NCERT Solutions Class 12 Mathematics Chapter 8

Q: Is it necessary for one to go over all the questions in NCERT Solutions Class 12 Mathematics Chapter 8 Application of Integrals?

Questions that get asked in a board test are highly unpredictable. Math professionals have carefully selected all of the sums in the NCERT Solutions Class 12 Mathematics Chapter 8 – Application of Integrals to provide students with a complete picture of the subject. As a result, if students wish to increase their chances of achieving a high grade on their exams, they must thoroughly practise all problems. They should also look at prior test papers when preparing for the boards.

Q: What is the significance of NCERT Solutions Class 12 Mathematics Chapter 8?

Class 12 is one of the essential grades in a student’s academic career. Therefore, it is equally crucial to ensure that students have developed a mathematical foundation that will carry them through this class and throughout their college careers. NCERT Solutions Class 12 Mathematics Chapter 8 provides a detailed explanation of every exercise problem, allowing students to clarify any doubts about the subject. As a result, youngsters will get clarity of the concepts that will last throughout their lives.

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Integral Calculus is related to Eudoxus’ (440 B.C.) and Archimedes’ (440 B.C.) exhaustion approach (300 B.C.). This method emerged from the need to solve problems estimating the areas of plane figures, surface areas, and solid body volumes. Newton began working on Calculus in 1665, which led to the formulation of concepts as we know them. NCERT solutions class 12 mathematics chapter 8 provides various hints, tactics, and intriguing facts to help students break up the monotony of studying and develop a mathematical base that will help them throughout their studies.

NCERT solutions class 12 mathematics chapter 8 – Application of Integrals is a continuation of the previous chapter, in which students learned about integration and how to solve problems using various methods. Students have learnt formulae for calculating the areas of triangles, rectangles, trapeziums, and circles, among other geometrical figures. These fundamental formulae are crucial in applying mathematics to a wide range of real-world issues. However, this method fails when estimating the areas encompassed by curves. Integral calculus is required to calculate the areas under a curve. NCERT solutions class 12 mathematics chapter 8 use integrals to get the area under curves. Curves such as circles, parabolas, and hyperbolas can be simple or complex.

Key Topics Covered in NCERT Solutions Class 12 Mathematics Chapter 8

All the topics covered in NCERT Solutions Class 12 Mathematics Chapter 8 are equally important. They go over many aspects of integration and how to calculate the area under a curve throughout multiple scenarios. Before attempting the practice questions, students must first grasp the underlying concepts. They should also be familiar with the preceding chapter and the formulas listed. The NCERT Solutions Class 12 Mathematics Chapter 8 broadly covers two main concepts of the chapter ‘Application of Integrals’ and ‘Integral Calculus’.

List of NCERT Solutions Class 12 Mathematics Chapter 8 Exercises

Students learned about definite integrals as the limit of a sum in Chapter 7 and how to evaluate them using the Fundamental Theorem of Calculus. This concept is used in NCERT solutions for class 12 mathematics chapter 8 to give students an easy approach to computing the area under curves. The area beneath the curve can be thought of as a collection of several very thin vertical stripes. We may now determine the total area necessary by integrating this region between some specific limitations. NCERT Solutions for class 12 mathematics chapter 8 Application of Integrals gives students a better grasp of integration using real-life examples. We’ll look at a range of examples, formulas, and theories in depth in this article to help students create a strong mathematical foundation.

Several formulas and theorems are used in this lesson. There is also a blend of current topics with those that the students learned in the previous chapter. If not approached systematically, integration can become a difficult and perplexing concept. Students must regularly review this lesson for all these reasons, particularly the formulas, to learn the material. With the help of examples and practice problems, all the tasks in NCERT Solutions Class 12 Mathematics Chapter 8 – Application of Integrals are well explained. Here are some NCERT solutions for students to practise –

Class 12 Mathematics Chapter 8 – Ex 8.1 Solutions – 13 Questions

Class 12 Mathematics Chapter 8 – Ex 8.2 Solutions – 7 Questions

Class 12 Mathematics Chapter 8 Miscellaneous Ex – 19 Questions

The area bounded by the curve of a function, the area between two curves, and the area under simple curves are all covered in NCERT solutions class 12 mathematics chapter 8.

NCERT Solutions Class 12 Mathematics Chapter 8 Formula List

NCERT solutions class 12 mathematics chapter 8 requires using formulas and a constant focus. Students must comprehend what the question requires and carry out the procedures stepwise. They should also divide the problem into smaller parts to make it easier to understand and avoid ambiguity, hence lowering the risk of errors. More significantly, children must understand the processes in the technique since they are the keys to answering a variety of questions.

The following are some formulas that are regularly used in NCERT solutions for class 12 mathematics chapter 8 –

Area of the region bounded by the curve y = f (x), x-axis and the lines x = a and x = b (b > a) : a∫by dx = a∫bf(x) dx.
The area of the region bounded by the curve x = φ (y), y-axis and the lines y = c, y = d : c∫d x dy = c∫d φ (y) dy.
The area of the region enclosed between two curves y = f (x), y = g (x) and the lines x = a, x = b : b∫a[f(x) – g(x)] dx.

Class 12 NCERT Mathematics Syllabus

Term – 1

Unit Name

Chapter Name

Relations and Function

Relations and Functions

Inverse Trigonometric Functions

Algebra

Matrices

Determinants

Calculus

Continuity and Differentiability

Application of Derivatives

Linear Programming

Term – 2

Unit Name	Chapter Name
Calculus	Integrals Application of Integrals Differential Equations
Vectors and Three-Dimensional Geometry	Vector Algebra Three Dimensional Geometry
Probability	Probability

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NCERT Mathematics Exam Pattern

Duration of Marks	3 hours 15 minutes
Marks for Internal	20 marks
Marks for Theory	80 marks
Total Number of Questions	38 Questions
Very short answer question	20 Questions
Short answer questions	7 Questions
Long Answer Questions (4 marks each)	7 Questions
Long Answer Questions (6 marks each)	4 Questions

Key Features of NCERT Solutions Class 12 Mathematics Chapter 8

To do well in CBSE and pass various competitive exams such as NEET and IIT, one must have a strong mathematical foundation. The NCERT books present you with difficult problems that improve your analytical ability and expose you to many questions that could appear in all of these exams. Our integrals class 12 solutions adhere to the most recent CBSE curriculum and are incredibly valuable to your preparation for the following reasons:

The answers are exceptional because teachers with deep subject knowledge create them.
You can revise all the significant points of a chapter in a short amount of time.
The solutions also prepare you for managing your stress and time, giving you a better chance of finishing your examinations on time.
NCERT solutions class 12 mathematics chapter 8 will assist you in grasping the fundamental concepts quickly and easily.

NCERT Exemplar Class 12 Mathematics

Here are some solutions and problems to help students prepare for their final exams. These example questions are a little more complex, and they cover a variety of concepts in each chapter of Class 12 Mathematics subject. Students will fully understand all the concepts covered in each chapter by practising these NCERT Exemplars for mathematics Class 12. Each question in these materials is connected to concepts covered in the CBSE Class 12 syllabus (2022-2023). Our experts provide the best solutions to challenges that students confront. To match the ideas taught in each class and provide the greatest practising materials or worksheets for students, all these questions reflect the question pattern found in NCERT books.

Q.1 Find the area of the region bounded by the curve y²= x and the lines x = 1, x = 4 and the x-axis.

Ans.

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.

$\begin{array}{l} A r e a of ABCD = \int_{1}^{4} y d x \\ = \int_{1}^{4} \sqrt{x} d x \\ = {[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}]}_{1}^{4} = \frac{2}{3} [4^{\frac{3}{2}} - 1^{\frac{3}{2}}] \\ = \frac{2}{3} (8 - 1) \\ = \frac{14}{3} u n i t s \end{array}$

Q.2 Find the area of the region bounded by y²= 9x, x = 2, x = 4 and the x-axis in the first quadrant.

