1. The document discusses calculating the area under curves using integral calculus. It provides examples of finding the area bounded by curves like circles, ellipses, parabolas, and lines using integrals.
2. The key approach discussed is treating the area under a curve as the limit of thin vertical or horizontal strips, and expressing the total area as the integral of the curve's equation between the bounds. Both vertical and horizontal strip approaches are demonstrated.
3. Several examples are provided of calculating areas bounded by standard curves like circles and ellipses using integrals, as well as areas between curves and lines. The document concludes with exercises for readers to practice similar area calculations.
The document discusses various applications of the definite integral, including finding the area under a curve, the area between two curves, and the volume of solids of revolution. It provides examples of calculating each of these, such as finding the area between the curves y=x and y=x5 from x=-1 to x=0. It also explains how to set up definite integrals to calculate volumes when rotating an area about the x- or y-axis. In conclusion, it states that integrals can represent areas or generalized areas and are fundamental objects in calculus, along with derivatives.
The document discusses different types of bounded regions and calculating their areas using integrals. It defines three types of regions: 1) bounded by two curves and vertical lines, 2) bounded by two curves, and 3) bounded by modulus functions where one curve is greater than the other over some intervals. Examples are provided for each type, such as finding the area between a parabola and line, two parabolas, a parabola and circle, and two circles. The key idea is that the region's area can be expressed as a definite integral of the differences between the bounding curves.
This document discusses integrals and their applications. It introduces integral calculus and its use in joining small pieces together to find amounts. It lists several types of integrals and mathematicians influential in integral calculus development like Euclid, Archimedes, Newton, and Riemann. The document also discusses applications of integration in business processes, automation tools for integrating disparate applications, and applications of very large scale integration circuit design.
Gaussian elimination is a method for solving systems of linear equations. It involves converting the augmented matrix into an upper triangular matrix using elementary row operations. There are three types of Gaussian elimination: simple elimination without pivoting, partial pivoting, and total pivoting. Partial pivoting interchanges rows to choose larger pivots, while total pivoting searches the whole matrix for the largest number to use as the pivot. Pivoting strategies help prevent zero pivots and reduce round-off errors.
The document discusses various applications of the definite integral, including finding the area under a curve, the area between two curves, and the volume of solids of revolution. It provides examples of calculating each of these, such as finding the area between the curves y=x and y=x5 from x=-1 to x=0. It also explains how to set up definite integrals to calculate volumes when rotating an area about the x- or y-axis. In conclusion, it states that integrals can represent areas or generalized areas and are fundamental objects in calculus, along with derivatives.
The document discusses different types of bounded regions and calculating their areas using integrals. It defines three types of regions: 1) bounded by two curves and vertical lines, 2) bounded by two curves, and 3) bounded by modulus functions where one curve is greater than the other over some intervals. Examples are provided for each type, such as finding the area between a parabola and line, two parabolas, a parabola and circle, and two circles. The key idea is that the region's area can be expressed as a definite integral of the differences between the bounding curves.
This document discusses integrals and their applications. It introduces integral calculus and its use in joining small pieces together to find amounts. It lists several types of integrals and mathematicians influential in integral calculus development like Euclid, Archimedes, Newton, and Riemann. The document also discusses applications of integration in business processes, automation tools for integrating disparate applications, and applications of very large scale integration circuit design.
Gaussian elimination is a method for solving systems of linear equations. It involves converting the augmented matrix into an upper triangular matrix using elementary row operations. There are three types of Gaussian elimination: simple elimination without pivoting, partial pivoting, and total pivoting. Partial pivoting interchanges rows to choose larger pivots, while total pivoting searches the whole matrix for the largest number to use as the pivot. Pivoting strategies help prevent zero pivots and reduce round-off errors.
The document discusses calculating volumes of revolution by rotating an area about the x-axis or y-axis. It provides the formulas for finding these volumes using integration, with examples of setting up the integrals to calculate specific volumes. It also covers cases where the curve needs to be rearranged in order to substitute it into the integral when rotating about the y-axis.
A matrix is a rectangular array of numbers arranged in rows and columns. The dimensions of a matrix are written as the number of rows x the number of columns. Each individual entry in the matrix is named by its position, using the matrix name and row and column numbers. Matrices can represent systems of equations or points in a plane. Operations on matrices include addition, multiplication by scalars, and dilation of points represented by matrices.
The document defines and provides examples of different types of matrices, including:
- Square matrices, where the number of rows equals the number of columns.
- Rectangular matrices, where the number of rows does not equal the number of columns.
- Row matrices, with only one row.
- Column matrices, with only one column.
- Null or zero matrices, with all elements equal to zero.
- Diagonal matrices, with all elements equal to zero except those on the main diagonal.
The document also discusses transpose, adjoint, and addition of matrices.
1. Addition and subtraction are basic matrix operations where corresponding elements are added or subtracted if the matrices have the same order.
2. Matrix multiplication is also a basic operation on matrices.
3. For matrix addition and subtraction to be defined, the matrices must have the same number of rows and columns. For matrix multiplication to be defined, the number of columns of the first matrix must be equal to the number of rows of the second matrix.
System of Homogeneous and Non-Homogeneous equations ppt nadi.pptxVinayKp11
The document summarizes homogeneous and non-homogeneous systems of linear equations. A homogeneous system has all constant terms equal to zero, so the general form is a11x1 + ... + amnxn = 0. Every homogeneous system has the trivial solution of all variables equal to zero, and may have non-trivial solutions. A non-homogeneous system includes constant terms not equal to zero, so the general form is a11x1 + ... + amnxn = b1. If the rank of the coefficient matrix A equals the rank of the augmented matrix AB, the system has a unique solution. Examples of solving both homogeneous and non-homogeneous systems are provided.
Basic Calculus 11 - Derivatives and Differentiation RulesJuan Miguel Palero
It is a powerpoint presentation that discusses about the lesson or topic of Derivatives and Differentiation Rules. It also encompasses some formulas, definitions and examples regarding the said topic.
1) The document discusses limits, properties of limits, one-sided limits, and continuity in functions. It provides examples of calculating limits as variables approach certain values.
2) One-sided limits are defined as left and right hand limits, depending on whether the variable approaches the point from the left or right.
3) For a function to be continuous at a point, its limit must exist at that point and be equal to the function value. Examples are given to demonstrate continuity.
This document discusses several applications of integrals including calculating displacement over time, total distance traveled, area between curves, and volume of solids of revolution using disc and washer methods. Specifically, it explains how to use integrals to find displacement given velocity over time, distance given velocity over time, and area between two curves by integrating the top curve minus the bottom curve between the boundaries. It also provides the formulas for finding the volume of a solid of revolution using the disc method which integrates the radius function squared or the washer method which integrates the outer radius squared minus the inner radius squared.
The document discusses partial ordered sets (POSETs). It begins by defining a POSET as a set A together with a partial order R, which is a relation on A that is reflexive, antisymmetric, and transitive. An example is given of the set of integers under the relation "greater than or equal to". It is shown that this relation satisfies the three properties of a partial order. The document emphasizes that a relation must satisfy all three properties - reflexive, antisymmetric, and transitive - to be considered a partial order. Some example relations on a set are provided and it is discussed which of these are partial orders.
This document provides an introduction and overview of MATLAB (Matrix Laboratory), an interactive program for numerical computation and visualization. It discusses basic MATLAB commands and functions for creating variables and matrices, performing mathematical operations, plotting graphs, and working with polynomials.
The document is a slide presentation on differential equations consisting of 5 slides. It includes definitions of ordinary and partial differential equations, classifications based on the number of independent variables, and examples of applications in fields like physics, engineering, and computer science. Real-life applications are discussed along with the use of software packages to study differential equations. The conclusion and references sections are also outlined. The presentation is attributed to a lecture by MD. Ashraful Islam from the Department of CSE at Dhaka International University.
Use of integral calculus in engineeringJunaid Ijaz
The document discusses several applications of integration in engineering, including:
1) Finding shear force and bending moment by integrating shear force.
2) Finding the volume of solids of revolution by slicing the solid and integrating the area of each slice.
3) Using integration to calculate moments of inertia, which are important for measuring structural resistance to bending and buckling.
4) Finding the centroid or center of mass of irregular shapes by dividing them into simpler shapes and taking integrals of the coordinates weighted by area.
The order of the given matrix is 2×3. So the maximum no. of elements is 2×3 = 6.
The correct option is B.
The element a32 belongs to 3rd row and 2nd column.
The correct option is B.
