NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals
Application of Integrals Chapter 8: NCERT Solutions
NCERT Solutions for Class 12 Maths Chapter 8 Applications of Integrals is a significant topic from the point of your upcoming boards mathematics examination. The exercises of the chapter have been prepared by experts around India to bring out the highest potentials of the students.
SimplyAcad has provided the best NCERT Solutions for Class 12 Application of Integrals Chapter 8 covering all the important exercises in an organised manner. Each question is explained in a detailed way to get students ready for the preparation of the upcoming boards as well as the several entrance examinations. The concepts are explained in a precise and simple language to avoid any confusions.
The chapter deals with a total of 2 exercises and one miscellaneous exercise, the answers of all are provided through a step-by-step approach for better understanding. Solutions provided here will help with your daily assignments and build command on the concepts related to the Application of Integrals.
Here are the main topics and sub-topics included in the Integrals chapter are the following:
&am | |
8.1 | Area under Simple Curves |
8.2 | Area between Two Curves |
NCERT Solutions for Class 12 Maths Chapter 8 Applications of Integrals – Exercise 8.1
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 area of the region bounded by the curve, y2 = x, the lines, x = 1 and x = 4, and the x-axis is the area ABCD.
Question 2. Find the area of the region bounded by y2 = 9x, x = 2, x = 4 and the x-axis in the first quadrant.
Solution :
The area of the region bounded by the curve, y2 = 9x, x = 2, and x = 4, and the x-axis is the area ABCD.
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 area of the region bounded by the curve, x2 = 4y, y = 2, and y = 4, and the y-axis is the area ABCD.
Question 4. Find the area of the region bounded by the ellipse
Solution :
The given equation of the ellipse, , can be represented as
It can be observed that the ellipse is symmetrical about x-axis and y-axis.
∴ Area bounded by ellipse = 4 × Area of OAB
Therefore, area bounded by the ellipse = 4 × 3π = 12π units
Question 5. Find the area of the region bounded by the ellipse
Solution :
The given equation of the ellipse can be represented as
It can be observed that the ellipse is symmetrical about x-axis and y-axis.
∴ Area bounded by ellipse = 4 × Area OAB
Therefore, area bounded by the ellipse = 4 x3π/2 = 6π units.
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 :
The area of the region bounded by the circle, x2 + y2 = 4,x = √3y and the x-axis is the area OAB.
The point of intersection of the line and the circle in the first quadrant is (√3,1).
Area OAB = Area ΔOCA + Area ACB
Question 7. Find the area of the smaller part of the circle x2 + y2 = a2 cut off by the line x = a/√2
Solution :
The area of the smaller part of the circle, x2 + y2 = a2, cut off by the line, x = a/√2, is the area ABCDA.
It can be observed that the area ABCD is symmetrical about x-axis.
∴ Area ABCD = 2 × Area ABC
Therefore, the area of smaller part of the circle, x2 + y2 = a2, cut off by the line,
Question8. 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 :
The line, x = a, divides the area bounded by the parabola and x = 4 into two equal parts.
∴ Area OAD = Area ABCD
It can be observed that the given area is symmetrical about x-axis.
⇒ Area OED = Area EFCD
Question9. Find the area of the region bounded by the parabola y = x2 and y = |x|
Solution :
The area bounded by the parabola, x2 = y,and the line,y = |x|, can be represented as
The given area is symmetrical about y-axis.
∴ Area OACO = Area ODBO
The point of intersection of parabola, x2 = y, and line, y = x, is A (1, 1).
Area of OACO = Area ΔOAM – Area OMACO
Therefore, required area = 2[1/6] = 1/3 units
Question10. Find the area bounded by the curve x2 = 4y and the line x = 4y – 2
Solution : The area bounded by the curve, x2 = 4y, and line, x = 4y – 2, is represented by the shaded area OBAO.
Let A and B be the points of intersection of the line and parabola.
Coordinates of point A are (-1, 1/4).
Coordinates of point B are (2, 1).
We draw AL and BM perpendicular to the x-axis.
It can be observed that,
Area OBAO = Area OBCO + Area OACO … (1)
Then, Area OBCO = Area OMBC – Area OMBO
Similarly, Area OACO = Area OLAC – Area OLAO
Therefore, required area = units
Question11. Find the area of the region bounded by the curve y2 = 4x and the line x = 3
Solution :
The region bounded by the parabola, y2 = 4x, and the line, x = 3, is the area OACO.
The area OACO is symmetrical about x-axis.
∴ Area of OACO = 2 (Area of OAB)
Therefore, the required area is 8√3 units.
Question12. Choose the correct answer:
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. π/2
C. π/3
D. π/4
Solution :
The area bounded by the circle and the lines, x = 0 and x = 2, in the first quadrant is represented as
Therefore, option (A) is correct.
