CBSE Class 12 Maths Revision Notes Chapter 8

Mathematics is an essential subject that forms the basis of all other subjects. It encourages logical reasoning and is necessary for business, finance and personal decision making. In this subject, strong basic knowledge is required to understand complex subjects and solve problems based on them.. The majority of students face issues and find it difficult to tackle problems in mathematics. To easily understand the concepts, methods and gain knowledge of the higher-level topics, students must learn from Class 12 Mathematics Chapter 8 notes and get an understanding of all the basic concepts.

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The Class 12 Chapter 8 Mathematics notes- Application of Integrals is a continuation of Chapter 7. Integration is an important concept in higher mathematics. This Chapter deals with the calculation of complex areas under curves, parabolas, ellipses, and intercepts. Students also learn about various integration techniques and formulas. With the help of the Class 12 Mathematics Chapter 8 notes, students can strengthen their foundation and solve problems based on the real world.

Extramarks make Mathematics interesting and fun with coloured illustrations and detailed information. The Class 12 Mathematics Chapter 8 notes help students to practice unlimited questions and attain high scores in the examination. Students can also access different practice tests, mock tests, reference books, etc., from the Extramarks web portal.

NCERT Class 12 Mathematics Chapter 8: Key Notes

The main topics covered in the Class 12 Mathematics Chapter 8 notes are:

Introduction
Definite integrals
Newton Leibnitz’s Theorem
Area under Simple Curves
Area between Two Curves.

Introduction:

The Chapter begins by recalling the concepts of finding areas bounded by the curve and definite integrals. It also introduces the various applications of integrals in this Chapter and topics like areas under simple curves. Between lines, and curves, parabola, and ellipses are also taught. Calculation of the average value of a function with the help of integration is included in the Chapter.

Real-life problems such as a record of rainfall in a day expressed in the form of a curve with specified limit x to limit y are included in the Class 12 Mathematics Chapter 8 notes.

Definite Integrals:

Let F(x) be the antiderivative of function f(x), then the definite integral of f(x) from a to b is given as F(b) – F(a), such that variable x has any two independent values a and b. It is denoted as abf(x) dx.

Therefore, we can say that abf(x) dx= F(b) – F(a). the values a and b are called the limits of integration.

Properties:

abf(x) dx = –baf(x) dx
abf(x) dx = baf(y) dy
abf(x) dx = acf(x) dx + cbf(x) dx, where a< c <b
0af(x) dx = abf(a-x) dx
abf(x) dx =abf(a+b-x) dx
0af(x)f(x)+f(a-x) dx = a2
abf(x)f(x)+f(a+b-x) dx = b-a2
02af(x) dx = 0af(x) dx + 0af(2a-x) dx

If f(x) is a periodic function, i.e. f(a+x) = f(x) then,

0naf(x) dx= n0af(x) dx
0naf(x) dx= (n-1) 0af(x) dx
0b+naf(x) dx= 0bf(x) dx
If f(x) 0 on [a, b], then abf(x) dx 0
If f(x) g(x) on [a, b], then abf(x) dx abg(x) dx
abf(x) dx abf(x) dx
abf(x) dx = f(c) (b – a), for a < c < b

Newton Leibnitz’s Theorem:

Consider two differentiable functions g(x) and h(x) for x [a, b], the function f is continuous in interval [a, b] then

ddxg(x)h(x)f(x) dx =ddxh(x) . f(h(x)) – ddxg(x) . f(g(x))

Definite Integral as a Limit of Sum:

The function f(x) is continuous on the interval [a, b] divided into n parts, then

abf(x) dx= nr=0n=1(b-a)n f(a+ (b-a) rn

Reduction Formulae in Definite Integrals

If In= o2sinnx dx then In=(n-1n) In-2

NOTE: In= o2sinnx dx = o2cosnx dx

If In= o4tannx dx then In+ In-2= 1n-1
If In= o2sinmx. cosnx dx then Im,n=(m-1m+n) Im-2, n

The area under the curves:

The total area A of the region bounded between the x-axis with co-ordinates x = a, x = b and the curve y = f (x) is given as abdA = aby dx =abf(x) dx.

If f(x)>0, ∀x∈[a,c) and f(x)<0 ∀x∈(c,b], then

Area = acf(x) dx+ cbf(x) dx = acf(x) dx – cbf(x) dx ∀ a < c < b

2. The area of the given region bounded between the curve x = g (y), y-axis and the lines y = c,

y = d is given as cddA = cdx dy =cdg(y) dx.

Area

NOTE: If the curve is below the x-axis, then f (x) < 0 from x = a to x = b. In this case, the area will be negative. But since only the numerical value is taken into consideration, we take the absolute value of the area, which is given by abf(x) dx

The area of the region bounded by a curve and a line:

The area of the region is bounded by a line and a circle, parabola, or an ellipse in their standard forms. Vertical stripes or horizontal stripes are used to calculate the area of the region.