3. A matrix whose each diagonal element is unity and all other elements are zero is called
A) Identity matrix B) Unit matrix C) Scalar matrix D) Diagonal matrix
4. A matrix whose each row sums to unity is called
A) Row matrix B) Column matrix C) Unit matrix D) Stochastic matrix
5. The sum of all the elements on the principal diagonal of a square
1. The document provides information on multiple integrals including double integrals, triple integrals, and integrals in spherical and cylindrical coordinates. It defines each type of integral and gives their general formulas.
2. Examples are provided for calculating double and triple integrals over different regions in rectangular, cylindrical, and spherical coordinate systems. The order of integration can be changed by considering strips or slices of the region.
3. Properties of the integrals include applying Fubini's theorem to change the order of integration, and relating the triple integral over a region to the double integral over the bounds and integrating over the third variable.
This document discusses integration in mathematics. It defines integration as the process opposite to differentiation, where integration finds the direct relationship between two variables given their rate of change. Several techniques for integration are described, including integration by parts and substitution. The document outlines the history of integration and its applications in fields like engineering, business, and its use in estimating important values.
The document discusses methods for calculating the surface area of 3D objects. It introduces the concept that the area of a parallelogram defined by two 3D vectors u and v can be calculated as |u x v|. It then describes a method to approximate the surface area of a smooth surface z=f(x,y) over a domain D by partitioning D into subrectangles, finding the area of the parallelogram in the tangent plane at each corner point, and taking the limit as the partitions approach zero. Two examples are outlined, one using implicit differentiation to find the surface area of a cylinder.
This document discusses the application of vector integration in various domains. It begins by defining vector calculus concepts like del, gradient, curl, and divergence. It then presents several theorems of vector integration. Next, it explains how vector integration can be used to find the rate of change of fluid mass and analyze fluid circulation, vorticity, and the Bjerknes Circulation Theorem regarding sea breezes. It also discusses using vector calculus concepts in electricity and magnetism.
The document discusses symmetric groups and permutations. Some key points:
- A symmetric group SX is the group of all permutations of a set X under function composition.
- Sn, the symmetric group of degree n, represents permutations of the set {1,2,...,n}.
- Permutations can be written using cycle notation, such as (1 2 3) or (1 4).
- Disjoint cycles commute; the product of two permutations is their composition as functions.
- Any permutation can be written as a product of disjoint cycles of length ≥2. Cycles of even length decompose into an odd number of transpositions, and vice versa for odd cycles.
This document provides examples of using integrals to calculate the area of bounded regions. It gives formulas for finding the area under a curve, between two curves, and bounded by curves and lines. It then lists 24 practice problems involving finding areas bounded by geometric shapes, curves, and regions defined with inequalities. The problems cover areas bounded by circles, ellipses, parabolas, lines, and combinations thereof. They require the use of integral calculus to calculate the exact area or to sketch the region and set up the appropriate integral.
El documento presenta 12 problemas resueltos sobre el cálculo de áreas de figuras planas limitadas por curvas. Se explican fórmulas para hallar el área cuando la región está limitada por una función, dos funciones, o curvas paramétricas. Los problemas aplican estas fórmulas para calcular áreas de regiones limitadas por funciones, curvas algebraicas como circunferencias e hipérbolas, y curvas como la cardioide y la astroide.
The document discusses calculating volumes of revolution by rotating an area about the x-axis or y-axis. It provides the formulas for finding these volumes using integration, with examples of setting up the integrals to calculate specific volumes. It also covers cases where the curve needs to be rearranged in order to substitute it into the integral when rotating about the y-axis.
A matrix is a rectangular array of numbers arranged in rows and columns. The dimensions of a matrix are written as the number of rows x the number of columns. Each individual entry in the matrix is named by its position, using the matrix name and row and column numbers. Matrices can represent systems of equations or points in a plane. Operations on matrices include addition, multiplication by scalars, and dilation of points represented by matrices.
The document defines and provides examples of different types of matrices, including:
- Square matrices, where the number of rows equals the number of columns.
- Rectangular matrices, where the number of rows does not equal the number of columns.
- Row matrices, with only one row.
- Column matrices, with only one column.
- Null or zero matrices, with all elements equal to zero.
- Diagonal matrices, with all elements equal to zero except those on the main diagonal.
The document also discusses transpose, adjoint, and addition of matrices.
1. Addition and subtraction are basic matrix operations where corresponding elements are added or subtracted if the matrices have the same order.
2. Matrix multiplication is also a basic operation on matrices.
3. For matrix addition and subtraction to be defined, the matrices must have the same number of rows and columns. For matrix multiplication to be defined, the number of columns of the first matrix must be equal to the number of rows of the second matrix.
System of Homogeneous and Non-Homogeneous equations ppt nadi.pptxVinayKp11
The document summarizes homogeneous and non-homogeneous systems of linear equations. A homogeneous system has all constant terms equal to zero, so the general form is a11x1 + ... + amnxn = 0. Every homogeneous system has the trivial solution of all variables equal to zero, and may have non-trivial solutions. A non-homogeneous system includes constant terms not equal to zero, so the general form is a11x1 + ... + amnxn = b1. If the rank of the coefficient matrix A equals the rank of the augmented matrix AB, the system has a unique solution. Examples of solving both homogeneous and non-homogeneous systems are provided.
Basic Calculus 11 - Derivatives and Differentiation RulesJuan Miguel Palero
It is a powerpoint presentation that discusses about the lesson or topic of Derivatives and Differentiation Rules. It also encompasses some formulas, definitions and examples regarding the said topic.
1) The document discusses limits, properties of limits, one-sided limits, and continuity in functions. It provides examples of calculating limits as variables approach certain values.
2) One-sided limits are defined as left and right hand limits, depending on whether the variable approaches the point from the left or right.
3) For a function to be continuous at a point, its limit must exist at that point and be equal to the function value. Examples are given to demonstrate continuity.
This document discusses several applications of integrals including calculating displacement over time, total distance traveled, area between curves, and volume of solids of revolution using disc and washer methods. Specifically, it explains how to use integrals to find displacement given velocity over time, distance given velocity over time, and area between two curves by integrating the top curve minus the bottom curve between the boundaries. It also provides the formulas for finding the volume of a solid of revolution using the disc method which integrates the radius function squared or the washer method which integrates the outer radius squared minus the inner radius squared.
The document discusses partial ordered sets (POSETs). It begins by defining a POSET as a set A together with a partial order R, which is a relation on A that is reflexive, antisymmetric, and transitive. An example is given of the set of integers under the relation "greater than or equal to". It is shown that this relation satisfies the three properties of a partial order. The document emphasizes that a relation must satisfy all three properties - reflexive, antisymmetric, and transitive - to be considered a partial order. Some example relations on a set are provided and it is discussed which of these are partial orders.
This document provides an introduction and overview of MATLAB (Matrix Laboratory), an interactive program for numerical computation and visualization. It discusses basic MATLAB commands and functions for creating variables and matrices, performing mathematical operations, plotting graphs, and working with polynomials.
The document is a slide presentation on differential equations consisting of 5 slides. It includes definitions of ordinary and partial differential equations, classifications based on the number of independent variables, and examples of applications in fields like physics, engineering, and computer science. Real-life applications are discussed along with the use of software packages to study differential equations. The conclusion and references sections are also outlined. The presentation is attributed to a lecture by MD. Ashraful Islam from the Department of CSE at Dhaka International University.
Use of integral calculus in engineeringJunaid Ijaz
The document discusses several applications of integration in engineering, including:
1) Finding shear force and bending moment by integrating shear force.
2) Finding the volume of solids of revolution by slicing the solid and integrating the area of each slice.
3) Using integration to calculate moments of inertia, which are important for measuring structural resistance to bending and buckling.
4) Finding the centroid or center of mass of irregular shapes by dividing them into simpler shapes and taking integrals of the coordinates weighted by area.
The order of the given matrix is 2×3. So the maximum no. of elements is 2×3 = 6.
The correct option is B.
The element a32 belongs to 3rd row and 2nd column.
The correct option is B.
3. A matrix whose each diagonal element is unity and all other elements are zero is called
A) Identity matrix B) Unit matrix C) Scalar matrix D) Diagonal matrix
4. A matrix whose each row sums to unity is called
A) Row matrix B) Column matrix C) Unit matrix D) Stochastic matrix
5. The sum of all the elements on the principal diagonal of a square
1. The document provides information on multiple integrals including double integrals, triple integrals, and integrals in spherical and cylindrical coordinates. It defines each type of integral and gives their general formulas.
2. Examples are provided for calculating double and triple integrals over different regions in rectangular, cylindrical, and spherical coordinate systems. The order of integration can be changed by considering strips or slices of the region.
3. Properties of the integrals include applying Fubini's theorem to change the order of integration, and relating the triple integral over a region to the double integral over the bounds and integrating over the third variable.