Ans.

The area of the region bounded by the curve, y² = 9x, the lines, x = 2 and x = 4, and the x-axis is the area ABCD.

$\begin{array}{l} A r e a of ABCD = \int_{2}^{4} y d x \\ = \int_{2}^{4} 3 \sqrt{x} d x \\ = 3 {[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}]}_{2}^{4} = 3 \times \frac{2}{3} [4^{\frac{3}{2}} - 2^{\frac{3}{2}}] \\ = 2 (8 - 2 \sqrt{2}) \\ = (16 - 4 \sqrt{2}) u n i t s \end{array}$

Q.3 Find the area of the region bounded by x²= 4y, y = 2, y = 4 and the y-axis in the first quadrant.

Ans.

The area of the region bounded by the curve, x² = 4y, the lines, y = 2 and y = 4, and the y-axis is the area ABCD.

$\begin{array}{l} A r e a of ABCD = \int_{2}^{4} x d y \\ = \int_{2}^{4} 2 \sqrt{y} d x \\ = 2 {[\frac{y^{\frac{3}{2}}}{\frac{3}{2}}]}_{2}^{4} = \frac{4}{3} [4^{\frac{3}{2}} - 2^{\frac{3}{2}}] \\ = \frac{4}{3} (8 - 2 \sqrt{2}) \\ = (\frac{32 - 8 \sqrt{2}}{3}) u n i t s \end{array}$

Q.4 Find the area of the region bounded by the ellipse (x²/16) + (y²/9) = 1.

Ans.

The given ellipse (x²/16) + (y²/9)=1 can be represented as given below:

Ellipse is symmetrical about x-axis and y-axis.
So, area bounded by ellipse = 4 x Area of OAB

$\begin{array}{l} A r e a of OAB = \int_{0}^{4} y d x \\ = \int_{0}^{4} 3 \sqrt{1 - \frac{x^{2}}{16}} d x \\ = \frac{3}{4} \int_{0}^{4} \sqrt{16 - x^{2}} d x \\ = \frac{3}{4} {[\frac{x}{2} \sqrt{16 - x^{2}} + \frac{16}{2} \sin^{- 1} \frac{x}{4}]}_{0}^{4} \\ = \frac{3}{4} [\frac{4}{2} \sqrt{16 - 4^{2}} + \frac{16}{2} \sin^{- 1} \frac{4}{4}] - \frac{3}{4} [\frac{0}{2} \sqrt{16 - 0^{2}} + \frac{16}{2} \sin^{- 1} \frac{0}{4}] \\ = \frac{3}{4} \times 8 \frac{π}{2} \\ = 3 π s q u a r e u n i t s \\ T h e r e f o r e, area bounded by ellipse \\ = 4 \times A r e a of OAB \\ = 4 \times 3 π s q u a r e u n i t s \\ = 12 π s q u a r e u n i t s \end{array}$

Q.5 Find the area of the region bounded by the ellipse (x²/4) + (y²/9) = 1.

Ans.

The given ellipse (x²/4) + (y²/9) =1 can be represented as given below:

$\begin{array}{l} T h e given equation of ellipse is \\ \frac{x^{2}}{4} + \frac{y^{2}}{9} = 1 \\ y = 3 \sqrt{1 - \frac{x^{2}}{4}} \\ I t can be observed that ellipse is symmetrical about x-axis and \\ y-axis . \\ ∴ Area bounded by ellipse = 4 \times A r e a OAB \\ ∴ Area of OAB = \int_{0}^{2} y d x \\ = \int_{0}^{2} 3 \sqrt{1 - \frac{x^{2}}{4}} d x \\ = \frac{3}{2} \int_{0}^{2} \sqrt{(4 - x^{2})} d x \\ = \frac{3}{2} {[\frac{x}{2} \sqrt{(4 - x^{2})} + \frac{4}{2} \sin^{- 1} \frac{x}{2}]}_{0}^{2} \end{array}$ $\begin{array}{l} = \frac{3}{2} [\frac{2}{2} \sqrt{(4 - 2^{2})} + \frac{4}{2} \sin^{- 1} \frac{2}{2}] - \frac{3}{2} [\frac{0}{2} \sqrt{(4 - 0^{2})} + \frac{4}{2} \sin^{- 1} \frac{0}{2}] \\ = \frac{3}{2} \times 2 \frac{π}{2} \\ = \frac{3}{2} π s q u a r e units \\ Therefore, area bounded by the ellipse \\ = 4 \times \frac{3}{2} π s q u a r e units \\ = 6 s q u a r e unit \end{array}$

Q.6

$\begin{array}{l} Findtheareaoftheregioninthefirst quadrantenclosed \\ byx - axis, linex = \sqrt{3} {yandthecirclex}^{2} + y^{2} = 4 . \end{array}$

Ans.

$\begin{array}{l} The area of the smaller part of the circle, x^{2} + y^{2} = a^{2}, cut off by \\ the line, x = \frac{a}{\sqrt{2}}, is the area ABCDA . \end{array}$

$\begin{array}{l} I t is observed that the area ABCD is symmetrical about x-axis . \\ ∴ area ABCD = 2 \times A r e a ABC \\ A r e a of ABC = \int_{\frac{a}{\sqrt{2}}}^{a} y d x \\ = \int_{\frac{a}{\sqrt{2}}}^{a} \sqrt{a^{2} - x^{2}} d x \\ = {[\frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} \sin^{- 1} \frac{x}{a}]}_{\frac{a}{\sqrt{2}}}^{a} \\ = [\frac{a}{2} \sqrt{a^{2} - a^{2}} + \frac{a^{2}}{2} \sin^{- 1} \frac{a}{a}] - [\frac{(\frac{a}{\sqrt{2}})}{2} \sqrt{a^{2} - {(\frac{a}{\sqrt{2}})}^{2}} + \frac{a^{2}}{2} \sin^{- 1} \frac{(\frac{a}{\sqrt{2}})}{a}] \\ = \frac{a^{2}}{2} \times \frac{π}{2} - \frac{a}{2 \sqrt{2}} \times \frac{a}{\sqrt{2}} - \frac{a^{2}}{2} (\frac{π}{4}) \\ = \frac{a^{2}}{2} \times \frac{π}{2} - \frac{a^{2}}{4} - \frac{a^{2}}{2} (\frac{π}{4}) \\ = \frac{a^{2}}{4} (π - 1 - \frac{π}{2}) \\ = \frac{a^{2}}{4} (\frac{π}{2} - 1) \\ ∴ A r e a ABCD = 2 {\frac{a^{2}}{4} (\frac{π}{2} - 1)} \\ = \frac{a^{2}}{2} (\frac{π}{2} - 1) s q u a r e u n i t s \\ T h e r e f o r e, {the area of the smaller part of the circle, x}^{2} {+y}^{2} = a^{2}, \\ cut off by the line, x= \frac{a}{\sqrt{2}} is \frac{a^{2}}{2} (\frac{π}{2} - 1) s q u a r e u n i t s . \end{array}$

Q.7 The area between x = y² and x = 4 is divided into two equal parts by the line x = a, find the value of a.

Ans.

The line x = a divides the area bounded by the parabola and line x= 4 into two equal parts.

$\begin{array}{l} A r e a O E D = \int_{0}^{a} y d x \\ = \int_{0}^{a} \sqrt{x} d x \\ = {[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}]}_{0}^{a} \\ = \frac{2}{3} a^{\frac{3}{2}} \\ A r e a of CDEF \\ = \int_{a}^{4} y d x \\ = \int_{a}^{4} \sqrt{x} d x \\ = {[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}]}_{a}^{4} \\ = \frac{2}{3} (4^{\frac{3}{2}} - a^{\frac{3}{2}}) \\ = \frac{2}{3} (8 - a^{\frac{3}{2}}) \\ A c c o r d i n g to equation (i), we have \\ \frac{2}{3} a^{\frac{3}{2}} = \frac{2}{3} (8 - a^{\frac{3}{2}}) \\ \Rightarrow a^{\frac{3}{2}} = (8 - a^{\frac{3}{2}}) \\ \Rightarrow 2 a^{\frac{3}{2}} = 8 \\ \Rightarrow a^{\frac{3}{2}} = 4 \\ \Rightarrow a = 4^{\frac{2}{3}} \\ T h e r e f o r e, the value of a is 4^{\frac{2}{3}} . \end{array}$

Q.8 Find the area of the region bounded by the parabola y = x² and y = |x|.