Question13. Choose the correct answer:
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
Solution :
The area bounded by the curve, y2 = 4x, y-axis, and y = 3 is represented as
Therefore, option (B) is correct.
NCERT Solutions for Class 12 Applications of Integrals Chapter 8- Exercise 8.2
Question 1. Find the area of the circle 4x2 + 4y2 = 9 which is interior to the parabola x2 = 4y
Solution :
The required area is represented by the shaded area OBCDO.
Solving the given equation of circle, 4x2 + 4y2 = 9, and parabola, x2 = 4y, we obtain the point of intersection as B(√2,1/2) and D (-√2,1/2).
It can be observed that the required area is symmetrical about the y-axis.
∴ Area OBCDO = 2 × Area OBCO
We draw BM perpendicular to OA.
Therefore, the coordinates of M are (√2,0).
Therefore, Area OBCO = Area OMBCO – Area OMBO
Therefore, the required area OBCDO is
Question 2. Find the area bounded by curves (x – 1)2 + y2 = 1 and x2 + y 2 = 1
Solution :
The area bounded by the curves, (x – 1)2 + y2 = 1 and x2 + y 2 = 1, is represented by the shaded area as
On solving the equations, (x – 1)2 + y2 = 1 and x2 + y 2 = 1, we obtain the point of intersection as A (1/2,√3/2) and B (1/2,√3/2) .
It can be observed that the required area is symmetrical about the x-axis.
∴ Area OBCAO = 2 × Area OCAO
We join AB, which intersects OC at M, such that AM is perpendicular to OC.
The coordinates of M are (1/2,0).
Therefore, required area OBCAO = units.
Question 3. Find the area of the region bounded by the curves y = x2 + 2, y = x, x = 0 and x = 3
Solution :
The area bounded by the curves, y = x2 + 2, y = x, x = 0, and x = 3, is represented by the shaded area OCBAO as
Then, Area OCBAO = Area ODBAO – Area ODCO
Question 4. Using integration, find the area of the region bounded by the triangle whose vertices are (–1, 0), (1, 3) and (3, 2).
Solution :
BL and CM are drawn perpendicular to x-axis.
It can be observed in the following figure that,
Area (ΔACB) = Area (ALBA) + Area (BLMCB) – Area (AMCA) … (1)
Therefore, from equation (1), we obtain
Area (ΔABC) = (3 + 5 – 4) = 4 units.
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 equations of sides of the triangle are y = 2x +1, y = 3x + 1, and x = 4.
On solving these equations, we obtain the vertices of triangle as A(0, 1), B(4, 13), and C (4, 9).
It can be observed that,
Area (ΔACB) = Area (OLBAO) –Area (OLCAO)
Question 6. Choose the correct answer:
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)
Solution :
The smaller area enclosed by the circle, x2 + y2 = 4, and the line, x + y = 2, is represented by the shaded area ACBA as
It can be observed that,
Area ACBA = Area OACBO – Area (ΔOAB)
Therefore, option (B) is correct.
Question 7. Choose the correct answer:
Area lying between the curves y2 = 4x and y = 2x is
(A) 2/3
(B) 1/3
(C) 1/4
(D) 3/4
Solution :
The area lying between the curve, y2 = 4x and y = 2x, is represented by the shaded area OBAO as
The points of intersection of these curves are O (0, 0) and A (1, 2).
We draw AC perpendicular to x-axis such that the coordinates of C are (1, 0).
∴ Area OBAO = Area (OCABO) – Area (ΔOCA)
Therefore, option (B) is correct.
NCERT Solutions for Class 12 Applications of Integrals Chapter 8 Miscellaneous Exercise
Question 1. Find the area under the given curves and given lines:
(i) y = x2, x = 1, x = 2 and x-axis
(ii) y = x4, x = 1, x = 5 and x –axis
Solution :
(i)The required area is represented by the shaded area ADCBA as
(ii)The required area is represented by the shaded area ADCBA as
Question 2. Find the area between the curves y = x and y = x2
Solution :
The required area is represented by the shaded area OBAO as
The points of intersection of the curves, y = x and y = x2, is A (1, 1).
We draw AC perpendicular to the x-axis.
∴ Area (OBAO) = Area (ΔOCA) – Area (OCABO) … (1)
Question 3. Find the area of the region lying in the first quadrant and bounded by y = 4x2, x = 0, y = 1 and y = 4
Solution : The area in the first quadrant bounded by y = 4x2, x = 0, y = 1, and y = 4 is represented by the shaded area ABCDA as
Question 4. Sketch the graph of y = |x + 3| and evaluate
Solution :
The given equation is y = |x + 3|
The corresponding values of x and y are given in the following table.
On plotting these points, we obtain the graph of y = |x + 3| as follows.