This document discusses integration in mathematics. It defines integration as the process opposite to differentiation, where integration finds the direct relationship between two variables given their rate of change. Several techniques for integration are described, including integration by parts and substitution. The document outlines the history of integration and its applications in fields like engineering, business, and its use in estimating important values.
The document discusses methods for calculating the surface area of 3D objects. It introduces the concept that the area of a parallelogram defined by two 3D vectors u and v can be calculated as |u x v|. It then describes a method to approximate the surface area of a smooth surface z=f(x,y) over a domain D by partitioning D into subrectangles, finding the area of the parallelogram in the tangent plane at each corner point, and taking the limit as the partitions approach zero. Two examples are outlined, one using implicit differentiation to find the surface area of a cylinder.
This document discusses the application of vector integration in various domains. It begins by defining vector calculus concepts like del, gradient, curl, and divergence. It then presents several theorems of vector integration. Next, it explains how vector integration can be used to find the rate of change of fluid mass and analyze fluid circulation, vorticity, and the Bjerknes Circulation Theorem regarding sea breezes. It also discusses using vector calculus concepts in electricity and magnetism.
The document discusses symmetric groups and permutations. Some key points:
- A symmetric group SX is the group of all permutations of a set X under function composition.
- Sn, the symmetric group of degree n, represents permutations of the set {1,2,...,n}.
- Permutations can be written using cycle notation, such as (1 2 3) or (1 4).
- Disjoint cycles commute; the product of two permutations is their composition as functions.
- Any permutation can be written as a product of disjoint cycles of length ≥2. Cycles of even length decompose into an odd number of transpositions, and vice versa for odd cycles.
This document provides examples of using integrals to calculate the area of bounded regions. It gives formulas for finding the area under a curve, between two curves, and bounded by curves and lines. It then lists 24 practice problems involving finding areas bounded by geometric shapes, curves, and regions defined with inequalities. The problems cover areas bounded by circles, ellipses, parabolas, lines, and combinations thereof. They require the use of integral calculus to calculate the exact area or to sketch the region and set up the appropriate integral.
El documento presenta 12 problemas resueltos sobre el cálculo de áreas de figuras planas limitadas por curvas. Se explican fórmulas para hallar el área cuando la región está limitada por una función, dos funciones, o curvas paramétricas. Los problemas aplican estas fórmulas para calcular áreas de regiones limitadas por funciones, curvas algebraicas como circunferencias e hipérbolas, y curvas como la cardioide y la astroide.
Materi kuliah tentang Aplikasi Integral. Cari lebih banyak mata kuliah Semester 1 di: http://muhammadhabibielecture.blogspot.com/2014/12/kuliah-semester-1-thp-ftp-ub.html
Integrals have many applications in engineering and science. They allow us to calculate areas, volumes, work, center of mass, moments of inertia, and probabilities. Integrals are essential for modeling real-world phenomena like distance, velocity, acceleration, and many other physical quantities that change with respect to another variable like time.
This document discusses the application of integration in business and economics. It provides examples of how integrals can be used to calculate total costs, profits, and areas between curves representing revenue and costs. Specifically:
1) Integrals allow calculating total costs by integrating marginal or incremental costs over quantity or time.
2) The difference between revenue and cost areas under their curves gives net profit over a range of quantities.
3) Investment profits over time can be found by taking the area under profit functions, allowing comparison of different investment options.
This document discusses numerical integration techniques, including:
- The need to approximate definite integrals numerically when exact solutions cannot be found.
- The trapezoidal rule, which approximates the area under a curve as a trapezoid, and generalizes this to multiple trapezoids to improve accuracy.
- Simpson's rule, which uses a parabolic interpolation between three points to better approximate curved areas.
- Examples are provided to demonstrate applying these rules to calculate integrals of simple functions.
The document discusses numerical methods and their applications. It provides definitions of numerical methods as procedures for solving problems with computable error estimates. Some common numerical methods are listed, including bisection, Newton-Raphson, iteration, and interpolation methods. Applications mentioned include root finding, profit/loss calculation, multidimensional root finding, and simulations. An example is given of using numerical methods for image deblurring. The document also discusses computational modeling, algorithm development and implementation, and limitations of computers in solving mathematical problems.
B.tech ii unit-3 material multiple integrationRai University
1. The document discusses multiple integrals and double integrals. It defines double integrals and provides two methods for evaluating them: integrating first with respect to one variable and then the other, or vice versa.
2. Examples are given of evaluating double integrals using these methods over different regions of integration in the xy-plane, including integrals over a circle and a hyperbolic region.
3. The document also discusses calculating double integrals over a region when the limits of integration are not explicitly given, but the region is described geometrically.
1. Se calcula la integral doble de una función en una región limitada transformando las coordenadas cartesianas a coordenadas polares. Se obtienen los límites de integración y se resuelve la integral.
2. Igualmente, se transforman coordenadas cartesianas a polares para calcular otra integral doble en una región limitada por circunferencias, encontrando los límites y resolviendo la integral.
3. De manera análoga, se calcula otra integral doble transformando a coordenadas polares y resolviendo.
Este documento describe el sistema de coordenadas polares. Define las coordenadas polares de un punto como la distancia (r) desde el origen y el ángulo (θ) medido desde el eje polar. Explica cómo convertir entre coordenadas polares y rectangulares y cómo trazar curvas dadas sus ecuaciones polares. También cubre conceptos como coordenadas polares generalizadas y ejercicios de conversión de sistemas.
Integration is used in physics to determine rates of change and distances given velocities. Numerical integration is required when the antiderivative is unknown. It involves approximating the definite integral of a function as the area under its curve between bounds. The Trapezoidal Rule approximates this area using straight lines between points, while Simpson's Rule uses quadratic or cubic functions, achieving greater accuracy with fewer points. Both methods involve dividing the area into strips and summing their widths multiplied by the function values at strip points.
Este documento presenta los conceptos y procedimientos para calcular el área de regiones planas utilizando la integral definida. Explica que el área de una región se puede obtener como la suma de áreas de elementos diferenciales infinitesimales, lo que equivale a evaluar una integral definida. Proporciona ejemplos detallados de cómo calcular el área entre curvas, bajo una curva, y de regiones simple-y. Concluye resumiendo los pasos a seguir para hallar el área de cualquier región plana mediante la integral.
Este documento presenta el sistema de coordenadas polares y cómo graficar ecuaciones en este sistema. Explica que las coordenadas polares (r, θ) describen un punto como la distancia r desde el origen y el ángulo θ. Luego detalla cómo graficar rectas, circunferencias, parábolas, elipses e hipérbolas mediante sus ecuaciones polares correspondientes. Finalmente, proporciona ejercicios para que el estudiante aplique estos conceptos.
This document summarizes numerical methods used in various fields including engineering, crime detection, scientific computing, finding roots, and solving heat equations. It discusses how numerical methods are widely used in engineering to model systems using mathematical equations when analytical solutions are not possible. Examples of applying numerical methods include structural analysis, fluid dynamics, image processing to deblur photos, and algorithms for finding roots of equations and solving differential equations.
Integral calculus deals with functions to be integrated. The integral is the reverse of differentiation and is equivalent to the area under a curve. There are several numerical methods for calculating integrals, including the trapezoidal rule, Simpson's 1st rule, and Simpson's 2nd rule. These rules approximate the integral of a function over an interval by summing areas of geometric objects.
Ejercicios resueltos integrales dobles y triples manoleter
Este documento presenta una serie de ejercicios resueltos sobre cálculo de integrales dobles en coordenadas rectangulares cartesianas. En total se presentan 7 problemas con sus respectivas soluciones, donde se calculan áreas, volúmenes y otras integrales dobles sobre diferentes regiones delimitadas por funciones.
This document discusses numerical integration techniques including the trapezoidal rule, Simpson's 1/3 rule, Simpson's 3/8 rule, and Gaussian integration formulas. It provides the formulas for calculating integration numerically using these methods and notes that accuracy increases with smaller interval widths h. Errors are estimated to be order h^2 for trapezoidal rule and order h^4 for Simpson's rules.
The document discusses Newton's Law of Cooling and its applications through differential equations.
- Newton's Law of Cooling states that the rate of change of an object's temperature is proportional to the difference between the object's temperature and the ambient temperature. This can be modeled as a first-order differential equation.
- The equation can be derived and solved using calculus techniques like separation of variables. The solution is an exponential decay function.
- Real-world applications include determining time of death from body temperature, designing efficient cooling systems for computer processors, and calculating heat transfer rates in devices like solar water heaters. Mathematical problems demonstrate using the law of cooling in investigations and engineering design.
this is the ppt on application of integrals, which includes-area between the two curves , volume by slicing , disk method , washer method, and volume by cylindrical shells,.
this is made by dhrumil patel and harshid panchal.