Ans.

The area bounded by the parabola y = x² and
y = |x| is shown as below:

Since the required area is symmetrical about y-axis.

So, area OACO = area OBDO

The points of intersection of parabola and line are A(1,1) and B(–1, 1).

$\begin{array}{l} A r e a of OACO = A r e a Δ AMO - a r e a O M A C O \\ = \frac{1}{2} \times O M \times A M - \int_{0}^{1} y d x \\ = \frac{1}{2} \times 1 \times 1 - \int_{0}^{1} x^{2} d x \\ = \frac{1}{2} - {[\frac{x^{3}}{3}]}_{0}^{1} \\ = \frac{1}{2} - [\frac{1^{3}}{3} - \frac{0^{3}}{3}] \\ = \frac{1}{2} - \frac{1}{3} \\ = \frac{1}{6} \\ ∴ Required area \\ = 2 (A r e a of OACO) \\ = 2 \times \frac{1}{6} \\ = \frac{1}{3} s q u a r e units \end{array}$

Q.9 Find the area of the region bounded by the curve x² = 4y and the line x = 4y – 2.

Ans.

The area bounded by the parabola x²= 4y and x = 4y – 2 is shown in figure.
The intersection points of line and parabola are A(–1, 1/4) and B(2,1).

$\begin{array}{l} H e r e A L ⊥ X - a x i s and BM ⊥ X - a x i s . T h e n \\ A r e a O B A O = A r e a O B C O + A r e a O A C O \\ = (A r e a O M B C O - A r e a O M B O) + (A r e a O L A C - A r e a O L A O) \\ = (\int_{0}^{2} \frac{x + 2}{4} d x - \int_{0}^{2} \frac{x^{2}}{4} d x) + (\int_{- 1}^{0} \frac{x + 2}{4} d x - \int_{- 1}^{0} \frac{x^{2}}{4} d x) \\ = \frac{1}{4} {[\frac{x^{2}}{2} + 2 x]}_{0}^{2} - \frac{1}{4} {[\frac{x^{3}}{3}]}_{0}^{2} + \frac{1}{4} {[\frac{x^{2}}{2} + 2 x]}_{- 1}^{0} - \frac{1}{4} {[\frac{x^{3}}{3}]}_{- 1}^{0} \\ = \frac{1}{4} [\frac{2^{2}}{2} + 2 (2)] - \frac{1}{4} [\frac{2^{3}}{3}] + 0 - \frac{1}{4} [\frac{{(- 1)}^{2}}{2} + 2 (- 1)] - 0 + \frac{1}{4} [\frac{{(- 1)}^{3}}{3}] \\ = \frac{1}{4} \times 6 - \frac{1}{4} \times \frac{8}{3} - \frac{1}{4} \times \frac{- 3}{2} + \frac{1}{4} \times \frac{- 1}{3} \\ = \frac{3}{2} - \frac{2}{3} + \frac{3}{8} - \frac{1}{12} \\ = \frac{36 - 16 + 9 - 2}{24} \\ = \frac{27}{24} = \frac{9}{8} s q u n i t s \\ T h u s, the area of OBAO is \frac{9}{8} s q . u n i t s . \end{array}$

Q.10 Find the area of the region bounded by the curve y² = 4x and the line x = 3.

Ans.

The region bounded by the parabola and line is given below:

The area COAC is symmetrical about x-axis. So,
Area OABO = Area OCBO
Then, Area OCAO = 2(area OABO)

$\begin{array}{l} A r e a O A C O = 2 \int_{0}^{3} y d x \\ = 2 \int_{0}^{3} 2 \sqrt{x} d x \\ = 4 {[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}]}_{0}^{3} \\ = \frac{2}{3} \times 4 [3^{\frac{3}{2}} - 0] = 8 \sqrt{3} \\ T h e r e f o r e, the required area is 8 \sqrt{3} s q u a r e u n i t s . \end{array}$

Q.11 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

$(a) π (b) \frac{π}{2} (c) \frac{π}{3} (d) \frac{π}{4}$

Ans.

The area bounded by the circle and the lines x = 0 and x = 2 lies in first quadrant.

$\begin{array}{l} ∴ A r e a O A B O = \int_{0}^{2} y d x \\ = \int_{0}^{2} \sqrt{4 - x^{2}} d x \\ = {[\frac{x}{2} \sqrt{4 - x^{2}} + \frac{4}{2} \sin^{- 1} (\frac{x}{2})]}_{0}^{2} \\ = [\frac{2}{2} \sqrt{4 - 2^{2}} + \frac{4}{2} \sin^{- 1} (\frac{2}{2})] - [\frac{0}{2} \sqrt{4 - 0^{2}} + \frac{4}{2} \sin^{- 1} (\frac{0}{2})] \\ = [0 + 2 \times \frac{π}{2}] - [0 + 0] = π \\ T h u s, t h e correct option is A . \end{array}$

Q.12 Choose the correct answer

Area of the region bounded by the curve y² = 4x, and y-axis and the line y = 3 is

(A) 2 (B) 9/4 (C) 9/3 (D) 9/2

Ans.

The area bounded by the parabola and the lines y = 0 and y = 3 lies in first quadrant.

$\begin{array}{l} ∴ A r e a O A B = \int_{0}^{3} x d y \\ = \int_{0}^{2} \frac{y^{2}}{4} d y \\ = \frac{1}{4} {[\frac{y^{3}}{3}]}_{0}^{3} \\ = \frac{1}{12} [3^{3} - 0^{3}] \\ = \frac{27}{12} = \frac{9}{4} \\ T h u s, t h e correct option is B . \end{array}$

Q.13 Find the area of the circle 4x² + 4y² = 9 which is interior to the parabola x² = 4y.

Ans.

The required area is represented by shaded area OBCDO.