Question 5. Find the area bounded by the curve y = sin x between x = 0 and x = 2π
Solution :
The graph of y = sin x can be drawn as
∴ Required area = Area OABO + Area BCDB
&amp;lt;p><strong>Question 6<</strong>/strong>. Find the area enclosed between the parabola <em>y</</em>em>2 = 4ax</em> and the line y = mxSolution :
The area enclosed between the parabola, y2 = 4ax, and the line, y = mx, is represented by the shaded area OABO as
The points of intersection of both the curves are (0, 0) and .
We draw AC perpendicular to the x-axis.
∴ Area OABO = Area OCABO – Area (ΔOCA)
Question 7. Find the area enclosed by the parabola 4y = 3x2 and the line 2y = 3x + 12
Solution :
The area enclosed between the parabola, 4y = 3x2, and the line, 2y = 3x + 12, is represented by the shaded area OBAO as
The points of intersection of the given curves are A (–2, 3) and (4, 12).
We draw AC and BD perpendicular to x-axis.
∴ Area OBAO = Area CDBA – (Area ODBO + Area OACO)
Question 8. Find the area of the smaller region bounded by the ellipse and the line
Solution :
The area of the smaller region bounded by the ellipse, = 1, and the line, , is represented by the shaded region BCAB as
∴ Area BCAB = Area (OBCAO) – Area (OBAO)
Question 9. Find the area of the smaller region bounded by the ellipse and the line
Solution :
The area of the smaller region bounded by the ellipse, , and the line, , is represented by the shaded region BCAB as
∴ Area BCAB = Area (OBCAO) – Area (OBAO)
Question 10. Find the area of the region enclosed by the parabola x2 = y, the line y = x + 2 and x-axis
Solution :
The area of the region enclosed by the parabola, x2 = y, the line, y = x + 2, and x-axis is represented by the shaded region OACO as
The point of intersection of the parabola, x2 = y, and the line, y = x + 2, is A (–1, 1) and C(2, 4).
Question 11.Using the method of integration, find the area enclosed 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 = 11]
Solution :
The area bounded by the curve, |x| + |y| = 1, is represented by the shaded region ADCB as
The curve intersects the axes at points A (0, 1), B (1, 0), C (0, –1), and D (–1, 0).
It can be observed that the given curve is symmetrical about x-axis and y-axis.
∴ Area ADCB = 4 × Area OBAO
Question 12. Find the area bounded by curves {(x, y) : y ≥ x2 and y = |x|}.
Solution :
The area bounded by the curves, {(x, y) : y ≥ x2 and y = |x|}., is represented by the shaded region as
It can be observed that the required area is symmetrical about y-axis.
Question 13.Using the method of integration, find the area of the triangle whose vertices are A (2, 0), B (4, 5) and C (6, 3).
Solution :
Vertices of the given triangle are A (2, 0), B (4, 5) and C (6, 3).
Equation of side AB is
Equation of side BC
Equation of side CA is
Area (ΔABC) = Area (ABLA) + Area (BLMCB) – Area (ACMA)
Question 14.Using the method of integration, find the area of the region bounded by the lines: 2x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0
Solution :
The given equations of lines are
2x + y = 4 … (1)
3x – 2y = 6 … (2)
And, x – 3y + 5 = 0 … (3)
The area of the region bounded by the lines is the area of ΔABC. AL and CM are the perpendiculars on x-axis.
Area (ΔABC) = Area (ALMCA) – Area (ALB) – Area (CMB)
Question 15. Find the area of the region {(x, y) : y2 ≤ 4x, 4×2 + 4y2 ≤ 9}.
Solution :
The area bounded by the curves, {(x, y) : y2 ≤ 4x, 4×2 + 4y2 ≤ 9}, is represented as
The points of intersection of both the curves are (1/2,√2) and (1/2, -√2).
The required area is given by OABCO.
It can be observed that the area OABCO is symmetrical about the x-axis.
∴ Area OABCO = 2 × Area OBC
Area OBCO = Area OMC + Area MBC
Therefore, the required area is units.
Question 16.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
Solution :
Therefore, option (D) is correct.
Question 17.Choose the correct answer:
The area bounded by the curve y = 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
Solution :
Therefore, option (C) is correct.
Question 18.Choose the correct answer:
The area of the circle x2 + y2 = 16 exterior to the parabola y2 = 6x is
Solution :
The given equations are
x2 + y2 = 16 … (1)
y2 = 6x … (2)
Area bounded by the circle and parabola
Thus, the correct answer is C.
Question 19.Choose the correct answer:
The area bounded by the y-axis, y = cos x and y = sin x when 0 ≤ x ≤ π/2.
(A) 2(√2 – 1)
(B) √2 – 1
(C) √2 + 1
(D) √2
Solution :
The given equations are
y = cos x … (1)
And, y = sin x … (2)
Required area = Area (ABLA) + area (OBLO)
Required area = Area (AB
Therefore, option (B) is correct.
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