1. The document provides instructions for a mathematics exam. It states that calculators may be used, full marks require showing working, and scale drawings will not be credited. It then lists various formulae that may be needed for the exam.
2. The exam consists of 11 multi-part questions testing a range of mathematics skills, including algebra, geometry, trigonometry, calculus and graph sketching. Candidates are advised to attempt all questions.
3. The document concludes by providing blank pages for working, followed by a notice that the exam has ended.
- The document discusses quadratic functions and their graphs. It explains that the graph of a quadratic function is a parabola, which is a U-shaped curve.
- It describes how to write quadratic functions in standard form and use that form to sketch the graph and find features like the vertex and axis of symmetry.
- Examples are provided to demonstrate how to graph quadratic functions in standard form and how to find the minimum or maximum value of a quadratic function by setting its derivative equal to zero.
1. This document provides instructions for completing Section A of a mathematics exam. Students must use a pencil and fill in answers on an answer sheet by making a horizontal line in the correct space.
2. The answer sheet contains the student's name, date of birth, and other identifying information. Students should check this is correct and report any errors.
3. There is one correct answer for each question, and rough work should not be done on the answer sheet.
This document discusses quadratic equations and functions. It explains how to solve quadratic equations by factoring, completing the square, and using the quadratic formula. It also discusses using the discriminant to determine the number and type of roots. Properties of quadratic functions such as the sum and product of roots are covered. Methods for constructing quadratic equations and functions given certain properties are provided. Finally, it briefly discusses sketching the graph of a quadratic function.
1. The set of all functions f: R → R with f(0) = 0 is a vector space, as the linear combination of such functions will also satisfy f(0) = 0.
2. The set of all odd functions is a vector space, as any linear combination of odd functions will also be odd.
3. The solution space to the differential equation y''(x) - 5y'(x) = 0 is 2-dimensional with basis {1, e^5x}, as the general solution is Ce^5x + D.
8.further calculus Further Mathematics Zimbabwe Zimsec Cambridgealproelearning
This document discusses techniques for approximating integrals, including the trapezium rule and Simpson's rule. The trapezium rule approximates the area under a curve as the sum of trapezoidal areas formed by the function values at the endpoints of subintervals. Simpson's rule approximates the area as the sum of triangular areas, weighted differently for even and odd terms, formed by the function values at three evenly spaced points in each subinterval. Examples are given to demonstrate applying these rules to approximate definite integrals. The Simpson's rule is generally more accurate because it approximates the curve by a quadratic rather than a straight line as in the trapezium rule.
The process of finding area of some plane region is called Quadrature. In this chapter we shall find the area bounded by some simple plane curves with the help of definite integral. For solving
(i) The area bounded by a cartesian curve y = f(x), x-axis and ordinates x = a and x = b is given by,
the problems on quadrature easily, if possible first draw the rough sketch of the required area.
b
Area = y dx
a
b
= f(x) dx
a
In chapter function, we have seen graphs of some simple elementary curves. Here we introduce some essential steps for curve tracing which will enable us to determine the required area.
(i) Symmetry
The curve f(x, y) = 0 is symmetrical
about x-axis if all terms of y contain even powers.
about y-axis if all terms of x contain even
powers.
about the origin if f (– x, – y) = f (x, y).
Examples
based on
Area Bounded by A Curve
For example, y2 = 4ax is symmetrical about x-axis, and x2 = 4ay is symmetrical about y- axis and the curve y = x3 is symmetrical about the origin.
(ii) Origin
Ex.1 Find the area bounded by the curve y = x3, x-axis and ordinates x = 1 and x = 2.
2
Sol. Required Area = ydx
1
If the equation of the curve contains no constant
2
= x3 dx =
Lx4 O2 15
MN PQ=
Ans.
term then it passes through the origin.
For example x2 + y2 + 2ax = 0 passes through origin.
(iii) Points of intersection with the axes
If we get real values of x on putting y = 0 in the equation of the curve, then real values of x and y = 0 give those points where the curve cuts the x-axis. Similarly by putting x = 0, we can get the points of intersection of the curve and y-axis.
1 4 1 4
Ex.2 Find the area bounded by the curve y = sec2x,
x-axis and the line x = 4
/4
Sol. Required Area = ydx
x0
For example, the curve x2/a2 + y2 /b2 = 1 intersects the axes at points (± a, 0) and (0, ± b) .
(iv) Region
/4
= sec2
0
xdx = tan x /4 = 1 Ans.
Write the given equation as y = f(x) , and find minimum and maximum values of x which determine the region of the curve.
For example for the curve xy2 = a2 (a – x)
Ex.3 Find the area bounded by the curve y = mx, x-axis and ordinates x = 1 and x = 2
2
Sol. Required area = y dx .
1
y = a
a x
x
2
= mxdx =
LMmx2 2
Now y is real, if 0 < x a , so its region lies between the lines x = 0 and x = a.
1 MN2
= m ( 4 – 1) =
2
PQ1
FGH3 JKm Ans.
Ex.4 Find the area bounded by the curve y = x (1– x)2 and x-axis.
Sol. Clearly the given curve meets the x-axis at (0,0) and (1,0) and for x = 0 to 1, y is positive
Ex.7 Find the area bounded between the curve y2 = 2y – x and y-axis.
Sol. The area between the given curve x = 2y – y2 and y-axis will be as shown in diagram.
so required area- Y
= xb1 xg2 dx
0
1
= ex 2x2 x3 jdx
0
O (0,0) (1,0) X
Lx2
2x3
x4 O1
1 2 1 1
= M P=
– + =
MN2 3 4 PQ0
2 3 4 12
Ans.
Required Area =
e2y y2 jdy
(i
This document discusses using integration to find the area between two curves by considering it as an accumulation process. It explains that we select a representative element, such as a rectangle, and use geometry formulas to relate the area of that element to the functions that define the curves. The area under each representative rectangle is summed to find the total area via integration. Two examples are provided, one finding the area between a parabola and the x-axis using vertical rectangles, and another using horizontal rectangles between two other curves.
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1) It derives the formula for finding the distance between two points in a plane as the square root of the sum of the squares of the differences of their x- and y-coordinates.
2) It derives the midpoint formula for finding the midpoint between two points as the average of their x-coordinates and the average of their y-coordinates.
3) It discusses the standard equation of a circle, gives methods for finding the center and radius from different forms of the circle equation, and notes degenerate cases where the equation does not represent a circle.
The document discusses integration and indefinite integrals. It covers determining integrals by reversing differentiation, integrating algebraic expressions like constants, variables, and polynomials. It also discusses determining the constant of integration and using integration to find equations of curves from their gradients. Examples are provided to illustrate integrating functions and finding volumes generated by rotating an area about an axis.
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Parabola direction , vertex ,roots, minimum and maximumNadeem Uddin
This document discusses quadratic functions of the form f(x) = ax^2 + bx + c. It provides details on:
1) The direction a parabola opens depending on whether a is positive or negative.
2) How to calculate the vertex (turning point) of a parabola using the formula (-b/2a, 4ac-b^2/4a).
3) How to calculate the roots (x-intercepts) of a quadratic function using the formula (-b ± √(b^2 - 4ac))/2a.
Several examples are provided to demonstrate finding the direction, vertex, roots, minimum/maximum values of various quadratic functions.
The document discusses graphing quadratic functions. It begins with reviewing key concepts like the vertex and axis of symmetry and how the a, b, and c coefficients affect the graph. Examples are provided for determining the width, direction opened, and vertical shift based on these coefficients. The remainder of the document provides step-by-step examples of graphing quadratic functions by finding the axis of symmetry, vertex, y-intercept, and other points to plot the parabolic curve.
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1) Vector-valued functions rA(θ) and rB(θ) are derived to describe points A and B on the curve parametrically.
2) These are combined to obtain the vector-valued function r(θ) for the overall Witch of Agnesi curve.
3) The rectangular equation y=(8a^3)/(x^2+4a^2) is then derived by eliminating
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The document discusses using the discriminant of a quadratic equation to determine the type of conic section represented by the graph of the equation. It defines the discriminant as B^2 - 4AC and explains that:
(a) If the discriminant is negative, the graph is an ellipse, circle, point, or has no graph.
(b) If the discriminant is positive, the graph is a hyperbola or intersecting lines.
(c) If the discriminant is 0, the graph is a parabola, line, parallel lines, or has no graph.