$\begin{array}{l} O n {solving the given equations 4x}^{2} {+ 4y}^{2} = {9 and parabola x}^{2} = 4y, \\ w e get the coordinates of point of intersection as B (\sqrt{2}, \frac{1}{2}) a n d \\ C (- \sqrt{2}, \frac{1}{2}) . T h i s area is symmetrical about y-axis . \\ ∴ A r e a O B C D = 2 (a r e a O B C O) \\ S i n c e, B M ⊥ X - a x i s . \\ S o, t h e coordinates of M are (\sqrt{2}, 0) . \\ T h e r e f o r e, a r e a OBCO = A r e a O M B C O - A r e a O M B O \\ = \int_{0}^{\sqrt{2}} \sqrt{\frac{9 - 4 x^{2}}{4}} d x - \int_{0}^{\sqrt{2}} \frac{x^{2}}{4} d x \\ = \frac{1}{2} \int_{0}^{\sqrt{2}} \sqrt{9 - 4 x^{2}} d x - \int_{0}^{\sqrt{2}} \frac{x^{2}}{4} d x \\ = \frac{1}{2} \int_{0}^{2 \sqrt{2}} \sqrt{9 - t^{2}} \frac{d t}{2} - \frac{1}{4} \int_{0}^{\sqrt{2}} x^{2} d x [\begin{array}{l} L e t t = 2 x \Rightarrow \frac{d t}{d x} = 2 \\ a n d t = 2 \times \sqrt{2}, (W h e n x = \sqrt{2}) \\ = 2 \sqrt{2} \\ t = 2 \times 0 (W h e n x = 0) \\ = 0 \end{array}] \\ = \frac{1}{4} {[\frac{t}{2} \sqrt{9 - t^{2}} + \frac{9}{2} \sin^{- 1} \frac{t}{3}]}_{0}^{2 \sqrt{2}} - \frac{1}{4} {[\frac{x^{3}}{3}]}_{0}^{\sqrt{2}} \\ = \frac{1}{4} [\frac{2 \sqrt{2}}{2} \sqrt{9 - {(2 \sqrt{2})}^{2}} + \frac{9}{2} \sin^{- 1} \frac{2 \sqrt{2}}{3}] - \\ \frac{1}{4} [(0) \sqrt{9 - 4 {(0)}^{2}} + \frac{9}{2} \sin^{- 1} \frac{2 (0)}{3}] - \frac{1}{4} [\frac{{(\sqrt{2})}^{3}}{3}] \\ = \frac{1}{4} (\sqrt{2} + \frac{9}{2} \sin^{- 1} \frac{2 \sqrt{2}}{3}) - \frac{\sqrt{2}}{6} \\ = \frac{\sqrt{2}}{4} + \frac{9}{8} \sin^{- 1} \frac{2 \sqrt{2}}{3} - \frac{\sqrt{2}}{6} \\ = \frac{\sqrt{2}}{12} + \frac{9}{8} \sin^{- 1} \frac{2 \sqrt{2}}{3} \\ = \frac{1}{2} (\frac{\sqrt{2}}{6} + \frac{9}{4} \sin^{- 1} \frac{2 \sqrt{2}}{3}) \\ T h e r e f o r e, area of shaded region is 2 \times \frac{1}{2} (\frac{\sqrt{2}}{6} + \frac{9}{4} \sin^{- 1} \frac{2 \sqrt{2}}{3}) \\ = (\frac{\sqrt{2}}{6} + \frac{9}{4} \sin^{- 1} \frac{2 \sqrt{2}}{3}) s q u a r e units . \end{array}$

Q.14 Find the area bounded by curves (x – 1)² + y² = 1 and x² + y² = 1.

Ans.

The required area bounded by (x – 1)² + y² = 1 and x² + y² = 1 is represented by shaded area ACBOA.

$\begin{array}{l} O n solving the given equations {(x - 1)}^{2} {+ y}^{2} = {1 and x}^{2} {+y}^{2} = 1, \\ w e get the coordinates of point of intersection as A (\frac{1}{2}, \frac{\sqrt{3}}{2}) a n d \\ B (\frac{1}{2}, - \frac{\sqrt{3}}{2}) . T h i s area is symmetrical about x-axis . \\ ∴ A r e a O B C A O = 2 (a r e a O C A O) \\ J o i n AB such that A M ⊥ O C . \\ S o, t h e coordinates of M are (\frac{1}{2}, 0) . \\ T h e r e f o r e, a r e a OCAO = A r e a O M A O + A r e a M C A M \\ = \int_{0}^{\frac{1}{2}} \sqrt{1 - {(x - 1)}^{2}} d x + \int_{\frac{1}{2}}^{1} \sqrt{1 - x^{2}} d x \\ = {[\frac{(x - 1)}{2} \sqrt{1 - {(x - 1)}^{2}} + \frac{1}{2} \sin^{- 1} (x - 1)]}_{0}^{\frac{1}{2}} \\ + {[\frac{x}{2} \sqrt{1 - x^{2}} + \frac{1}{2} \sin^{- 1} x]}_{\frac{1}{2}}^{1} \\ = [\frac{(\frac{1}{2} - 1)}{2} \sqrt{1 - {(\frac{1}{2} - 1)}^{2}} + \frac{1}{2} \sin^{- 1} (\frac{1}{2} - 1)] \\ - [\frac{(0 - 1)}{2} \sqrt{1 - {(0 - 1)}^{2}} + \frac{1}{2} \sin^{- 1} (0 - 1)] + [\frac{1}{2} \sqrt{1 - 1^{2}} + \frac{1}{2} \sin^{- 1} 1] \\ - [\frac{\frac{1}{2}}{2} \sqrt{1 - {(\frac{1}{2})}^{2}} + \frac{1}{2} \sin^{- 1} (\frac{1}{2})] \\ = [- \frac{\sqrt{3}}{8} + \frac{1}{2} (- \frac{π}{6}) - \frac{1}{2} (- \frac{π}{2})] + \frac{1}{2} (\frac{π}{2}) - \frac{\sqrt{3}}{8} - \frac{1}{2} (\frac{π}{6}) \\ = - \frac{\sqrt{3}}{8} - \frac{π}{12} + \frac{π}{4} + \frac{π}{4} - \frac{\sqrt{3}}{8} - \frac{π}{12} \\ = - \frac{\sqrt{3}}{4} - \frac{π}{6} + \frac{π}{2} = \frac{2 π}{6} - \frac{\sqrt{3}}{4} \\ T h e r e f o r e, required area of OBCAO \\ = 2 (\frac{2 π}{6} - \frac{\sqrt{3}}{4}) = (\frac{2 π}{3} - \frac{\sqrt{3}}{2}) square units \end{array}$

Q.15 Find the area of the region bounded by the curves y = x² + 2, y = x, x = 0 and x = 3.

Ans.

The area bounded by y = x²+2, y = x, x = 0 and x = 3 is represented by the shaded area

$\begin{array}{l} T h e n, a r e a O C B A O = a r e a O D B A O - a r e a O D C O \\ = \int_{0}^{3} (x^{2} + 2) d x - \int_{0}^{3} x d x \\ = {[\frac{x^{3}}{3} + 2 x]}_{0}^{3} - {[\frac{x^{2}}{2}]}_{0}^{3} \\ = [\frac{3^{3}}{3} + 2 (3) - 0 - 0] - [\frac{3^{2}}{2} - 0] \\ = 15 - \frac{9}{2} = \frac{21}{2} u n i t s \\ T h u s, the required area of shaded region is \frac{21}{2} s q u n i t s . \end{array}$

Q.16 Using integration find the area of region bounded by the triangle whose vertices are (–1, 0), (1, 3) and (3, 2).

Ans.

Here, BL and CM are perpendicular to x-axis.

$\begin{array}{l} E q u a t i o n of line AB is \\ y - 0 = \frac{3 - 0}{1 + 1} (x + 1) \\ y = \frac{3}{2} (x + 1) \\ E q u a t i o n of line BC is \\ y - 3 = \frac{2 - 3}{3 - 1} (x - 1) \\ y = \frac{- 1}{2} (x - 1) + 3 \\ = \frac{1}{2} (- x + 7) \\ E q u a t i o n of line CA is \\ y - 2 = \frac{2 - 0}{3 + 1} (x - 3) \\ y = \frac{1}{2} (x - 3) + 2 \\ = \frac{1}{2} (x + 1) \\ A r e a Δ A C B = A r e a (A L B A) + A r e a (B L M C B) - A r e a (Δ A M C A) \\ = \int_{- 1}^{1} \frac{3}{2} (x + 1) d x + \int_{1}^{3} \frac{1}{2} (- x + 7) d x - \int_{- 1}^{3} \frac{1}{2} (x + 1) d x \\ = \frac{3}{2} {[\frac{x^{2}}{2} + x]}_{- 1}^{1} + \frac{1}{2} {[\frac{- x^{2}}{2} + 7 x]}_{1}^{3} - \frac{1}{2} {[\frac{x^{2}}{2} + x]}_{- 1}^{3} \\ = \frac{3}{2} [\frac{1^{2}}{2} + 1 - \frac{{(- 1)}^{2}}{2} - (- 1)] + \frac{1}{2} [\frac{- 3^{2}}{2} + 7 (3) + \frac{1^{2}}{2} - 7 (1)] \\ - \frac{1}{2} [\frac{3^{2}}{2} + 3 - \frac{{(- 1)}^{2}}{2} - (- 1)] \\ = \frac{3}{2} (2) + \frac{1}{2} (- \frac{9}{2} + 21 + \frac{1}{2} - 7) - \frac{1}{2} (\frac{9}{2} + 3 - \frac{1}{2} + 1) \\ = 3 + 5 - 4 \end{array}$

=4 square units

Q.17 Using integration find the area of the triangular region whose sides have the equations y = 2x + 1, y = 3x + 1 and x = 4.

