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Chapters
Chapter 2: Inverse Trigonometric Functions
Chapter 3: Matrices
Chapter 4: Determinants
Chapter 5: Continuity and Differentiability
Chapter 6: Application of Derivatives
Chapter 7: Integrals
Chapter 8: Application of Integrals
Chapter 9: Differential Equations
Chapter 10: Vector Algebra
Chapter 11: Three Dimensional Geometry
Chapter 12: Linear Programming
Chapter 13: Probability
Solutions for Chapter 8: Application of Integrals
Below listed, you can find solutions for Chapter 8 of CBSE, Karnataka Board PUC NCERT for Class 12 Maths.
NCERT solutions for Class 12 Maths Chapter 8 Application of Integrals Exercise 8.1 [Pages 365 - 366]
Find the area of the region bounded by the curve y2 = x and the lines x = 1, x = 4 and the x-axis.
Find the area of the region bounded by y2 = 9x, x = 2, x = 4 and the x-axis in the first quadrant.
Find the area of the region bounded by x2 = 4y, y = 2, y = 4 and the y-axis in the first quadrant.
Find the area of the region bounded by the ellipse `x^2/16 + y^2/9 = 1`
Find the area of the region bounded by the ellipse `x^2/4 + y^2/9 = 1`
Find the area of the region in the first quadrant enclosed by x-axis, line x = `sqrt3` y and the circle x2 + y2 = 4.
Find the area of the smaller part of the circle x2 + y2 = a2 cut off by the line `x = a/sqrt2`
The area between x = y2 and x = 4 is divided into two equal parts by the line x = a, find the value of a.
Find the area of the region bounded by the parabola y = x2 and y = |x| .
Find the area bounded by the curve x2 = 4y and the line x = 4y – 2
Find the area of the region bounded by the curve y2 = 4x and the line x = 3
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. `pi/2`
C. `pi/3`
D. `pi/4`
Area of the region bounded by the curve y2 = 4x, y-axis and the line y = 3 is
A. 2
B. 9/4
C. 9/3
D. 9/2
NCERT solutions for Class 12 Maths Chapter 8 Application of Integrals Exercise 8.2 [Pages 371 - 372]
Find the area of the circle 4x2 + 4y2 = 9 which is interior to the parabola x2 = 4y
Find the area bounded by curves (x – 1)2 + y2 = 1 and x2 + y 2 = 1
Find the area of the region bounded by the curves y = x2 + 2, y = x, x = 0 and x = 3
Using integration finds the area of the region bounded by the triangle whose vertices are (–1, 0), (1, 3) and (3, 2).
Using integration find the area of the triangular region whose sides have the equations y = 2x +1, y = 3x + 1 and x = 4.
Smaller area enclosed by the circle x2 + y2 = 4 and the line x + y = 2 is
A. 2 (π – 2)
B. π – 2
C. 2π – 1
D. 2 (π + 2)
Area lying between the curve y2 = 4x and y = 2x is
A. 2/3
B. 1/3
C. 1/4
D. 3/4
NCERT solutions for Class 12 Maths Chapter 8 Application of Integrals Exercise 8.3 [Pages 375 - 376]
Find the area under the given curves and given lines:
y = x2, x = 1, x = 2 and x-axis
Find the area under the given curves and given lines:
y = x4, x = 1, x = 5 and x –axis
Find the area between the curves y = x and y = x2
Find the area of the region lying in the first quadrant and bounded by y = 4x2, x = 0, y = 1 and y = 4
Sketch the graph of y = |x + 3| and evaluate `int_(-6)^0 |x + 3|dx`
Find the area bounded by the curve y = sin x between x = 0 and x = 2π
Find the area enclosed between the parabola y2 = 4ax and the line y = mx
Find the area enclosed by the parabola 4y = 3x2 and the line 2y = 3x + 12
Find the area of the smaller region bounded by the ellipse `x^2/9 + y^2/4` and the line `x/3 + y/2 = 1`
Find the area of the smaller region bounded by the ellipse `x^2/a^2 + y^2/b^2 = 1` and the line `x/a + y/b = 1`
Find the area of the region enclosed by the parabola x2 = y, the line y = x + 2 and x-axis
Using the method of integration find the area bounded by the curve |x| + |y| = 1 .
[Hint: The required region is bounded by lines x + y = 1, x– y = 1, – x + y = 1 and
– x – y = 1].
Find the area bounded by curves {(x, y) : y ≥ x2 and y = |x|}.
Using the method of integration find the area of the triangle ABC, coordinates of whose vertices are A(2, 0), B (4, 5) and C (6, 3).
Using the method of integration find the area of the region bounded by lines: 2x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0
Find the area of the region {(x, y) : y2 ≤ 4x, 4x2 + 4y2 ≤ 9}
Choose the correct answer Area bounded by the curve y = x3, the x-axis and the ordinates x = –2 and x = 1 is
A. – 9
B. `-15/4`
C. `15/4`
D. `17/4`
Choose the correct answer The area bounded by the curve y = x | x| ,, x-axis and the ordinates x = –1 and x = 1 is given by
[Hint: y = x2 if x > 0 and y = –x2 if x < 0]
A. 0
B. `1/3`
C. `2/3`
D. `4/3`
Choose the correct answer The area of the circle x2 + y2 = 16 exterior to the parabola y2 = 6x is
A. `4/3 (4pi - sqrt3)`
B. `4/3 (4pi + sqrt3)`
C. `4/3 (8pi - sqrt3)`
D.`4/3 (4pi + sqrt3)`
The area bounded by the y-axis, y = cos x and y = sin x when 0 <= x <= `pi/2`
(A) 2 ( 2 −1)
(B) `sqrt2 -1`
(C) `sqrt2 + 1`
D. `sqrt2`
Solutions for Chapter 8: Application of Integrals
NCERT solutions for Class 12 Maths chapter 8 - Application of Integrals
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Concepts covered in Class 12 Maths chapter 8 Application of Integrals are Area of the Region Bounded by a Curve and a Line, Area Between Two Curves, Area Under Simple Curves.
Using NCERT Class 12 Maths solutions Application of Integrals exercise by students is an easy way to prepare for the exams, as they involve solutions arranged chapter-wise and also page-wise. The questions involved in NCERT Solutions are essential questions that can be asked in the final exam. Maximum CBSE, Karnataka Board PUC Class 12 Maths students prefer NCERT Textbook Solutions to score more in exams.
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