For, e.g. Consider the figure given below. We have to find the area bounded by the ellipse in its standard form and the ordinates x = 0 and x = ae, where b2= a2(1- e2) and e < 1. The area of the highlighted region is enclosed by the lines x = 0 and x = ae and the eclipse. Therefore using the formula for area, we will integrate and find the solution.

Example of the area bounded by a curve and a line

The area of the region between two curves:

Area between two curves

Here total area = ab[f(x)-g(x)]dx i.e.,

Area A= [area bounded by the curve y = f (x), x-axis and lines x = a, x = b] – [area bounded by the curve

y = g (x), x-axis and lines x = a, x = b]

A = abf(x)dx –abg(x)dx = ab[f(x)-g(x)]dx where f(x) > g(x) in the interval [a, b]

Area of the region

In this case, the total area A= Area of the ACBDA region+ Area of the BPRQB region.

Area = ac[f(x)-g(x)]dx + cb[g(x)-f(x)]dx, where a < c < b and f (x) ≥ g (x) in interval [a, c] and f (x) ≤ g (x) in the interval [c, b].

Curve Tracing:

To locate the area of a region, it is necessary to draw a rough sketch. Consider the curve f(x,y) = 0. To find the area of the curve, follow the steps given below:

Step 1: Symmetry

The curve is said to be symmetric about the x-axis if all the powers of y in the equation are even.
The curve will be symmetric about the y-axis if all the powers of x in the equation are even.
The curve is symmetric about line y = x if the given equation remains unchanged on interchanging the value of x and y.
The curve will be symmetrical in opposite quadrants if the given equation remains unchanged when x and y values are replaced by -x and -y.

Step 2: Origin

If the constant term is absent in the given equation, then we can say that the curve passes through the origin (0,0).
Then calculate the tangents at the point (0, 0) by equating the terms having the lowest degree in the given equation to zero.

Step 3: Intersection with Co-ordinates Axes

Find values of x by substituting y=0 to estimate the intersecting points of the curve with an x-axis
Find values of y by substituting x=0 to estimate the intersecting points of the curve with the y-axis

Step 4: Asymptotes

Compare the coefficient of the highest power of variable y in the given algebraic equation to zero to find out the vertical asymptotes.
Compare the coefficient of the highest power of variable x in the given algebraic equation to zero to find out the horizontal asymptotes.

Step 5: Region

Solve the given algebraic equation for x in terms of y or vice versa to determine the regions in which the curve doesn’t lie.

Step 6: Critical Points

Differentiate the value of y with respect to x and find out which values of x satisfy ddxy = 0

Step 7: Trace the given curve

Chapter 8 Mathematics Class 12 Notes: Exercises & Answer Solutions

The Class 12 Mathematics Chapter 8 notes ensure detailed and apt information of all the concepts for students to get a clear understanding. Students learn to calculate the area of different regions bounded by curves, lines, parabolas and ellipses. With the help of the Class 12 Mathematics Chapter 8 notes, students can get all important definitions, formulas, properties, and theorems in one place to enable quick revision to clear their doubts and provide them with a solid foundation..

Click on the links given below to gain access to the Extramarks Questions & Answers of this chapter.

Extramarks, an online learning platform, aims to provide a fun and engaging learning experience. The Class 12 Mathematics Chapter 8 notes are prepared by academic experts by analysing various CBSE sample papers and CBSE previous year question papers. Extramarks provides various study material and CBSE revision notes to help students in the exam preparation.

NCERT Exemplar Class 12 Mathematics

The Extramarks platform provides the best study materials such as the NCERT Exemplar and other NCERT books to help students in their Class 12 board exams as well as other competitive examinations. The NCERT Exemplar includes an unlimited set of CBSE extra questions and miscellaneous problems for students to gain an in-depth knowledge of all concepts included in the Class 12 Mathematics Chapter 8 notes.

The NCERT Exemplar is based on the latest CBSE syllabus. Students must regularly practice all the important questions to attain high scores in the examination. It also teaches several shortcut techniques to tackle complex problems easily in no time. With the help of academic notes such as NCERT Exemplar and Extramarks Class 12 Mathematics Chapter 8 notes, students can study hassle-free according to CBSE pattern.

NCERT Class 12 Mathematics Chapter 8 Notes: Key Features

The key features of Extramarks Class 12 Mathematics Chapter 8 notes are as under.

The notes provide all concepts included in the CBSE syllabus in a detailed and lucid manner.
It is prepared by subject matter experts at Extramarks
Students get an idea of the marking system, weightage and exam pattern of the CBSE examinations.
The Class 12 mathematics notes Chapter 8 provides authentic knowledge and helps to clarify all doubts or queries way ahead of the exam to put the students at ease.
Inadvertently students develop time management and problem-solving skills.