Ans.

The coordinates of intersection points of the given lines are A(0, 1), B(4, 13) and C(4, 9).

$\begin{array}{l} A r e a Δ A B C = A r e a O L B A O - A r e a O L C A O \\ = \int_{0}^{4} (3 x + 1) d x - \int_{0}^{4} (2 x + 1) d x \\ = {[3 \frac{x^{2}}{2} + x]}_{0}^{4} - {[2. \frac{x^{2}}{2} + x]}_{0}^{4} \\ = [3 \frac{{(4)}^{2}}{2} + 4 - 3 \frac{{(0)}^{2}}{2} - 0] - [2. \frac{4^{2}}{2} + 4 - 2. \frac{0^{2}}{2} + 0] \\ = (24 + 4) - (16 + 4) \\ = 28 - 20 \\ = 8 square units \end{array}$

Q.18 Choose the correct answer

Smaller area enclosed by the circle x² + y² = 4 and the line x + y =2 is

$\begin{array}{l} (a) 2 (π - 2) \\ (b) π - 2 \\ (c) 2 π - 1 \\ (d) 2 (π + 2) \end{array}$

Ans.

The smaller area enclosed by the circle, x² + y² = 4 and line x + y = 2 is given by shaded region ABCA.

$\begin{array}{l} A r e a A B C A = A r e a O A C B O - A r e a Δ A O B \\ = \int_{0}^{2} \sqrt{(4 - x^{2})} d x - \int_{0}^{2} (2 - x) d x \\ = {[\frac{x}{2} \sqrt{4 - x^{2}} + \frac{4}{2} \sin^{- 1} (\frac{x}{2})]}_{0}^{2} - {[2 x - \frac{x^{2}}{2}]}_{0}^{2} \\ = [\frac{2}{2} \sqrt{4 - 2^{2}} + 2 \sin^{- 1} (\frac{2}{2})] - [2 (2) - \frac{2^{2}}{2}] \\ = (2 \times \frac{π}{2}) - (4 - 2) \\ = (π - 2) square units \\ Therefore, correct option is B . \end{array}$

Q.19 Area lying between the curves y² = 4x and y = 2x is
(A) 2/3 (B) 1/3 (C) 1/4 (D) 3/4

Ans.

The area lying between y²= 4x and y = 2x is shown by shaded area given graph.

$\begin{array}{l} A r e a O B A O = a r e a O C A B O - a r e a O C A O \\ = \int_{0}^{1} 2 \sqrt{x} d x - \int_{0}^{1} 2 x d x \\ = 2 {[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}]}_{0}^{1} - 2 {[\frac{x^{2}}{2}]}_{0}^{1} \\ = \frac{4}{3} [1 - 0] - [1^{2} - 0] \\ = \frac{4}{3} - 1 = \frac{1}{3} s q u a r e units \end{array}$

Therefore, correct option is B.

Q.20 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

Ans.

(i) The required area ADCBA is represented by shaded region in graph.

$\begin{array}{l} A r e a A B C D A = \int_{1}^{2} x^{2} d x \\ = {[\frac{x^{3}}{3}]}_{1}^{2} \\ = \frac{1}{3} [2^{3} - 1^{3}] \\ = \frac{7}{3} s quare units \end{array}$

(ii) The required area ADCBA is represented by shaded region in graph.

$\begin{array}{l} A r e a A B C D A = \int_{1}^{5} x^{4} d x \\ = {[\frac{x^{5}}{5}]}_{1}^{5} \\ = \frac{1}{5} [5^{5} - 1^{5}] \\ = \frac{3124}{5} \\ = 624.8 s quare units \end{array}$

Q.21 Find the area between the curves y = x and y = x².

Ans.

The point of intersection of curve y = x² and line y = x is (1, 1). Here AC is drawn as perpendicular to x-axis.

$\begin{array}{l} ∴ A r e a O B A O = A r e a (Δ O C A) - A r e a (O C A B O) \\ = \int_{0}^{1} x d x - \int_{0}^{1} x^{2} d x \\ = {[\frac{x^{2}}{2}]}_{0}^{1} - {[\frac{x^{3}}{3}]}_{0}^{1} \\ = [\frac{1^{2}}{2} - 0] - [\frac{1^{3}}{3} - 0] \\ = \frac{1}{2} - \frac{1}{3} \\ = \frac{1}{6} square units \end{array}$

Q.22 Find the area of the region lying in the first quadrant and bounded by y = 4x², x = 0, y = 1 and y = 4.

Ans.

The area of the region lying in the first quadrant and bounded by y = 4x², x = 0, y = 1 and y = 4 is shown by shaded region ABCDA.

$\begin{array}{l} ∴ A r e a A B C D = \int_{1}^{4} \frac{\sqrt{y}}{2} d y [∵ y = 4 x^{2}] \\ = \frac{1}{2} \int_{1}^{4} y^{\frac{1}{2}} d y \\ = \frac{1}{2} {[\frac{y^{\frac{3}{2}}}{\frac{3}{2}}]}_{1}^{4} \\ = \frac{1}{3} [4^{\frac{3}{2}} - 1^{\frac{3}{2}}] \\ = \frac{1}{3} (8 - 1) = \frac{7}{3} square units \end{array}$

Q.23

$S k e t c h t h e g r a p h o f y = | x + 3 | a n d e v a l u a t e \int_{- 6}^{0} | x + 3 | d x .$

Ans.

The graph of y = |x + 3| is as shown below:

$\begin{array}{l} ∵ y = | x + 3 | = {\begin{cases} - (x + 3), x + 3 < 0 \Rightarrow x < - 3 \\ (x + 3), x + 3 \geq 0 \Rightarrow x \geq - 3 \end{cases} \\ ∴ \int_{- 6}^{0} | (x + 3) | d x = \int_{- 6}^{- 3} - (x + 3) d x + \int_{- 3}^{0} (x + 3) d x \\ = - {[\frac{x^{2}}{2} + 3 x]}_{- 6}^{- 3} + {[\frac{x^{2}}{2} + 3 x]}_{- 3}^{0} \\ = - [\frac{{(- 3)}^{2}}{2} + 3 (- 3) - \frac{{(- 6)}^{2}}{2} - 3 (- 6)] + [0 - \frac{{(- 3)}^{2}}{2} - 3 (- 3)] \\ = - (\frac{9}{2} - 9 - 18 + 18) + (- \frac{9}{2} + 9) \\ = \frac{9}{2} + \frac{9}{2} = 9 square units \end{array}$

Q.24

$\begin{array}{l} Findtheareaboundedbythecurvey = sinxbetweenx = 0 \\ andx = 2 π . \end{array}$

Ans.

The graph of y = sin x is shown below:

$\begin{array}{l} Required area = Area OABO + Area BCDB \\ = \int_{0}^{π} sinxdx + | \int_{π}^{2 π} sinxdx | \\ = {[- cosx]}_{0}^{π} + | {[- cosx]}_{π}^{2 π} | \\ = - [cosπ + \cos 0] + | \cos 2 π - cosπ | \\ = - [- 1 - 1] + (1 + 1) \\ = 4 square units \end{array}$

Q.25 Find the area enclosed between the parabola y² = 4ax and the line y = mx.

Ans.

Solving y² = 4ax and y = mx, we get x = (4a/m²)
and y = (4a/m).