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Finding the opening of the parabola, vertex, axis of symmetry, y-intercept, x- intercept, domain, range, and the minimum/maximum value including the illustration of the graph
1. APPLICATION OF INTEGRALS 359
Chapter 8
APPLICATION OF INTEGRALS
One should study Mathematics because it is only through Mathematics that
nature can be conceived in harmonious form. – BIRKHOFF
8.1 Introduction
In geometry, we have learnt formulae to calculate areas
of various geometrical figures including triangles,
rectangles, trapezias and circles. Such formulae are
fundamental in the applications of mathematics to many
real life problems. The formulae of elementary geometry
allow us to calculate areas of many simple figures.
However, they are inadequate for calculating the areas
enclosed by curves. For that we shall need some concepts
of Integral Calculus.
In the previous chapter, we have studied to find the
area bounded by the curve y = f (x), the ordinates x = a,
x = b and x-axis, while calculating definite integral as the
limit of a sum. Here, in this chapter, we shall study a specific
application of integrals to find the area under simple curves, A.L. Cauchy
area between lines and arcs of circles, parabolas and (1789-1857)
ellipses (standard forms only). We shall also deal with finding
the area bounded by the above said curves.
8.2 Area under Simple Curves
In the previous chapter, we have studied
definite integral as the limit of a sum and
how to evaluate definite integral using
Fundamental Theorem of Calculus. Now,
we consider the easy and intuitive way of
finding the area bounded by the curve
y = f (x), x-axis and the ordinates x = a and
x = b. From Fig 8.1, we can think of area
under the curve as composed of large
number of very thin vertical strips. Consider
an arbitrary strip of height y and width dx,
then dA (area of the elementary strip) = ydx,
where, y = f (x). Fig 8.1
2. 360 MATHEMATICS
This area is called the elementary area which is located at an arbitrary position
within the region which is specified by some value of x between a and b. We can think
of the total area A of the region between x-axis, ordinates x = a, x = b and the curve
y = f (x) as the result of adding up the elementary areas of thin strips across the region
PQRSP. Symbolically, we express
b b b
A= ∫ a
dA = ∫ ydx = ∫ f ( x ) dx
a a
The area A of the region bounded by
the curve x = g (y), y-axis and the lines y = c,
y = d is given by
∫ xdy = ∫ g ( y) dy
d d
A= c c
Here, we consider horizontal strips as shown in
the Fig 8.2 Fig 8.2
Remark If the position of the curve under consideration is below the x-axis, then since
f (x) < 0 from x = a to x = b, as shown in Fig 8.3, the area bounded by the curve, x-axis
and the ordinates x = a, x = b come out to be negative. But, it is only the numerical
value of the area which is taken into consideration. Thus, if the area is negative, we
∫
b
take its absolute value, i.e., f ( x ) dx .
a
Fig 8.3
Generally, it may happen that some portion of the curve is above x-axis and some is
below the x-axis as shown in the Fig 8.4. Here, A1 < 0 and A2 > 0. Therefore, the area
A bounded by the curve y = f (x), x-axis and the ordinates x = a and x = b is given
by A = | A1 | + A2.
3. APPLICATION OF INTEGRALS 361
Fig 8.4
Example 1 Find the area enclosed by the circle x2 + y2 = a2.
Solution From Fig 8.5, the whole area enclosed
by the given circle
= 4 (area of the region AOBA bounded by
the curve, x-axis and the ordinates x = 0 and
x = a) [as the circle is symmetrical about both
x-axis and y-axis]
a
= 4 ∫ 0 ydx (taking vertical strips)
a
= 4∫ a 2 − x 2 dx
0
Since x2 + y2 = a2 gives y = ± a 2 − x2 Fig 8.5
As the region AOBA lies in the first quadrant, y is taken as positive. Integrating, we get
the whole area enclosed by the given circle
a
⎡x 2 a2 x⎤
= 4⎢ a − x2 + sin –1 ⎥
⎣2 2 a ⎦0
⎡⎛ a a2 −1 ⎞
⎤ ⎛ a2 ⎞ ⎛ π ⎞
= 4 ⎢ ⎜ × 0 + sin 1⎟ − 0⎥ = 4 ⎜
2
⎝2 2 ⎠ ⎟ ⎜ ⎟ = πa
⎣ ⎦ ⎝ 2 ⎠ ⎝ 2⎠
4. 362 MATHEMATICS
Alternatively, considering horizontal strips as shown in Fig 8.6, the whole area of the
region enclosed by circle
a
= 4 ∫ xdy = 4 ∫
a
a 2 − y 2 dy (Why?)
0 0
a
⎡y 2 a2 y⎤
= 4⎢ a − y2 + sin −1 ⎥
⎣ 2 2 a ⎦0
⎡⎛ a a2 ⎞ ⎤
= 4 ⎢⎜ × 0 + sin −1 1⎟ − 0⎥
⎣⎝ 2 2 ⎠ ⎦
a2 π
= 4 = πa 2
2 2
Fig 8.6
x2 y2
Example 2 Find the area enclosed by the ellipse 2 + 2 = 1
a b
Solution From Fig 8.7, the area of the region ABA′B′A bounded by the ellipse
⎛ area of the region AOBA in the first quadrant bounded ⎞
= 4⎜ ⎟
⎝ by the curve, x − axis and the ordinates x = 0, x = a ⎠
(as the ellipse is symmetrical about both x-axis and y-axis)
a
= 4 ∫ 0 ydx (taking verticalstrips)
x2 y 2 b 2
Now 2 + 2 = 1 gives y = ± a − x 2 , but as the region AOBA lies in the first
a b a
quadrant, y is taken as positive. So, the required area is
a b 2
= 4∫ 0 a − x 2 dx
a
a
4b ⎡ x 2 2 a 2 –1 x ⎤
= ⎢ a − x + sin ⎥ (Why?)
a ⎣2 2 a ⎦0
4b ⎡⎛ a a2 −1 ⎞
⎤
= ⎢⎜ × 0 + sin 1⎟ − 0 ⎥
a ⎢⎝ 2
⎣ 2 ⎠ ⎥ ⎦
4b a 2 π
= = π ab
a 2 2 Fig 8.7
5. APPLICATION OF INTEGRALS 363
Alternatively, considering horizontal strips as
shown in the Fig 8.8, the area of the ellipse is
b
a
= 4 ∫ xdy = 4
b
b∫
b2 − y 2 dy (Why?)
0
0
b
4a ⎡ y 2 b2 y⎤
= ⎢ b − y 2 + sin –1 ⎥
b ⎣2 2 b ⎦0
4 a ⎡⎛ b b2 –1 ⎞
⎤ Fig 8.8
= b ⎢⎜ 2 × 0 + 2 sin 1⎟ − 0 ⎥
⎢⎝
⎣ ⎠ ⎥ ⎦
4a b2 π
= = πab
b 2 2
8.2.1 The area of the region bounded by a curve and a line
In this subsection, we will find the area of the region bounded by a line and a circle,
a line and a parabola, a line and an ellipse. Equations of above mentioned curves will be
in their standard forms only as the cases in other forms go beyond the scope of this
textbook.
Example 3 Find the area of the region bounded
by the curve y = x2 and the line y = 4.
Solution Since the given curve represented by
the equation y = x2 is a parabola symmetrical
about y-axis only, therefore, from Fig 8.9, the
required area of the region AOBA is given by
4
2∫ xdy =
0
Fig 8.9
⎛ area of the region BONB bounded by curve, y − axis ⎞
2⎜ ⎟
⎝ and the lines y = 0 and y = 4 ⎠
4
2⎡ 2⎤
3
4 4 32
= 2∫ ydy = 2 × ⎢ y ⎥ = ×8= (Why?)
0 3⎢ ⎥
⎣ ⎦ 3 3
0
Here, we have taken horizontal strips as indicated in the Fig 8.9.
6. 364 MATHEMATICS
Alternatively, we may consider the vertical
strips like PQ as shown in the Fig 8.10 to
obtain the area of the region AOBA. To this
end, we solve the equations x2 = y and y = 4
which gives x = –2 and x = 2.
Thus, the region AOBA may be stated as
the region bounded by the curve y = x2, y = 4
and the ordinates x = –2 and x = 2.
Therefore, the area of the region AOBA
2 Fig 8.10
= ∫ −2 ydx
[ y = ( y-coordinate of Q) – (y-coordinate of P) = 4 – x2 ]
( )
2
= 2 ∫ 0 4 − x dx
2
(Why?)
2
⎡ x3 ⎤ ⎡ 8 ⎤ 32
= 2 ⎢4 x − ⎥ = 2 ⎢4 × 2 − ⎥ =
⎣ 3 ⎦0 ⎣ 3⎦ 3
Remark From the above examples, it is inferred that we can consider either vertical
strips or horizontal strips for calculating the area of the region. Henceforth, we shall
consider either of these two, most preferably vertical strips.