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

Ans

$\begin{array}{l} y^{2} = x trepresents a parabola with vertex at (0, 0) . \\ x = 1 and x = 4 represent the straight lines parallel to y - axis . \\ The required area = \int_{1}^{4} \sqrt{x} dx \\ = {[\frac{2 x^{\frac{3}{2}}}{3}]}_{1}^{4} \\ = \frac{2}{3} [4 \sqrt{4} - 1 \sqrt{1}] \\ = \frac{14}{3} sq . units \end{array}$

Q.2 Find the area of the region bounded by the line y = 3x + 2, the x-axis and the ordinates x = – 1 and x = 1.

Ans

$\begin{array}{l} The given line meets x - axis at x = - \frac{2}{3} and its graph lies below x - axis for x \in (- 1, \frac{2}{3}) \\ and above x - axis for x \in (- \frac{2}{3}, 1) . \\ The required area = area of the region ACBA + area of the region ADEA \\ = |\int_{- 1}^{- \frac{2}{3}} (3 x + 2) dx| + \int_{- \frac{2}{3}}^{1} (3 x + 2) dx \\ = |{[\frac{3 x^{2}}{2} + 2 x]}_{01}^{- \frac{2}{3}}| + {[\frac{3 x^{2}}{2} + 2 x]}_{- \frac{2}{3}}^{1} \\ = \frac{1}{6} + \frac{25}{6} \\ = \frac{13}{3} sq . units \end{array}$

Q.3 Using integration, find the area bounded by the curves x² + y² = 1 and y² = (x +1).

Ans

$\begin{array}{l} x^{2} + y^{2} = 1 represents circle with centre (0, 0) and radius = 1. \\ y^{2} = (x + 1) represents parabola with vertex at (- 1, 0) . \\ From x^{2} + y^{2} + 1 = 1 and y^{2} = (x + 1), we get \\ x^{2} + x + 1 = 1 \\ or x^{2} + x = 0 \\ or x^{2} (x + 1) = 0 \\ o x = 0, - 1 \\ The required area = 2 x [\int_{- 1}^{0} \sqrt{1 - x^{2}} dx - \int_{- 1}^{0} \sqrt{x - 1} dx] \\ The required area = 2 x {[\frac{x}{2} \sqrt{1 - x^{2}} + \frac{1}{2} \sin^{- 1} x - \frac{2}{3} {(x + 1)}^{\frac{3}{2}}]}_{- 1}^{0} \\ The required area = 2 x [- \frac{2}{3} - (- \frac{π}{4})] \\ The required area = \frac{π}{2} - \frac{4}{3} sq . units \end{array}$

Q.4 Find the area bounded by curves x² + y² = 1 and (x – 1)² + y² = 1.

Ans

$\begin{array}{l} Equations of the given circles are as follows : \\ x^{2} + y^{2} = 1 . \dots. (1) \\ and {(x - 1)}^{2} + y^{2} = 1 . \dots. (2) \\ Equation (1) is a circle with centra = (0, 0) and radius = 1. \\ Equation (2) is a circle with centra = (1, 0) and radius = 1. \\ x^{2} + 1 - {(x - 1)}^{2} = 1 \\ or x^{2} - x^{2} - 1 + 2 x = 0 \\ or x = \frac{1}{2} \\ Required area = 2 x [\int_{0}^{\frac{1}{2}} \sqrt{1 - {(x - 1)}^{2}} dx + \int_{0}^{\frac{1}{2}} \sqrt{1 - (1 - x^{2})} dx] \\ = 2 x \{{[\frac{x - 1}{2} \sqrt{1 - {(x - 1)}^{2} + \frac{1}{2}} \sin^{- 1} (x - 1)]}_{0}^{\frac{1}{2}} + {[\frac{x}{2} \sqrt{1 - x^{2} + \frac{1}{2}} \sin^{- 1}]}_{\frac{1}{2}}^{1}\} \\ = 2 x \{{[(- \frac{\sqrt{3}}{8} - \frac{π}{12}) + \frac{π}{4}]}_{}^{} + {[\frac{π}{4} - (\frac{\sqrt{3}}{8} + \frac{π}{12})]}_{}^{}\} \\ = 2 x \{- \frac{\sqrt{3}}{8} - \frac{π}{12} + \frac{π}{4} + \frac{π}{4} + \frac{\sqrt{3}}{8} - \frac{π}{12}\} \\ = 2 x \{- \frac{\sqrt{3}}{4} - \frac{π}{6} + \frac{π}{2}\} \\ = 2 x \{- \frac{\sqrt{3}}{4} + \frac{2 π}{6}\} \\ = \{- \frac{2 π}{3} - \frac{\sqrt{3}}{2}\} sq . units . \end{array}$

Q.5

$Using integration, find the area of ΔABC where A is (2, 3), B is (4, 7) and C is (6, 2) .$