$\begin{array}{l} T h e points of intersection of both the curves are (0, 0) and \\ (\frac{4 a}{m^{2}}, \frac{4 a}{m}) . \\ ∴ A r e a O B A O = A r e a O C A B O - A r e a (Δ O C A) \\ = \int_{0}^{\frac{4 a}{m^{2}}} 2 \sqrt{a x} d x - \int_{0}^{\frac{4 a}{m^{2}}} m x d x \\ = 2 \sqrt{a} {[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}]}_{0}^{\frac{4 a}{m^{2}}} - m {[\frac{x^{2}}{2}]}_{0}^{\frac{4 a}{m^{2}}} \\ = \frac{4}{3} \sqrt{a} {{(\frac{4 a}{m^{2}})}^{\frac{3}{2}} - 0} - \frac{m}{2} {{(\frac{4 a}{m^{2}})}^{2} - 0} \\ = \frac{4}{3} \sqrt{a} (\frac{8 a \sqrt{a}}{m^{3}}) - \frac{m}{2} (\frac{16 a^{2}}{m^{4}}) \\ = \frac{4}{3} \times \frac{8 a^{2}}{m^{3}} - \frac{8 a^{2}}{m^{3}} \\ = \frac{8 a^{2}}{3 m^{3}} \end{array}$

Q.26 Find the area enclosed between the parabola 4y = 3x² and the line 2y = 3x + 12.

Ans.

The area enclosed between parabola 4y = 3x² and line 2y = 3x + 12 is shown by shaded part AOBA in figure and point of intersection of the given curves are A(–2, 3) and B(4, 12). Two perpendiculars AC and BD are drawn on x-axis.

$\begin{array}{l} ∴ A r e a AOBA = A r e a C D B A - (A r e a A C O A + A r e a O D B O) \\ = \int_{- 2}^{4} \frac{3 x + 12}{2} d x - \int_{- 2}^{4} \frac{3 x^{2}}{4} d x \\ = \frac{1}{2} {[3 \frac{x^{2}}{2} + 12 x]}_{- 2}^{4} - \frac{3}{4} {[\frac{x^{3}}{3}]}_{- 2}^{4} \\ = \frac{1}{2} {\frac{3}{2} \times 4^{2} + 12 \times 4 - \frac{3}{2} \times {(- 2)}^{2} - 12 \times (- 2)} \\ - \frac{1}{4} {4^{3} - {(- 2)}^{3}} \\ = \frac{1}{2} (24 + 48 - 6 + 24) - \frac{1}{4} (64 + 8) \\ = \frac{1}{2} \times 90 - 18 \\ = 45 - 18 \\ = 27 s q u a r e u n i t s \end{array}$

Q.27

$\begin{array}{l} F i n d t h e a r e a o f t h e s m a l l e r r e g i o n b o u n d e d b y t h e e l l i p s e \\ \frac{x^{2}}{9} + \frac{y^{2}}{4} = 1 a n d t h e l i n e \frac{x}{3} + \frac{y}{2} = 1 . \end{array}$

Ans.

The area of the smaller region bounded by the hyperbola and line is represented by shaded region ABCA.

$\begin{array}{l} ∴ A r e a (A B C A) = A r e a (O B C A O) - A r e a (O B A O) \\ = \int_{0}^{3} 2 \sqrt{1 - \frac{x^{2}}{9}} d x - \int_{0}^{3} 2 (1 - \frac{x}{3}) d x \\ = \frac{2}{3} \int_{0}^{3} \sqrt{9 - x^{2}} d x - \frac{2}{3} \int_{0}^{3} (3 - x) d x \\ = \frac{2}{3} {[\frac{x}{2} \sqrt{9 - x^{2}} + \frac{9}{2} \sin^{- 1} (\frac{x}{3})]}_{0}^{3} - \frac{2}{3} {[3 x - \frac{x^{2}}{2}]}_{0}^{3} \\ = \frac{2}{3} {\frac{3}{2} \sqrt{9 - 3^{2}} + \frac{9}{2} \sin^{- 1} (\frac{3}{3}) - \frac{0}{2} \sqrt{9 - 0^{2}} - \frac{9}{2} \sin^{- 1} (\frac{0}{3})} \\ - \frac{2}{3} {3 (3) - \frac{3^{2}}{2} - 3 (0) + \frac{0^{2}}{2}} \\ = \frac{2}{3} {\frac{9}{2} \times \frac{π}{2}} - \frac{2}{3} {9 - \frac{9}{2}} \\ = \frac{3 π}{2} - 3 = \frac{3}{2} (π - 2) s q u a r e units \end{array}$

Q.28

$\begin{array}{l} F i n d t h e a r e a o f t h e s m a l l e r r e g i o n b o u n d e d b y t h e e l l i p s e \\ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 a n d t h e l i n e \frac{x}{a} + \frac{y}{b} = 1 . \end{array}$

Ans.

The area of the smaller region bounded by the hyperbola and line is represented by shaded region ABCA.

$\begin{array}{l} ∴ A r e a (A B C A) = A r e a (O B C A O) - A r e a (O B A O) \\ = \int_{0}^{a} b \sqrt{1 - \frac{x^{2}}{a^{2}}} d x - \int_{0}^{a} b (1 - \frac{x}{a}) d x \\ = \frac{b}{a} \int_{0}^{a} \sqrt{a^{2} - x^{2}} d x - \frac{b}{a} \int_{0}^{a} (a - x) d x \\ = \frac{b}{a} {[\frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} \sin^{- 1} (\frac{x}{a})]}_{0}^{a} - \frac{b}{a} {[a x - \frac{x^{2}}{2}]}_{0}^{a} \\ = \frac{b}{a} {\frac{a}{2} \sqrt{a^{2} - a^{2}} + \frac{a^{2}}{2} \sin^{- 1} (\frac{a}{a}) - \frac{0}{2} \sqrt{a^{2} - 0^{2}} - \frac{a^{2}}{2} \sin^{- 1} (\frac{0}{a})} \\ - \frac{b}{a} {a (a) - \frac{a^{2}}{2} - 3 (0) + \frac{0^{2}}{2}} \\ = \frac{b}{a} {\frac{a^{2}}{2} \times \frac{π}{2}} - \frac{b}{a} {a^{2} - \frac{a^{2}}{2}} \\ = \frac{b}{a} (\frac{π a^{2}}{4} - \frac{a^{2}}{2}) \\ = \frac{b a^{2}}{2 a} (\frac{π}{2} - 1) \\ = \frac{a b}{4} (π - 2) s q u a r e units \end{array}$

Q.29 Find the area of the region enclosed by the parabola x² = y and the line y = x + 2 and x-axis.

Ans.

The area of the region enclosed by the hyperbola and line is represented by shaded region OABCO.
The point of intersection of the parabola, x² = y and the line, y = x + 2, is A(–1, –1).

$\begin{array}{l} ∴ A r e a (O C A O) = \int_{- 1}^{2} (x + 2) d x - \int_{- 1}^{2} x^{2} d x \\ = {[\frac{x^{2}}{2} + 2 x]}_{- 1}^{2} - {[\frac{x^{3}}{3}]}_{- 1}^{2} \\ = [{\frac{{(2)}^{2}}{2} + 2 (2)} - {\frac{{(- 1)}^{2}}{2} + 2 (- 1)}] - {\frac{2^{3}}{3} - \frac{{(- 1)}^{3}}{3}} \\ = (2 + 4 - \frac{1}{2} + 2) - (\frac{8}{3} + \frac{1}{3}) \\ = \frac{15}{2} - 3 \\ = \frac{9}{2} s q u a r e units \end{array}$

Q.30 Using the method of integration find the area bounded by the curve |x| + |y| = 1.

Ans.

The area bounded by the curve |x| + |y| = 1 can be obtained by tracing curves of the lines x + y = 1,
x – y =1, –x – y= 1 and –x + y = 1. The shaded portion ABCD is showing the required area in the graph.

The curves are intersecting the axes at the points
A (0, 1), B (1, 0), C (0, –1) and D (–1, 0).
The given curve is symmetrical about x-axis and y-axis.