Example 4 Find the area of the region in the first quadrant enclosed by the x-axis,
the line y = x, and the circle x2 + y2 = 32.
Y
Solution The given equations are
y= x ... (1) y=x
and x + y = 32
2 2
... (2) B
(4,4)
Solving (1) and (2), we find that the line
and the circle meet at B(4, 4) in the first
quadrant (Fig 8.11). Draw perpendicular A
BM to the x-axis. X' X
M
O (4 2 ,0)
Therefore, the required area = area of
the region OBMO + area of the region
BMAB.
Now, the area of the region OBMO
4 4
= ∫ 0 ydx = ∫ 0 xdx ... (3)
Y'
1 2 4
= ⎡ x ⎤0 = 8 Fig 8.11
2⎣ ⎦
7. APPLICATION OF INTEGRALS 365
Again, the area of the region BMAB
4 2 4 2
= ∫4 ydx = ∫4 32 − x 2 dx
4 2
⎡1 1 x ⎤
= ⎢ x 32 − x 2 + × 32 × sin –1 ⎥
⎣ 2 2 4 2 ⎦4
⎛1 1 –1 ⎞ ⎛4 1 1 ⎞
= ⎜ 4 2 × 0 + × 32 × sin 1 ⎟ − ⎜ 32 − 16 + × 32 × sin –1 ⎟
⎝2 2 ⎠ ⎝2 2 2⎠
= 8 π – (8 + 4π) = 4π – 8 ... (4)
Adding (3) and (4), we get, the required area = 4π.
x2 y 2
Example 5 Find the area bounded by the ellipse 2 + 2 =1 and the ordinates x = 0
a b
and x = ae, where, b2 = a2 (1 – e2) and e < 1.
Solution The required area (Fig 8.12) of the region BOB′RFSB is enclosed by the
ellipse and the lines x = 0 and x = ae. Y
Note that the area of the region BOB′RFSB
B x = ae
ae
S
= 2∫ 0 ydx = 2 b
ae
∫0
2 2
a − x dx
a
F (ae, o)
X′ X
ae O
2b ⎡ x 2 2 a x⎤ 2
= ⎢ a −x + sin –1 ⎥
a ⎣2 2 a ⎦0
R
B'
2b ⎡ 2 2 2 2 –1 ⎤
=
2a ⎢ae a − a e + a sin e ⎥
⎣ ⎦ Y′
Fig 8.12
⎡ 2 –1 ⎤
= ab ⎢e 1 − e + sin e ⎥
⎣ ⎦
EXERCISE 8.1
1. Find the area of the region bounded by the curve y2 = x and the lines x = 1,
x = 4 and the x-axis.
2. Find the area of the region bounded by y2 = 9x, x = 2, x = 4 and the x-axis in the
first quadrant.
8. 366 MATHEMATICS
3. Find the area of the region bounded by x2 = 4y, y = 2, y = 4 and the y-axis in the
first quadrant.
x2 y2
4. Find the area of the region bounded by the ellipse + =1 .
16 9
x2 y2
5. Find the area of the region bounded by the ellipse + =1 .
4 9
6. Find the area of the region in the first quadrant enclosed by x-axis, line x = 3 y
and the circle x2 + y2 = 4.
a
7. Find the area of the smaller part of the circle x2 + y2 = a2 cut off by the line x =
.
2
8. The area between x = y2 and x = 4 is divided into two equal parts by the line
x = a, find the value of a.
9. Find the area of the region bounded by the parabola y = x2 and y = x .
10. Find the area bounded by the curve x2 = 4y and the line x = 4y – 2.
11. Find the area of the region bounded by the curve y2 = 4x and the line x = 3.
Choose the correct answer in the following Exercises 12 and 13.
12. Area lying in the first quadrant and bounded by the circle x2 + y2 = 4 and the lines
x = 0 and x = 2 is
π π π
(A) π (B) (C) (D)
2 3 4
13. Area of the region bounded by the curve y = 4x, y-axis and the line y = 3 is
2
9 9 9
(A) 2 (B) (C) (D)
4 3 2
8.3 Area between Two Curves
Intuitively, true in the sense of Leibnitz, integration is the act of calculating the area by
cutting the region into a large number of small strips of elementary area and then
adding up these elementary areas. Suppose we are given two curves represented by
y = f (x), y = g (x), where f (x) ≥ g(x) in [a, b] as shown in Fig 8.13. Here the points of
intersection of these two curves are given by x = a and x = b obtained by taking
common values of y from the given equation of two curves.
For setting up a formula for the integral, it is convenient to take elementary area in
the form of vertical strips. As indicated in the Fig 8.13, elementary strip has height
9. APPLICATION OF INTEGRALS 367
f (x) – g (x) and width dx so that the elementary area
Fig 8.13
dA = [f (x) – g(x)] dx, and the total area A can be taken as
b
A= ∫ a [f ( x) − g ( x)] dx
Alternatively,
A = [area bounded by y = f (x), x-axis and the lines x = a, x = b]
– [area bounded by y = g (x), x-axis and the lines x = a, x = b]
∫ a f ( x) dx − ∫ a g ( x) dx = ∫ a [ f ( x) − g ( x)] dx, where f (x) ≥ g (x) in [a, b]
b b b
=
If f (x) ≥ g (x) in [a, c] and f (x) ≤ g (x) in [c, b], where a < c < b as shown in the
Fig 8.14, then the area of the regions bounded by curves can be written as
Total Area = Area of the region ACBDA + Area of the region BPRQB
∫ a [ f ( x) − g ( x)] dx + ∫ c [ g ( x) − f ( x)] dx
c b
=
y = g (x)
Y y = f ( x) P
C
B R
A
D y = g (x) Q y = f (x)
x=a x =c x =b
X′ O X
Y′ Fig 8.14
10. 368 MATHEMATICS
Example 6 Find the area of the region bounded by the two parabolas y = x2 and y2 = x.
Solution The point of intersection of these two
parabolas are O (0, 0) and A (1, 1) as shown in
the Fig 8.15.
Here, we can set y 2 = x or y = x = f(x) and y = x2
= g (x), where, f (x) ≥ g (x) in [0, 1].
Therefore, the required area of the shaded region
1
= ∫ 0 [ f ( x) − g ( x)] dx
1
⎡ 3 3⎤
⎡ x − x dx = ⎢ 2 x 2 − x ⎥
1
= ∫0⎣
2⎤
⎦
⎢3
⎣ 3⎥⎦0 Fig 8.15
2 1 1
= − =
3 3 3
Example 7 Find the area lying above x-axis and included between the circle
x2 + y2 = 8x and the parabola y2 = 4x.
Solution The given equation of the circle x 2 + y 2 = 8x can be expressed as
(x – 4)2 + y2 = 16. Thus, the centre of the Y
circle is (4, 0) and radius is 4. Its intersection
P (4, 4)
with the parabola y2 = 4x gives
x2 + 4x = 8x
or x2 – 4x = 0
or x (x – 4) = 0
X X
or x = 0, x = 4 O C (4, 0) Q (8, 0)
Thus, the points of intersection of these
two curves are O(0, 0) and P(4,4) above the
x-axis.
From the Fig 8.16, the required area of
the region OPQCO included between these Fig 8.16
two curves above x-axis is Y
= (area of the region OCPO) + (area of the region PCQP)
4 8
= ∫ 0 ydx + ∫ 4 ydx
4 8
= 2∫ x dx + ∫ 42 − ( x − 4) 2 dx (Why?)
0 4
11. APPLICATION OF INTEGRALS 369
4
2⎡ 2⎤
3 4
= 2 × ⎢ x ⎥ + ∫ 4 − t dt , where, x − 4 = t
2 2
(Why?)
3⎢ ⎥ 0
⎣ ⎦ 0
4
32 ⎡ t 1 t⎤
= + ⎢ 42 − t 2 + × 42 × sin –1 ⎥
3 ⎣2 2 4 ⎦0
32 ⎡ 4 1 ⎤ 32 ⎡ π ⎤ 32 4
= + ⎢ × 0 + × 42 × sin –1 1⎥ = + ⎢0 + 8 × ⎥ = + 4π = (8 + 3π)
3 ⎣2 2 ⎦ 3 ⎣ 2⎦ 3 3
Example 8 In Fig 8.17, AOBA is the part of the ellipse 9x2 + y2 = 36 in the first
quadrant such that OA = 2 and OB = 6. Find the area between the arc AB and the
chord AB.
x2 y2
Solution Given equation of the ellipse 9x + y = 36 can be expressed as
2 2 + = 1 or
4 36
x2 y2
+ = 1 and hence, its shape is as given in Fig 8.17.