Ans

$\begin{array}{l} The equation of side AB is \\ y -3 = \frac{7 - 3}{4 - 2} (x - 2) \\ or y =2 x -1 . \dots.. (i) \\ The equation of side BC is \\ y -7 = \frac{2 - 7}{6 - 4} (x - 4) \\ or y = - \frac{5}{2} x + 17 . \dots (ii) \\ The equation of side AC is \\ y -3 = \frac{2 - 3}{6 - 2} (x - 2) \\ or y = \frac{- 1}{4} x + \frac{7}{2} . \dots. (iii) \\ Area of the triangle = \int_{2}^{4} y_{AB} dx + \int_{4}^{6} y_{BC} dx - \int_{2}^{6} y_{AC} dx \\ = \int_{2}^{4} (2 x - 1) dx + \int_{4}^{6} (- \frac{5}{2} x + 17) dx - \int_{2}^{6} (\frac{- 1}{4} x + \frac{7}{2}) dx \\ = {[x^{2} - x]}_{2}^{4} + {[- \frac{5}{2} x^{2} + 17 x]}_{4}^{6} - {[- \frac{x^{2}}{8} + \frac{7}{2} x]}_{4}^{6} \\ = [12 - 2] + [(- 45 + 102) - (20 + 68)] - [(- \frac{9}{2} + 21) - (- \frac{1}{2} + 7)] \\ = [10] + [57 - 48] - [- 4 + 14] \\ = 10 + 9 - 10 \\ = 9 sq . units \end{array}$

Q.6

$Find the area of the region \{x, y : 0 \leq y \leq x^{2} + 1, 0 \leq y \leq y \leq x + 1, 0 \leq x \leq 2\} .$

Ans

$\begin{array}{l} y = x^{2} + 1 represents a parabola with vertex . \\ y = x +1 represents a straight line . \\ From y = x^{2} + 1 and y = x + 1, we get \\ x + 1 = x^{2} + 1 \\ or x - x^{2} = 0 \\ or x (1 - x) = 0 \\ or x = 0, 1 \\ The required area = \int_{0}^{1} (x^{2} + 1) dx + \int_{0}^{1} (x + 1) dx \\ = {[\frac{x^{3}}{3} + x]}_{0}^{1} + {[\frac{{(x + 1)}^{2}}{2}]}_{1}^{2} \\ = [\frac{1}{3} + 1] + [\frac{9}{2} - \frac{4}{2}] \\ = \frac{4}{5} + \frac{5}{2} \\ = \frac{8 + 15}{6} \\ = \frac{23}{6} sq . units \\ Find the area of the region {x, y : 0 y x^{2} + 1, 0 y y x + 1, 0 x 2} . \\ Find the area of the region \{x, y : 0 \leq y \leq x^{2} + 1, 0 \leq y \leq y \leq x + 1, 0 \leq x \leq 2\} . \end{array}$

Q.7 Find the area of the region bounded by y² = x and x² = y.

Ans

$\begin{array}{l} y^{2} = x and x^{2} = y represents parabola with vertex at (0, 0) . \\ From y^{2} = x and x^{2} = y, we get \\ x^{4} = x \\ or x^{4} - x = 0 \\ or x (x^{3} - 1) = 0 \\ or x = 0, 1 \\ The required area = \int_{0}^{1} x dx - \int_{0}^{1} x^{2} dx \\ = {[\frac{2 x^{2}}{2} - \frac{x^{3}}{3}]}_{0}^{1} \\ = \frac{2}{3} - \frac{1}{3} \\ = \frac{1}{3} sq . units \end{array}$

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

Ans

$\begin{array}{l} From y = |x| and x^{2} = y, we get \\ x = 1, -1 \\ The required area = 2 x [\int_{0}^{1} y_{L} dx - \int_{0}^{1} y_{p} dx] \\ = 2 x [\int_{0}^{1} xdx - \int_{0}^{1} x^{2} dx] \\ = 2 x \frac{x^{2}}{2} - {\frac{x^{3}}{3}}_{0}^{1} \\ = 2 x [\frac{1}{2} - \frac{1}{3}] \\ = 2 x [\frac{1}{6}] \\ = \frac{1}{3} sq . units \end{array}$

Q.9 Find the area of the circle x² + y² = 16 exterior to the parabola y² = 6x.

