NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.2 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.2.
|Chapter Name||Application of Integrals|
|Number of Questions Solved||7|
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.2
Find the area of the circle 4x² + 4y² = 9 which is interior to the parabola x² = 4y.
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 =
Find the area bounded by curves (x – 1)² + y² = 1 and x² + y² = 1.
Given circles are x² + y² = 1 …(i)
and (x – 1)² + y² = 1 …(ii)
Centre of (i) is O (0,0) and radius = 1
Find the area of the region bounded by the curves y = x² + 2, y = x, x = 0 and x = 3.
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
Using integration find the area of region bounded by the triangle whose vertices are (-1,0), (1,3) and (3,2).
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
Using integration find the area of the triangular region whose sides have the equations y = 2x + 1,y = 3x + 1 and x = 4.
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),
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)
(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)
Area lying between the curves y² = 4x and y = 2x.
(b) The curve is y² = 4x …(1)
and the line is y = 2x …(2)
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