$\begin{array}{l} So, Area ABCD = 4 (Area AOB) \\ = 4 \int_{0}^{1} (1 - x) d x \\ = 4 {[x - \frac{x^{2}}{2}]}_{0}^{1} \\ = 4 (1 - \frac{1^{2}}{2} - 0 + \frac{0^{2}}{2}) \\ = 4 (\frac{1}{2}) = 2 s q u a r e units \end{array}$

Q.31

$F i n d t h e a r e a b o u n d e d b y c u r v e s {(x, y) : y \geq x^{2} and y = | x |} .$

Ans.

null

The required area is symmetrical about y-axis.

$\begin{array}{l} S o, required area = 2 {A r e a (O C A O) - A r e a (O C A D O)} \\ = 2 [\int_{0}^{1} x d x - \int_{0}^{1} x^{2} d x] \\ = 2 {[\frac{x^{2}}{2}]}_{0}^{1} - 2 {[\frac{x^{3}}{3}]}_{0}^{1} \\ = 2 [\frac{1^{2}}{2} - 0] - 2 [\frac{1^{3}}{3} - 0] \\ = 2 (\frac{1}{2} - \frac{1}{3}) \\ = 2 (\frac{1}{6}) = \frac{1}{3} square units \end{array}$

Q.32 Using 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).

Ans.

The vertices of triangle ABC are A(2, 0), B(4, 5) and C(6, 3).

$\begin{array}{l} E q u a t i o n of line AB is \\ y - 0 = \frac{5 - 0}{4 - 2} (x - 2) \\ y = \frac{5}{2} (x - 2) \\ E q u a t i o n of line BC is \\ y - 5 = \frac{3 - 5}{6 - 4} (x - 4) \\ y = \frac{- 2}{2} (x - 4) + 5 = - x + 9 \\ E q u a t i o n of line CA is \\ y - 3 = \frac{0 - 3}{2 - 6} (x - 6) \\ y = \frac{3}{4} (x - 6) + 3 \\ = \frac{3}{4} (x - 2) \\ A r e a (Δ A B C) = A r e a (Δ A B L A) + A r e a (Δ B L M C B) - A r e a (Δ A C M A) \\ = \int_{2}^{4} \frac{5}{2} (x - 2) d x + \int_{4}^{6} (- x + 9) d x - \int_{2}^{6} \frac{3}{4} (x - 2) d x \\ = \frac{5}{2} {[\frac{x^{2}}{2} - 2 x]}_{2}^{4} + {[- \frac{x^{2}}{2} + 9 x]}_{4}^{6} - \frac{3}{4} {[\frac{x^{2}}{2} - 2 x]}_{2}^{6} \\ = \frac{5}{2} [\frac{4^{2}}{2} - 2 (4) - \frac{2^{2}}{2} + 2 (2)] + [- \frac{6^{2}}{2} + 9 (6) + \frac{4^{2}}{2} - 9 (4)] \\ - \frac{3}{4} [\frac{6^{2}}{2} - 2 (6) - \frac{2^{2}}{2} + 2 (2)] \\ = \frac{5}{2} (8 - 8 - 2 + 4) + (- 18 + 54 + 8 - 36) - \frac{3}{4} (18 - 12 - 2 + 4) \\ = \frac{5}{2} (2) + 8 - \frac{3}{4} (8) \\ = 5 + 8 - 6 \\ = 7 squre units \end{array}$

Q.33 Using method of integration find the area of the region bounded by lines: 2x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0

Ans.

The equations of given lines are:
2x + y = 4 … (i)
3x – 2y= 6 … (ii)
x – 3y +5 =0 … (iii)
The graphical representation of these lines is given in below. Here AL and CM are perpendicular to x-axis.

$\begin{array}{l} Area (ΔABC) = Area (ALMCA) - Area (ALB) - Area (CMB) \\ = \int_{1}^{4} \frac{x + 5}{3} dx - \int_{1}^{2} (4 - 2 x) dx - \int_{2}^{4} \frac{3 x - 6}{2} dx \\ = \frac{1}{3} {[\frac{x^{2}}{2} + 5 x]}_{1}^{4} - {[4 x - x^{2}]}_{1}^{2} - \frac{1}{2} {[3 \frac{x^{2}}{2} - 6 x]}_{2}^{4} \\ = \frac{1}{3} [\frac{4^{2}}{2} + 5 (4) - \frac{1^{2}}{2} - 5 (1)] - [4 (2) - 2^{2} - 4 (1) + 1^{2}] \\ - \frac{1}{2} [3 \frac{4^{2}}{2} - 6 (4) - 3 \frac{2^{2}}{2} + 6 (2)] \\ = \frac{1}{3} (8 + 20 - \frac{1}{2} - 5) - (8 - 4 - 4 + 1) - \frac{1}{2} (24 - 24 - 6 + 12) \\ = \frac{1}{3} (\frac{45}{2}) - 1 - 3 \\ = \frac{15}{2} - 4 \\ = \frac{7}{2} square units \end{array}$

Q.34

$F i n d t h e a r e a o f t h e r e g i o n {(x, y) : y^{2} \geq 4 x, 4 x^{2} + 4 y^{2} \leq 9} .$

Ans.

$\begin{array}{l} The intersection points of both the curves are (\frac{1}{2}, \sqrt{2}) and (\frac{1}{2}, - \sqrt{2}) . \\ T h e required area is given by OABCO . \\ OABCO is symmetrical about x-axis . \\ ∴ Area OABCO = 2 \times A r e a O A B \\ A r e a O A B O = A r e a O M A O + A r e a A M B A \\ = \int_{0}^{\frac{1}{2}} 2 \sqrt{x} + \int_{\frac{1}{2}}^{\frac{3}{2}} \frac{1}{2} \sqrt{9 - 4 x^{2}} d x \\ = 2 {[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}]}_{0}^{\frac{1}{2}} + \frac{1}{4} {[\frac{2 x}{2} \sqrt{9 - 4 x^{2}} + \frac{9}{2} \sin^{- 1} (\frac{2 x}{3})]}_{\frac{1}{2}}^{\frac{3}{2}} \\ = \frac{4}{3} [{(\frac{1}{2})}^{\frac{3}{2}} - 0] + \frac{1}{4} [\begin{array}{l} \frac{2 (\frac{3}{2})}{2} \sqrt{9 - 4 {(\frac{3}{2})}^{2}} + \frac{9}{2} \sin^{- 1} (\frac{2 \times \frac{3}{2}}{3}) \\ - \frac{2 \times \frac{1}{2}}{2} \sqrt{9 - 4 {(\frac{1}{2})}^{2}} - \frac{9}{2} \sin^{- 1} (\frac{2 \times \frac{1}{2}}{3}) \end{array}] \\ = \frac{4}{3} \times \frac{1}{2 \sqrt{2}} + \frac{1}{4} {0 + \frac{9}{2} \times \frac{π}{2} - \frac{1}{2} \times 2 \sqrt{2} - \frac{9}{2} \times \sin^{- 1} (\frac{1}{3})} \\ = \frac{\sqrt{2}}{3} + \frac{9 π}{16} - \frac{\sqrt{2}}{4} - \frac{9}{8} \sin^{- 1} (\frac{1}{3}) \end{array}$

$\begin{array}{l} = \frac{9 π}{16} - \frac{9}{8} \sin^{- 1} (\frac{1}{3}) + \frac{\sqrt{2}}{12} \\ = \frac{9 π}{16} - \frac{9}{8} \sin^{- 1} (\frac{1}{3}) + \frac{\sqrt{2}}{12} \\ T h u s, the required area of shaded region is \\ 2 \times (\frac{9 π}{16} - \frac{9}{8} \sin^{- 1} (\frac{1}{3}) + \frac{\sqrt{2}}{12}) square units \\ i .e ., (\frac{9 π}{8} - \frac{9}{4} \sin^{- 1} (\frac{1}{3}) + \frac{\sqrt{2}}{6}) square units . \end{array}$

Q.35 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

Ans.

