In this chapter, we provide NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.2 for English medium students, Which will very helpful for every student in their exams. Students can download the latest NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.2 pdf, free NCERT solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.2 book pdf download.
| Board | CBSE |
| Textbook | NCERT |
| Class | Class 12 |
| Subject | Maths |
| Chapter | Chapter 8 |
| Chapter Name | Application of Integrals |
| Exercise | Ex 8.2 |
| Number of Questions Solved | 7 |
| Category | NCERT Solutions |
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.2
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)
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