Ans

$\begin{array}{l} x^{2} + y^{2} = 16 represents a circle with centre at (0, 0) and radius = 4 units . \\ y^{2} = 6 x represents a parabola with vertex = (0, 0) . \\ From the above equations x^{2} + y^{2} = 16 and y^{2} = 6 x, we get \\ x^{2} + 6 x - 16 = 0 \\ or x^{2} + 8 x - 2 x - 16 = 0 \\ or (x + 8) (x + 2) = 0 \\ \Rightarrow x = 2, or - 8 (neglected) \\ Thus the required area = 16 π - 2 \times [\int_{0}^{2} \sqrt{6 xdx} + \int_{2}^{4} \sqrt{16 - x^{2}} dx] \\ = 16 π - 2 \times \{\sqrt{6} {\frac{2 x^{\frac{3}{2}}}{3}}_{0}^{2} + {[\frac{x}{2} \sqrt{16 - x^{2} + 8 \sin^{- 1} \frac{x}{4}}]}_{2}^{4}\} \\ = 16 π - 2 \times \{{\frac{4 \sqrt{12}}{3}}_{0}^{2} + {[4 π - (\sqrt{12} + \frac{8 π}{6})]}_{2}^{4}\} \\ = 16 π - 2 \times \{\frac{8 \sqrt{3}}{3} - 2 \sqrt{3} + 4 π + \frac{4 π}{3}\} \\ = 16 π - 2 \times \{\frac{2 \sqrt{3}}{3} + \frac{8 π}{3}\} \\ = 16 π - \frac{4 \sqrt{3}}{3} - \frac{16 π}{3} \\ = \frac{32 π}{3} - \frac{4 \sqrt{3}}{3} \\ = \frac{4}{3} (8 π - \sqrt{3}) sq . units \end{array}$

Q.10 Find the area of the region bounded by ellipse

$\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$

.

Ans

$\begin{array}{l} \frac{x^{2}}{9} + \frac{y^{2}}{4} = 1 represents ellipse with centre at (0, 0) and a = 3, b = 2. \\ The required area = 4 \times \frac{2}{3} \int_{0}^{3} \sqrt{3 \frac{2}{} - x^{2}} dx \\ = 4 \times \frac{2}{3} {[\frac{x}{2} \sqrt{3^{2} - x^{2} + \frac{9}{2} \sin^{- 1} \frac{x}{3}}]}_{0}^{3} \\ = 4 \times \frac{2}{3} [\frac{9 π}{4}] \\ = 6 sq . units \end{array}$

Q.11 Using the method of integration find the area bounded by the curve |x| + |y| = 1.

Ans

$\begin{array}{l} |x| + |y| = 1 or \\ x + y = 1 if x, y \geq 0 \\ x - y = 1 if x \geq 0, y < 0 \\ - x + y = 1 if x <0, y \geq 0 \\ - x - y = 1 if x, y < 0 \\ Area of shaded region = 4 x [\int_{0}^{1} (1 - x) dx] \\ = 4 x [\int_{0}^{1} (1 - x) dx] \\ = 4 x {\frac{{(1 - x)}^{2}}{- 2}}_{0}^{1} \\ = 4 x [0 + \frac{1}{2}] = \frac{4}{2} \\ = 2 sq . units . \end{array}$

Q.12 Find the area bounded by the curve x² = 4y and the line x = 4y – 2.

Ans

$\begin{array}{l} x = 4 y -2 represents a straight line . \\ x^{2} = 4 y represents a parabola with vertex at (0, 0) . \\ From x = 4 y -2 and x^{2} = 4, we get \\ x = x^{2} - 2 \\ or x^{2} - x - 2 = 0 \\ or (x + 1) (x - 2) = 0 \\ or x = - 1, 2 \\ The required area = \int_{- 1}^{2} \frac{x + 2}{4} dx - \int_{- 1}^{2} \frac{x^{2}}{4} dx \\ = \frac{1}{4} {[\frac{{(x + 2)}^{2}}{2} - \frac{x^{3}}{3}]}_{1}^{2} \\ = \frac{1}{4} [\frac{15}{2} - \frac{9}{3}] = \frac{1}{4} [\frac{45 - 18}{6}] = \frac{1}{4} [\frac{27}{6}] \\ = \frac{9}{8} sq . units \end{array}$

Q.13 Using integration, find the area of the region bounded by the line 3y = 2x +4, x-axis and the line x = 1 and x = 3.

Ans

$\begin{array}{l} 3 y = 2 x +4 represents a straight line intersecting x - axis and y - axis . \\ Also, x = 1 and x = 3 are straight lines parallel to y - axis . \\ So, the required area = \int_{1}^{3} \frac{2 x + 4}{3} \\ = \frac{2}{3} \int_{1}^{3} (x + 2) dx \\ = \frac{2}{3} \int_{1}^{3} \frac{x^{2}}{2} + 2 x_{1}^{3} \\ = \frac{2}{3} [\frac{9}{2} + 6 - \frac{1}{2} - 2] \\ = \frac{16}{3} sq . units \end{array}$

Q.14

$Find area of the region given by {x, y : x^{2} \leq y \leq x} using integration .$

Ans

$\begin{array}{l} x^{2} = y represents a parabola with vertex at (0, 0) . \\ y = x is a line passing through the origin and making an angle of 45 ° with the x - axis . \\ Solving x^{2} = y and x = y, we get x = 1. \\ So, the required area = [\int_{0}^{1} xdx - \int_{0}^{1} x^{2} dx] \\ = {[\frac{x^{2}}{2} - \frac{x^{3}}{3}]}_{0}^{1} \\ = [\frac{1}{2} - \frac{1}{3}] \\ = \frac{1}{6} sq . units \end{array}$

Q.15 Find the area of the region bounded by exponential function from x = 0 and x = 1.

