 Class 12 Maths Question 8.
The area between x = y² and x = 4 is divided into two equal parts by the line x = a, find the value of a.
Solution:
Graph of the curve x = y² is a parabola as given in the figure. Its vertex is O and axis is x-axis. QR is the ordinate along x = 4

Ex 8.1 Class 12 Maths Question 9.
Find the area of the region bounded by the parabola y = x² and y = |x|.
Solution:
Clearly x² = y represents a parabola with vertex at (0, 0) positive direction of y-axis as its axis it opens upwards.
y = |x| i.e., y = x and y = -x represent two lines passing through the origin and making an angle of 45° and 135° with the positive direction of the x-axis.
The required region is the shaded region as shown in the figure. Since both the curve are symmetrical about y-axis. So,
Required area = 2 (shaded area in the first quadrant)

Ex 8.1 Class 12 Maths Question 10.
Find the area bounded by the curve x² = 4y and the line x = 4y – 2
Solution:
Given curve is x² = 4y …(i)
which is an upward parabola with vertex at (0,0) and it is symmetrical about y-axis
Equation of the line is x = 4y – 2 …(ii)
Solving (i) and (ii) simultaneously, we get; (4y – 2)² = 4y
⇒ 16y² – 16y + 4 = 4y
⇒ 4y² – 5y + 1 = 0


Ex 8.1 Class 12 Maths Question 11.
Find the area of the region bounded by the curve y² = 4x and the line x = 3.
Solution:
The curve y² = 4x is a parabola as shown in the figure. Axis of the parabola is x-axis. The area of the region bounded by the curve y² = 4x and the line x = 3 is
A = Area of region OPQ = 2 (Area of the region OLQ)

Choose the correct answer in the following Exercises 12 and 13:

Ex 8.1 Class 12 Maths Question 12.
Area lying in the first quadrant and bounded by the circle x² + y² = 4 and the lines x = 0 and x = 2 is
(a) π
(b) 


(c) 


(d) 


Solution:
(a) x² + y² = 4. It is a circle at the centre (0,0) and r=2

Ex 8.1 Class 12 Maths Question 13.
Area of the region bounded by the curve y² = 4x, y-axis and the line y = 3 is
(a) 2
(b) 


(c) 


(d) 


Solution:
(b) y² = 4x is a parabola

Ex 8.2 Class 12 Maths Question 1.
Find the area of the circle 4x² + 4y² = 9 which is interior to the parabola x² = 4y.
Solution:
Area is bounded by the circle 4x² + 4y² = 9 and interior of the parabola x² = 4y.
Putting x² = 4y in x² + y² = 


We get 4y + y² = 


Ex 8.2 Class 12 Maths Question 2.
Find the area bounded by curves (x – 1)² + y² = 1 and x² + y² = 1.
Solution:
Given circles are x² + y² = 1 …(i)
and (x – 1)² + y² = 1 …(ii)
Centre of (i) is O (0,0) and radius = 1


Ex 8.2 Class 12 Maths Question 3.
Find the area of the region bounded by the curves y = x² + 2, y = x, x = 0 and x = 3.
Solution:
Equation of the parabola is y = x² + 2 or x² = (y – 2)
Its vertex is (0,2) axis is y-axis.
Boundary lines are y = x, x = 0, x = 3.
Graphs of the curve and lines have been shown in the figure.
Area of the region PQRO = Area of the region OAQR-Area of region OAP

Ex 8.2 Class 12 Maths Question 4.
Using integration find the area of region bounded by the triangle whose vertices are (-1,0), (1,3) and (3,2).
Solution:
The points A (-1,0), B( 1,3) and C (3,2) are plotted and joined.
Area of ∆ABC = Area of ∆ ABL + Area of trap. BLMC – Area of ∆ACM …(i)
The equation of the line joining the points

Ex 8.2 Class 12 Maths Question 5.
Using integration find the area of the triangular region whose sides have the equations y = 2x + 1,y = 3x + 1 and x = 4.
Solution:
The given lines are y = 2x + 1 …(i)
y = 3x + 1 …(ii)
x = 4 …(iii)
Subtract (i) from eq (ii) we get x = 0, Putting x = 0 in eq(i) y = 1
∴ Lines (ii) and (i) intersect at A (0,1) putting x = 4 in eq. (2) =>y = 12 + 1 = 13
The lines (ii) and (iii) intersect at B (4,13) putting x=4ineq. (i):y = 8 + 1 = 9
∴ Lines (i) and (ii); Intersect at C (4,9),

Ex 8.2 Class 12 Maths Question 6.
Smaller area bounded by the circle x² + y² = 4 and the line x + y = 2
(a) 2 (π – 2)
(b) π – 2
(c) 2π – 1
(d) 2(π + 2)
Solution:
(b) A circle of radius 2 and centre at O is drawn.The line AB: x + y = 2 is passed through (2,0) and (0,2). Area of the region ACB
= Area of quadrant OAB – Area of ∆OAB …(i)

Ex 8.2 Class 12 Maths Question 7.
Area lying between the curves y² = 4x and y = 2x.
(a) 


(b) 


(c) 


(d) 


Solution:
(b) The curve is y² = 4x …(1)
and the line is y = 2x …(2)

NCERT Solutions for 12th Class Maths: Chapter 8: Download PDF

NCERT Solutions for 12th Class Maths: Chapter 8-Application of Integrals

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