22 62
Accordingly, the equation of the chord AB is
6−0
y–0= ( x − 2)
0−2
or y = – 3(x – 2)
or y = – 3x + 6
Area of the shaded region as shown in the Fig 8.17.
2 2
= 3∫ 4 − x 2 dx − ∫ (6 − 3x )dx (Why?)
0 0
2 2 Fig 8.17
⎡x 2 4 –1 x ⎤ ⎡ 3x 2 ⎤
= 3⎢ 4 − x + sin − ⎢6 x − ⎥
⎣2 2 2 ⎥0 ⎣
⎦ 2 ⎦0
⎡2 ⎤ ⎡ 12 ⎤ π
= 3 ⎢ × 0 + 2sin –1 (1) ⎥ − ⎢12 − ⎥ = 3 × 2 × − 6 = 3π – 6
⎣2 ⎦ ⎣ 2⎦ 2
12. 370 MATHEMATICS
Example 9 Using integration find the area of region bounded by the triangle whose
vertices are (1, 0), (2, 2) and (3, 1).
Solution Let A (1, 0), B (2, 2) and C (3, 1) be
the vertices of a triangle ABC (Fig 8.18).
Area of ΔABC
= Area of ΔABD + Area of trapezium
BDEC – Area of ΔAEC
Now equation of the sides AB, BC and
CA are given by
Fig 8.18
1
y = 2 (x – 1), y = 4 – x, y = (x – 1), respectively.
2
2 3 3 x −1
Hence, area of Δ ABC = ∫ 1 2 ( x − 1) dx + ∫ 2 (4 − x) dx − ∫ 1 2
dx
2 3 3
⎡ x2 ⎤ ⎡ x2 ⎤ 1 ⎡ x2 ⎤
= 2 ⎢ − x⎥ + ⎢4x − ⎥ − ⎢ − x⎥
⎣2 ⎦1 ⎣ 2 ⎦2 2 ⎣ 2 ⎦1
⎡⎛ 2 2 ⎞ ⎛ 1 ⎞ ⎤ ⎡⎛ 32 ⎞ ⎛ 22 ⎞ ⎤ 1 ⎡⎛ 32 ⎞ ⎛ 1 ⎞⎤
= 2 ⎢ ⎜ − 2 ⎟ − ⎜ − 1 ⎟ ⎥ + ⎢⎜ 4 × 3 − ⎟ − ⎜ 4 × 2 − ⎟ ⎥ – ⎢⎜ − 3 ⎟ − ⎜ − 1⎟ ⎥
⎣⎝ 2 ⎠ ⎝ 2 ⎠ ⎦ ⎣⎝ 2⎠ ⎝ 2 ⎠⎦ 2 ⎣⎝ 2 ⎠ ⎝ 2 ⎠⎦
3
=
2
Example 10 Find the area of the region enclosed between the two circles: x2 + y2 = 4
and (x – 2)2 + y2 = 4.
Solution Equations of the given circles are
x2 + y2 = 4 ... (1)
and (x – 2)2 + y2 = 4 ... (2)
Equation (1) is a circle with centre O at the
origin and radius 2. Equation (2) is a circle with
centre C (2, 0) and radius 2. Solving equations
(1) and (2), we have
(x –2)2 + y2 = x2 + y2
or x2 – 4x + 4 + y2 = x2 + y2
or x = 1 which gives y = ± 3
Thus, the points of intersection of the given
circles are A(1, 3 ) and A′(1, – 3 ) as shown in
Fig 8.19
the Fig 8.19.
13. APPLICATION OF INTEGRALS 371
Required area of the enclosed region O ACA′O between circles
= 2 [area of the region ODCAO] (Why?)
= 2 [area of the region ODAO + area of the region DCAD]
= 2 ⎡ ∫ 0 y dx + ∫ 1 y dx ⎤
1 2
⎢
⎣ ⎥
⎦
= 2 ⎡ ∫ 0 4 − ( x − 2) dx + ∫ 1 4 − x dx ⎤
1 2 2 2
⎢ ⎥ (Why?)
⎣ ⎦
1
⎡1 1 ⎛ x − 2 ⎞⎤
= 2 ⎢ ( x − 2) 4 − ( x − 2) 2 + × 4sin –1 ⎜ ⎟
⎣2 2 ⎝ 2 ⎠ ⎥0 ⎦
2
⎡1 1 x⎤
+ 2 ⎢ x 4 − x 2 + × 4sin –1 ⎥
⎣2 2 2 ⎦1
1 2
⎡ ⎛ x − 2 ⎞⎤ ⎡ –1 x ⎤
= ⎢ ( x − 2) 4 − ( x − 2) 2 + 4sin –1 ⎜ 2
⎟ ⎥ + ⎢ x 4 − x + 4sin ⎥
⎣ ⎝ 2 ⎠⎦ 0 ⎣ 2 ⎦1
⎡⎛ –1 ⎛ −1 ⎞ ⎞ −1 ⎤ ⎡ –1 −1 1 ⎤
= ⎢ ⎜ − 3 + 4sin ⎜ ⎟ ⎟ − 4sin ( −1) ⎥ + ⎢ 4sin 1 − 3 − 4sin ⎥
⎣⎝ ⎝ 2 ⎠⎠ ⎦ ⎣ 2⎦
⎡⎛ π⎞ π⎤ ⎡ π π⎤
= ⎢ ⎜ − 3 − 4× ⎟ + 4× ⎥ + ⎢ 4× − 3 − 4× ⎥
⎣⎝ 6⎠ 2⎦ ⎣ 2 6⎦
⎛ 2π ⎞ ⎛ 2π ⎞
= ⎜− 3 − + 2 π ⎟ + ⎜ 2π − 3 − ⎟
⎝ 3 ⎠ ⎝ 3 ⎠
8π
= −2 3
3
EXERCISE 8.2
1. Find the area of the circle 4x2 + 4y2 = 9 which is interior to the parabola x2 = 4y.
2. Find the area bounded by curves (x – 1)2 + y2 = 1 and x2 + y2 = 1.
3. Find the area of the region bounded by the curves y = x2 + 2, y = x, x = 0 and
x = 3.
4. Using integration find the area of region bounded by the triangle whose vertices
are (– 1, 0), (1, 3) and (3, 2).
5. Using integration find the area of the triangular region whose sides have the
equations y = 2x + 1, y = 3x + 1 and x = 4.
14. 372 MATHEMATICS
Choose the correct answer in the following exercises 6 and 7.
6. 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)
7. Area lying between the curves y = 4x and y = 2x is
2
2 1 1 3
(A) (B) (C) (D)
3 3 4 4
Miscellaneous Examples
Example 11 Find the area of the parabola y2 = 4ax bounded by its latus rectum.
Solution From Fig 8.20, the vertex of the parabola Y
L
y2 = 4ax is at origin (0, 0). The equation of the
latus rectum LSL′ is x = a. Also, parabola is
symmetrical about the x-axis.
The required area of the region OLL′O
= 2 (area of the region OLSO) (a, 0)
X′ X
a a O S
= 2∫ 0 ydx = 2 ∫ 0 4ax dx
a
= 2× 2 a ∫0 xdx
a
L'
2⎡ ⎤
3
= 4 a × ⎢x2 ⎥ Y′
3⎢ ⎥
⎣ ⎦
Fig 8.20
0
8 ⎡ 3⎤ 8 2
= a ⎢a 2 ⎥ = a
3 ⎢ ⎥
⎣ ⎦ 3
Example 12 Find the area of the region bounded
by the line y = 3x + 2, the x-axis and the ordinates
x = –1 and x = 1.
Solution As shown in the Fig 8.21, the line
−2
y = 3x + 2 meets x-axis at x = and its graph
3
⎛ −2 ⎞
lies below x-axis for x ∈⎜ −1, ⎟ and above
⎝ 3⎠
⎛ −2 ⎞ Fig 8.21
x-axis for x ∈⎜ ,1⎟ .
⎝ 3 ⎠
15. APPLICATION OF INTEGRALS 373
The required area = Area of the region ACBA + Area of the region ADEA
−2
1
= ∫ 3 (3 x +
−1
2)dx + ∫ −2 (3 x + 2)dx
3
−2
1
⎡ 3x 2 ⎤ 3 ⎡ 3x 2 ⎤ 1 25 13
= ⎢ + 2x⎥ + ⎢ + 2x⎥ = + =
⎣ 2 ⎦ −1 ⎣ 2 ⎦ −2 6 6 3
3
Example 13 Find the area bounded by
the curve y = cos x between x = 0 and
x = 2π.
Solution From the Fig 8.22, the required
area = area of the region OABO + area
of the region BCDB + area of the region
DEFD.