Ans

$\begin{array}{l} The area of the bounded region = \int_{0}^{1} ydx \\ = \int_{0}^{1} e^{x} dx \\ = {[e^{x}]}_{0}^{1} \\ = [e^{1} - e^{0}] \\ = (e - 1) square unit \end{array}$

Q.16 Find the area of the region bounded by Line
x – y=0 , X-axis and ordinates x=1.

Ans

$\begin{array}{l} The area of the bounded region = \int_{0}^{1} ydx \\ = \int_{0}^{1} x dx \\ = \frac{1}{2} {[x^{2}]}_{0}^{1} \\ = \frac{1}{2} [{(1)}^{2} - 0] \\ = \frac{1}{2} square unit \end{array}$

Q.17

$Find the area of the regionbounded by sine function from x = 0 and x = \frac{π}{2} .$

Ans

$\begin{array}{l} The area of the bounded region = \int_{0}^{\frac{π}{2}} ydx \\ = \int_{0}^{\frac{π}{2}} sinx dx \\ = - {[cosx]}_{0}^{\frac{π}{2}} \\ = - [\cos \frac{π}{2} - coso] \\ = - (0 - 1) = 1 square unit \end{array}$

Q.18

$Find the area of the region bounded by cosine function from x = 0 and x = \frac{π}{2} .$

Ans

$\begin{array}{l} The area of the bounded region = \int_{0}^{\frac{π}{2}} ydx \\ = \int_{0}^{\frac{π}{2}} cosx dx \\ = {[sinx]}_{0}^{\frac{π}{2}} \\ = [\sin \frac{π}{2} - sino] \\ = (1 - 0) = 1 square unit \end{array}$

Q.19

$Find the area of the region bounded by cosine function from x = 0 and x = π \frac{π}{2} .$

Ans

Q.20 Find the area of the region bounded by curve xy = 1 from x = 1 and x = e.

Ans

$\begin{array}{l} The area of the bounded region = \int_{1}^{e} ydx \\ = \int_{1}^{e} \frac{dx}{x} [∵ xy = 1 \Rightarrow y = \frac{1}{x}] \\ = {[logx]}_{1}^{e} \\ = [loge - \log 1] \\ = (1 - 0) = 1 square unit \end{array}$

Q.21 Find the area of the region bounded by curve
y = x², X-axis and ordinates x = 1.

Ans

$\begin{array}{l} The area of the bounded region = \int_{0}^{1} ydx \\ = \int_{0}^{1} x^{2} dx \\ = \frac{1}{2} {[x^{3}]}_{0}^{1} \\ = \frac{1}{2} [{(1)}^{3} - 0] \\ = \frac{1}{2} square unit \end{array}$

Q.22 Find the area of the region bounded by line
y = 2x+1, X-axis, Y-axis and ordinates x = 1.

Ans

$\begin{array}{l} The area of the bounded region = \int_{0}^{1} ydx \\ = \int_{0}^{1} (2 x + 1) dx \\ = {[2 \frac{x^{2}}{2} + x]}_{0}^{1} \\ = [(1 + 1) - 0] \\ = 2 square units \end{array}$

Q.23 Find the area of the region bounded by curve y=f(x), ordinates x=a and x=b, as shown below.

Ans

$\begin{matrix} The area of the bounded region = \int_{a}^{b} ydx \\ = \int_{a}^{b} f (x) dx \end{matrix}$

Q.24 Find the area of the region bounded by curve y = f(x), y = g(x) and ordinates x = a and x = b, as shown below.

Ans

$\begin{array}{l} The area of the bounded region = \int_{a}^{b} [upper - lower curve] dx \\ = \int_{a}^{b} [f (x) - g (x)] dx \end{array}$

Q.25 Find the area of the region bounded by parabola y² = x and ordinates x=1.

Ans

$\begin{array}{l} The area of the bounded region = 2 \int_{0}^{1} ydx \\ = 2 \int_{0}^{1} \sqrt{x} dx [y^{2} = x \Rightarrow y = \sqrt{x}] \\ = \frac{4}{3} {[{(x)}^{\frac{3}{2}}]}_{0}^{1} \\ = \frac{4}{3} [{(1)}^{\frac{3}{2}} - {(0)}^{\frac{3}{2}}] \\ = \frac{4}{3} [1 - 0] = \frac{4}{3} square units \end{array}$

Q.26 Find the area of the region bounded by parabola y² = x and ordinates x = 1 and x = 4.