The required area is shaded in the figure.

$\begin{array}{l} Re q u i r e d area = | \int_{- 2}^{0} x^{3} d x | + \int_{0}^{1} x^{3} d x \\ = | {[\frac{x^{4}}{4}]}_{- 2}^{0} | + {[\frac{x^{4}}{4}]}_{0}^{1} \\ = | \frac{1}{4} (0 - 16) | + \frac{1}{4} (1 - 0) \\ = \frac{16}{4} + \frac{1}{4} \\ = \frac{17}{4} \\ T h u s, correct option is D . \end{array}$

Q.36 The area bounded by the curve y = x |x|, x-axis and the ordinates x = –1 and x = 1 is given by

(A) 0 (B) 1/3 (C) 2/3 (D) 4/3

Ans.

$\begin{array}{l} W e have, \\ y = x | x | = {\begin{cases} - x^{2}, x < 0 \\ x^{2}, x \geq 0 \end{cases} \end{array}$

$\begin{array}{l} Intersection points of curve x | x | and lines x = 1 a n d x = - 1 are \\ (1, 1) and (- 1, - 1) respectively . \\ ∴ Required area = \int_{- 1}^{1} x | x | d x \\ = | \int_{- 1}^{0} - x^{2} d x | + \int_{0}^{1} x^{2} d x \\ = | {[- \frac{x^{3}}{3}]}_{- 1}^{0} | + {[\frac{x^{3}}{3}]}_{0}^{1} \\ = | (0 + \frac{1}{3}) | + (\frac{1}{3} - 0) \\ = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} square units \\ Thus, the correct option is C . \end{array}$

Q.37

$\begin{array}{l} The area of the circle x^{2} + y^{2} = 16 exterior to the parabola \\ y^{2} = 6 x is \\ (A) \frac{4}{3} (4 π - \sqrt{3}) (B) \frac{4}{3} (4 π + \sqrt{3}) \\ (C) \frac{4}{3} (8 π - \sqrt{3}) (D) \frac{4}{3} (8 π + \sqrt{3}) \end{array}$

Ans.

$\begin{array}{l} The intersection points of parabola and circle are A (2, 2 \sqrt{3}) and \\ B (2, - 2 \sqrt{3}) . Area of circle which is exterior to parabola is shown \\ by shaded area . \end{array}$

$\begin{array}{l} ∴ U n s h a d e d A r e a bounded by circle and parabola \\ = 2 (\int_{0}^{2} \sqrt{6 x} d x + \int_{2}^{4} \sqrt{16 - x^{2}} d x) \\ = 2 (\sqrt{6} {[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}]}_{0}^{2} + {[\frac{x}{2} \sqrt{16 - x^{2}} + \frac{16}{2} \sin^{- 1} (\frac{x}{4})]}_{2}^{4}) \\ = 2 (\frac{2 \sqrt{6}}{3} [2^{\frac{3}{2}} - 0] + {[\begin{array}{l} 0 + 8 \sin^{- 1} (1) - \sqrt{12} \\ - 8 \sin^{- 1} (\frac{1}{2}) \end{array}]}_{2}^{4}) \\ = 2 (\frac{8 \sqrt{3}}{3} + 8 \times \frac{π}{2} - 2 \sqrt{3} - 8 \times \frac{π}{6}) \\ = 2 (\frac{2 \sqrt{3}}{3} + \frac{8 π}{3}) = \frac{4 \sqrt{3}}{3} + \frac{16 π}{3} \\ A r e a of shaded region = A r e a of circle - (\frac{4 \sqrt{3}}{3} + \frac{16 π}{3}) \\ = π {(4)}^{2} - (\frac{4 \sqrt{3}}{3} + \frac{16 π}{3}) \\ = 16 π - \frac{16 π}{3} - \frac{4 \sqrt{3}}{3} \\ = \frac{32 π}{3} - \frac{4 \sqrt{3}}{3} \\ = \frac{4}{3} (8 π - \sqrt{3}) square units \\ Thus, correct option is C . \end{array}$

Q.38

$\begin{array}{l} The area bounded by the y-axis, y = cosx and y = sinx \\ when 0 \leq x \leq \frac{π}{2} is \\ (A) 2 (\sqrt{2} - 1) (B) \sqrt{2} - 1 (C) \sqrt{2} +1 (D) \sqrt{2} \end{array}$

Ans.

$\begin{array}{l} G i v e n curves are y = sinx and y = cosx \\ Intersection point of both given curves is B (\frac{π}{4}, \frac{1}{\sqrt{2}}) . \\ ∴ A r e a of shaded region = A r e a (A M B A) + A r e a (O M B O) \\ = \int_{0}^{\frac{1}{\sqrt{2}}} x d y + \int_{\frac{1}{\sqrt{2}}}^{1} x d y \\ = \int_{0}^{\frac{1}{\sqrt{2}}} \sin^{- 1} y d y + \int_{\frac{1}{\sqrt{2}}}^{1} \cos^{- 1} y d y \\ = \int_{0}^{\frac{1}{\sqrt{2}}} \sin^{- 1} y .1 d y + \int_{\frac{1}{\sqrt{2}}}^{1} \cos^{- 1} y .1 d y \\ I n t e g r a t i n g by parts, we get \\ = {[y \sin^{- 1} y + \sqrt{1 - y^{2}}]}_{0}^{\frac{1}{\sqrt{2}}} + {[y \cos^{- 1} y - \sqrt{1 - y^{2}}]}_{\frac{1}{\sqrt{2}}}^{1} \\ = [\frac{1}{\sqrt{2}} \sin^{- 1} (\frac{1}{\sqrt{2}}) + \sqrt{1 - {(\frac{1}{\sqrt{2}})}^{2}} - 0 - 1] \\ + [1 \cos^{- 1} (1) - \sqrt{1 - {(1)}^{2}} - \frac{1}{\sqrt{2}} \cos^{- 1} \frac{1}{\sqrt{2}} + \sqrt{1 - {(\frac{1}{\sqrt{2}})}^{2}}] \\ = [\frac{1}{\sqrt{2}} \frac{π}{4} + \frac{1}{\sqrt{2}} - 1] + [0 - 0 - \frac{1}{\sqrt{2}} \frac{π}{4} + \frac{1}{\sqrt{2}}] \\ = \frac{1}{\sqrt{2}} \frac{π}{4} + \frac{1}{\sqrt{2}} - 1 - \frac{1}{\sqrt{2}} \frac{π}{4} + \frac{1}{\sqrt{2}} \\ = \frac{2}{\sqrt{2}} - 1 = \sqrt{2} - 1 \\ T h u s, the correct option is B . \end{array}$

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Chapter 3 - Matrices

Chapter 4 - Determinants

Chapter 5 - Continuity and Differentiability

Chapter 6 - Application of Derivatives

Chapter 7 - Integrals

Chapter 9 - Differential Equations

Chapter 10 - Vector Algebra

Chapter 11 - Three Dimensional Geometry

Chapter 12 - Linear Programming

Chapter 13 - Probability

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Chapter 5 - Continuity and Differentiability

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Chapter 7 - Integrals

Chapter 9 - Differential Equations

Chapter 10 - Vector Algebra

Chapter 11 - Three Dimensional Geometry

Chapter 12 - Linear Programming

Chapter 13 - Probability

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Related Chapters

Chapter 1 - Relations and Functions

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 9 - Differential Equations

Chapter 10 - Vector Algebra

Chapter 11 - Three Dimensional Geometry

Chapter 12 - Linear Programming

Chapter 13 - Probability

FAQs (Frequently Asked Questions)

1. Is it necessary for one to go over all the questions in NCERT Solutions Class 12 Mathematics Chapter 8 Application of Integrals?

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3. What is the significance of NCERT Solutions Class 12 Mathematics Chapter 8?

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