Fig 8.22
Thus, we have the required area
π 3π
2π
∫ 0 cos x dx + ∫ π cos x dx + ∫ 3π cos x dx
2 2
=
2 2
π 3π
2π
= [ sin x ]0 + [ sin x ] + [ sin x ] 3π
2 2
π
2 2
=1+2+1=4
Example 13 Prove that the curves y2 = 4x and x2 = 4y
divide the area of the square bounded by x = 0, x = 4,
y = 4 and y = 0 into three equal parts.
Solution Note that the point of intersection of the
parabolas y2 = 4x and x2 = 4y are (0, 0) and (4, 4) as Fig 8.23
16. 374 MATHEMATICS
shown in the Fig 8.23.
Now, the area of the region OAQBO bounded by curves y2 = 4x and x2 = 4y.
4
4⎛ x2 ⎞ ⎡ 2 2 x3 ⎤
3
= ∫ 0 ⎜ 2 x − ⎟ dx = ⎢ 2 × x − ⎥
⎝ 4⎠ ⎢
⎣ 3 12 ⎥
⎦0
32 16 16
− = = ... (1)
3 3 3
Again, the area of the region OPQAO bounded by the curves x2 = 4y, x = 0, x = 4
and x-axis
x2 1 3 4 16
4
∫ 0 4 dx = 12 ⎡ x ⎤0 = 3
= ⎣ ⎦ ... (2)
Similarly, the area of the region OBQRO bounded by the curve y2 = 4x, y-axis,
y = 0 and y = 4
4 4 y2 1 4 16
= ∫0 xdy = ∫
0 4
dy = ⎡ y 3 ⎤ =
12 ⎣ ⎦0 3 ... (3)
From (1), (2) and (3), it is concluded that the area of the region OAQBO = area of
the region OPQAO = area of the region OBQRO, i.e., area bounded by parabolas
y2 = 4x and x2 = 4y divides the area of the square in three equal parts.
Example 14 Find the area of the region Y
R
{(x, y) : 0 ≤ y ≤ x + 1, 0 ≤ y ≤ x + 1, 0 ≤ x ≤ 2}
2
Q(1, 2)
Solution Let us first sketch the region whose area is to
be found out. This region is the intersection of the x=2
following regions. P (0,1) x = 1
A1 = {(x, y) : 0 ≤ y ≤ x2 + 1}, X' X
O T S
A2 = {(x, y) : 0 ≤ y ≤ x + 1}
Y'
and A3 = {(x, y) : 0 ≤ x ≤ 2}
Fig 8.24
The points of intersection of y = x2 + 1 and y = x + 1 are points P(0, 1) and Q(1, 2).
From the Fig 8.24, the required region is the shaded region OPQRSTO whose area
= area of the region OTQPO + area of the region TSRQT
1 2
∫ 0 (x + 1) dx + ∫ ( x + 1) dx
2
= 1 (Why?)
17. APPLICATION OF INTEGRALS 375
1 2
⎡⎛ x3 ⎞ ⎤ ⎡⎛ x 2 ⎞⎤
= ⎢ ⎜ + x ⎟ ⎥ + ⎢⎜ + x ⎟ ⎥
⎣⎝ 3 ⎠ ⎦ 0 ⎣⎝ 2 ⎠ ⎦1
⎡⎛ 1 ⎞ ⎤ ⎡ ⎛ 1 ⎞⎤ 23
= ⎢ ⎜ + 1 ⎟ − 0 ⎥ + ⎢( 2 + 2 ) − ⎜ + 1 ⎟ ⎥ =
⎣ ⎝3 ⎠ ⎦ ⎣ ⎝ 2 ⎠⎦ 6
Miscellaneous Exercise on Chapter 8
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
2. Find the area between the curves y = x and y = x2.
3. Find the area of the region lying in the first quadrant and bounded by y = 4x2,
x = 0, y = 1 and y = 4.
0
4. Sketch the graph of y = x + 3 and evaluate ∫− 6 x + 3 dx .
5. Find the area bounded by the curve y = sin x between x = 0 and x = 2π.
6. Find the area enclosed between the parabola y2 = 4ax and the line y = mx.
7. Find the area enclosed by the parabola 4y = 3x2 and the line 2y = 3x + 12.
x2 y 2
8. Find the area of the smaller region bounded by the ellipse + = 1 and the
9 4
x y
line + =1 .
3 2
x2 y 2
9. Find the area of the smaller region bounded by the ellipse 2 + 2 = 1 and the
a b
x y
line + = 1 .
a b
10. Find the area of the region enclosed by the parabola x2 = y, the line y = x + 2 and
the x-axis.
11. 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].
18. 376 MATHEMATICS
12. Find the area bounded by curves {(x, y) : y ≥ x2 and y = | x |}.
13. 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).
14. 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
15. Find the area of the region {(x, y) : y2 ≤ 4x, 4x2 + 4y2 ≤ 9}
Choose the correct answer in the following Exercises from 16 to 20.
16. Area bounded by the curve y = x3, the x-axis and the ordinates x = – 2 and x = 1 is
−15 15 17
(A) – 9 (B) (C) (D)
4 4 4
17. The area bounded by the curve y = x | x | , x-axis and the ordinates x = – 1 and
x = 1 is given by
1 2 4
(A) 0 (B) (C) (D)
3 3 3
[Hint : y = x2 if x > 0 and y = – x2 if x < 0].
18. The area of the circle x2 + y2 = 16 exterior to the parabola y2 = 6x is
4 4 4 4
(A) (4π − 3) (B) (4π + 3) (C) (8π − 3) (D) (8π + 3)
3 3 3 3
π
19. The area bounded by the y-axis, y = cos x and y = sin x when 0 ≤ x ≤ is
2
(A) 2 ( 2 − 1) (B) 2 −1 (C) 2 +1 (D) 2
Summary
The area of the region bounded by the curve y = f (x), x-axis and the lines
b b
x = a and x = b (b > a) is given by the formula: Area = ∫ ydx = ∫ f ( x) dx .
a a
The area of the region bounded by the curve x = φ (y), y-axis and the lines
d d
y = c, y = d is given by the formula: Area = ∫ xdy = ∫ φ ( y)dy .
c c
19. APPLICATION OF INTEGRALS 377
The area of the region enclosed between two curves y = f (x), y = g (x) and
the lines x = a, x = b is given by the formula,
[ f ( x) − g ( x)] dx , where, f (x) ≥ g (x) in [a, b]
b
Area = ∫
a
If f (x) ≥ g (x) in [a, c] and f (x) ≤ g (x) in [c, b], a < c < b, then
[ f ( x) − g ( x)] dx + ∫ c [ g ( x) − f ( x)] dx .
c b
Area = ∫
a
Historical Note
The origin of the Integral Calculus goes back to the early period of development
of Mathematics and it is related to the method of exhaustion developed by the
mathematicians of ancient Greece. This method arose in the solution of problems
on calculating areas of plane figures, surface areas and volumes of solid bodies
etc. In this sense, the method of exhaustion can be regarded as an early method
of integration. The greatest development of method of exhaustion in the early
period was obtained in the works of Eudoxus (440 B.C.) and Archimedes
(300 B.C.)
Systematic approach to the theory of Calculus began in the 17th century.
In 1665, Newton began his work on the Calculus described by him as the theory
of fluxions and used his theory in finding the tangent and radius of curvature at
any point on a curve. Newton introduced the basic notion of inverse function
called the anti derivative (indefinite integral) or the inverse method of tangents.
During 1684-86, Leibnitz published an article in the Acta Eruditorum
which he called Calculas summatorius, since it was connected with the summation
of a number of infinitely small areas, whose sum, he indicated by the symbol ‘∫’.
In 1696, he followed a suggestion made by J. Bernoulli and changed this article to
Calculus integrali. This corresponded to Newton’s inverse method of tangents.
Both Newton and Leibnitz adopted quite independent lines of approach which
was radically different. However, respective theories accomplished results that
were practically identical. Leibnitz used the notion of definite integral and what is
quite certain is that he first clearly appreciated tie up between the antiderivative
and the definite integral.
Conclusively, the fundamental concepts and theory of Integral Calculus
and primarily its relationships with Differential Calculus were developed in the
work of P.de Fermat, I. Newton and G. Leibnitz at the end of 17th century.
20. 378 MATHEMATICS
However, this justification by the concept of limit was only developed in the
works of A.L. Cauchy in the early 19th century. Lastly, it is worth mentioning the
following quotation by Lie Sophie’s:
“It may be said that the conceptions of differential quotient and integral which
in their origin certainly go back to Archimedes were introduced in Science by the
investigations of Kepler, Descartes, Cavalieri, Fermat and Wallis .... The discovery
that differentiation and integration are inverse operations belongs to Newton
and Leibnitz”.
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