Ans

$We have parabola y^{2} = x with vertex at origin .$

$\begin{array}{l} The area of the bounded region = 2 \int_{1}^{4} ydx \\ = 2 \int_{1}^{4} \sqrt{x} dx [y^{2} = x \Rightarrow y = \sqrt{x}] \\ = \frac{4}{3} {[{(x)}^{\frac{3}{2}}]}_{1}^{4} \\ = \frac{4}{3} [{(4)}^{\frac{3}{2}} - {(1)}^{\frac{3}{2}}] \\ = \frac{4}{3} [8 - 1] = \frac{28}{3} square units \end{array}$

Q.27 Find the area of the region bounded by curve y²= x and straight line y = x.

Ans

$\begin{array}{l} We know that curve y^{2} = x represents a parabola with vertex at \\ origin, symmetric about X - axis and open towards right hand side . \\ The y = x is a straight line passing through origin inclined \\ at 45° [∵ m = tanθ = 1]. \end{array}$

$\begin{array}{l} On solving these two equations, we get \\ x = x^{2} \\ \Rightarrow x (x - 1) = 0 \\ \Rightarrow x = 0 or x =1 \\ The area of the bounded region = \int_{0}^{1} [Y_{upper} - Y_{lower curve}] dx \\ = \int_{0}^{1} [\sqrt{x} - x] dx \\ = {[\frac{2}{3} {(x)}^{\frac{3}{2}} - \frac{x^{2}}{2}]}_{0}^{1} \\ = [(\frac{2}{3} - \frac{1}{2}) - 0] \\ = \frac{1}{6} square unit \end{array}$

Q.28 Find the area of the region bounded by curve y=x² and x=y².

Ans

$\begin{array}{l} We know that curve y^{2} = x represents a parabola with vertex at \\ origin, symmetric about X - axis and open towards right hand side . \\ The curve x^{2} = y represents equation of parabola with vertex \\ at origin, symmetric about Y - axis and open towards upward . \end{array}$

$\begin{array}{l} On solving these two equations, we get \\ x = x^{4} \\ \Rightarrow x (x^{3} - 1) = 0 \\ \Rightarrow x = 0 or x =1 \\ The area of the bounded region = \int_{0}^{1} [Y_{upper} - Y_{lower curve}] dx \\ = \int_{0}^{1} [\sqrt{x} - x^{2}] dx \\ = {[\frac{2}{3} {(x)}^{\frac{3}{2}} - \frac{x^{3}}{3}]}_{0}^{1} \\ = [(\frac{2}{3} - \frac{1}{3}) - 0] \\ = \frac{1}{3} square units \end{array}$

Q.29

$Find the area of the region bounded by curve x = \sqrt{y} and straight line y = x+2; \forall x \geq 0.$

Ans

$\begin{array}{l} We know that curve x = \sqrt{y} (x^{2} = y) represents parabola with \\ vertex at origin, symmetric about Y - axis and open towards upwards . \\ ∵ x \geq 0 for \forall y \in R \\ ∴ There is no portion of curve that lies in the LHS of Y - axis . \\ We have, y = x +2 \\ at x =0, y =2 \\ at y =0, x =-2 \\ The y = x +2 is a straight line passing through (0, 2) and (- 2, 0) \\ inclined at 45°. [∵ m = \tan 45 °=1] \end{array}$

$\begin{array}{l} On solving these two equations, we get \\ x^{2} = x + 2 \\ \Rightarrow x^{2} - x - 2 = 0 \\ \Rightarrow (x + 1) (x - 2) = 0 \\ \Rightarrow x = - 1 or x =2 \\ ∵ Shaded portion lies in x \geq 0 and x \leq 2 . \\ The area of the bounded region = \int_{0}^{2} [Y_{upper} - Y_{lower curve}] dx \\ = \int_{0}^{1} [(x + 2) - x^{2}] dx \\ = {[\frac{x^{2}}{2} + 2 x - \frac{x^{3}}{3}]}_{0}^{1} \\ = [(\frac{1}{2} + 2 - \frac{1}{3}) - 0] \\ = 2 + \frac{1}{6} = \frac{13}{6} sq . units \end{array}$

Q.30 Using the method of integration, find the area of the region bounded by inequation 2x+y8, x+2y8, x 0 and y0.

Ans

$\begin{array}{l} x+2y \leq 8 . .. (1) \\ Let x+2y=8 \\ Put x=0, we get y=4, that is point (0, 4) . \\ Put y=0, we get x=8, that is point (8, 0) . \\ We shall draw a line between these two points. \\ Now, put origin (x = 0, y = 0) in (1), we get \\ 0+0 \leq 8 \\ \Rightarrow 0 \leq 8 \\ which is true, so we will shade region containing origin. \end{array}$ $\begin{array}{l} 2 x + y \leq 8 . .. (2) \\ Let 2 x + y =8 \\ Put x =0, we get y =8, that is point (0, 8) . \\ Put y =0, we get x =4, that is point (4, 0) . \\ We shall draw a line through these two points . \\ Now, put origin (x = 0, y = 0) in (2), we get \\ 0 +0 \leq 8 \\ \Rightarrow 0 \leq 8 \\ which is true, so we will shade region containing origin . \\ Since x \geq 0 and y \geq 0, represent the region of First Quadrant . \end{array}$

$\begin{array}{l} On solving the equation, \\ 2 x + y =8 and x +2 y =8, we get x = \frac{8}{3} \\ Thus, we can divide the shaded region in two parts at C . \\ The bounded area = Area under line AB + Area under line BC \\ = \int_{0}^{\frac{8}{3}} Y_{AB} dx + \int_{\frac{8}{3}}^{4} Y_{BC} dx \\ = \int_{0}^{\frac{8}{3}} (\frac{8 - x}{2}) dx + \int_{\frac{8}{3}}^{4} (8 - 2 x) dx \\ = {[4 x - \frac{1}{2} . \frac{x^{2}}{2}]}^{\frac{8}{3}}_{0} + {[8 x - 2 \frac{x^{2}}{2}]}^{4}_{\frac{8}{3}} \\ = [(\frac{32}{3} - \frac{16}{9}) - 0] + [(32 - 16) - (\frac{64}{3} - \frac{64}{9})] \\ = \frac{80}{9} + 16 - \frac{128}{9} \\ = 16 - \frac{48}{9} \\ = 16 - \frac{16}{3} \\ = \frac{32}{3} sq . units \end{array}$

Q.31 Find the area of the region bounded by the curve y² = 4x and the line y = 4x – 2.

Ans

The given curve is: y² = 4x …(i)

And the given line is: y = 4x – 2 …(ii)

On solving both the equations:

(4x – 2)² = 4x

16x² – 16x + 4 = 4x

16x² – 20x + 4 = 0

Or 4x² – 5x + 1 = 0

4x² – 4x – x + 1 = 0

(4x – 1)(x – 1) = 0

Or x = 1/4, 1

And y = 4(1/4) – 2 or 4(1) – 2

Or y = –1 or 2

Therefore, the points of intersection are (1/4, –1) and (1, 2).

Coordinate of C:

y = 0 or 4x – 2 = 0 or x = ½

So, coordinates of C are (1/2, 0).

And coordinates of E are (1/4, 0).

The area bounded by the parabola y² = 4x and line y = 4x – 2 is given below:

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Extramarks, an online learning platform, provides chapter-wise solutions such as the Class 12 Mathematics Chapter 8 notes to help students in their studies. All the crucial topics included in the syllabus are well-explained in a detailed and stepwise manner. Students of Class 12 are advised to study with the help of these academic notes to get a clear idea of the exam pattern, marking system, weightage as well as important concepts from each chapter.

Chapter 1: Relations and Functions

Chapter 2: Inverse Trigonometric Functions

Chapter 3: Matrices

Chapter 4: Determinants

Chapter 5: Continuity and Differentiability

Chapter 6: Applications 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

CBSE Class 12 Maths Revision Notes Chapter 8

NCERT Class 12 Mathematics Chapter 8 Notes

NCERT Class 12 Mathematics Chapter 8: Key Notes

Introduction:

Definite Integrals:

Newton Leibnitz’s Theorem:

Definite Integral as a Limit of Sum:

Reduction Formulae in Definite Integrals

The area under the curves:

The area of the region bounded by a curve and a line:

The area of the region between two curves:

Curve Tracing:

Step 1: Symmetry

Step 2: Origin

Step 3: Intersection with Co-ordinates Axes

Step 4: Asymptotes

Step 5: Region

Step 6: Critical Points

Step 7: Trace the given curve

Chapter 8 Mathematics Class 12 Notes: Exercises & Answer Solutions

NCERT Exemplar Class 12 Mathematics

NCERT Class 12 Mathematics Chapter 8 Notes: Key Features

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1. Why should I refer to Chapter 8 Mathematics notes by Extramarks?

2. Which chapters are included in the Extramarks NCERT Solutions of Class 12 Mathematics?

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CBSE Class 12 Maths Revision Notes Chapter 8

NCERT Class 12 Mathematics Chapter 8 Notes

NCERT Class 12 Mathematics Chapter 8: Key Notes

Introduction:

Definite Integrals:

Newton Leibnitz’s Theorem:

Definite Integral as a Limit of Sum:

Reduction Formulae in Definite Integrals

The area under the curves:

The area of the region bounded by a curve and a line:

The area of the region between two curves:

Curve Tracing:

Step 1: Symmetry

Step 2: Origin

Step 3: Intersection with Co-ordinates Axes

Step 4: Asymptotes

Step 5: Region

Step 6: Critical Points

Step 7: Trace the given curve

Chapter 8 Mathematics Class 12 Notes: Exercises & Answer Solutions

NCERT Exemplar Class 12 Mathematics

NCERT Class 12 Mathematics Chapter 8 Notes: Key Features

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