Golden Integral Calculus By N P Bali

July 10, 2023 bernbart

Golden Integral Calculus by N P Bali: A Comprehensive Guide

If you are looking for a book that covers all the topics of integral calculus in a clear and concise manner, then Golden Integral Calculus by N P Bali is the book for you. This book is written by a renowned author who has more than four decades of experience in teaching mathematics to graduate students. The book meets the requirements of students of degree and honours classes of Indian universities, as well as those preparing for various competitive examinations.

Golden Integral Calculus by N P Bali covers the following topics:

Integration as inverse of differentiation
Integration by substitution
Integration by parts
Integration of rational functions
Integration of irrational functions
Integration of trigonometric functions
Reduction formulae
Definite integrals
Quadrature
Volumes and surfaces of solids of revolution
Differentiation under the integral sign
Improper integrals, beta and gamma functions, theorems of Pappus and Guldin
Double and triple integrals
Centre of gravity
Rectification

The book also provides a list of important formulae, numerous solved examples, exercises with answers, and objective type questions for practice. The book is written with the conviction that a good book needs minimum guidance from the teacher and is self-sufficient for clarity of basic concepts. The book also deals with some additional topics that are not usually found in other books on integral calculus, such as differentiation under the integral sign, improper integrals, double and triple integrals, and application of integration to centre of gravity.

In this article, we will explore some of the topics covered in Golden Integral Calculus by N P Bali in more detail. We will see how to apply the techniques of integration to solve various problems involving areas, lengths, volumes, surfaces, and centres of gravity. We will also learn some useful formulae and theorems that can simplify the calculations and enhance our understanding of integral calculus.

Integration as Inverse of Differentiation

One of the fundamental concepts of integral calculus is that integration is the inverse operation of differentiation. This means that if we have a function f(x) that is differentiable, then we can find another function F(x) such that F'(x) = f(x). The function F(x) is called an antiderivative or an indefinite integral of f(x), and we write F(x) = ∫f(x)dx. The symbol ∫ is called the integral sign, and dx indicates the variable of integration.

For example, if f(x) = x², then we can find an antiderivative F(x) = x³/3 + C, where C is an arbitrary constant. We can check that F'(x) = (x³/3 + C)’ = x² + 0 = f(x), so F(x) is indeed an indefinite integral of f(x). We write F(x) = ∫x²dx = x³/3 + C.

The constant C is necessary because there can be infinitely many antiderivatives of a given function. For example, if we add or subtract any constant to F(x), we get another antiderivative of f(x). This is because the derivative of a constant is zero. Therefore, we usually write the indefinite integral with a general constant C to indicate that there are many possible antiderivatives.

The process of finding an antiderivative or an indefinite integral is called integration. Integration is the opposite of differentiation, and it can be used to undo the effect of differentiation. For example, if we know that the derivative of a function is 2x + 3, then we can find the original function by integrating: ∫(2x + 3)dx = x² + 3x + C.

Integration by Substitution

Sometimes, finding an antiderivative or an indefinite integral of a function can be difficult or complicated. In such cases, we can use a technique called integration by substitution to simplify the problem. Integration by substitution is based on the chain rule of differentiation, which states that if f and g are differentiable functions, then (f ∘ g)'(x) = f'(g(x))g'(x).

Integration by substitution reverses the chain rule and allows us to replace a complicated function with a simpler one. The idea is to find a suitable function g(x) such that f(x) = h(g(x)), where h is another function. Then we can substitute u = g(x) and du = g'(x)dx, and rewrite the integral as ∫f(x)dx = ∫h(g(x))g'(x)dx = ∫h(u)du. The new integral may be easier to evaluate than the original one.

For example, suppose we want to find an antiderivative of f(x) = sin(2x + 1). This function looks complicated because it involves a sine function with an argument that is not just x. However, we can notice that f(x) = h(g(x)), where h(u) = sin(u) and g(x) = 2x + 1. Then we can substitute u = 2x + 1 and du = 2dx, and rewrite the integral as ∫sin(2x + 1)dx = (1/2)∫sin(u)du. The new integral is much simpler to evaluate: (1/2)∫sin(u)du = -(1/2)cos(u) + C = -(1/2)cos(2x + 1) + C.

Therefore, an antiderivative of f(x) = sin(2x + 1) is F(x) = -(1/2)cos(2x + 1) + C.

the partial fractions of f(x), we can integrate each fraction separately and then add them up to get the antiderivative of f(x). For example, ∫(x² + 1)/(x³ – 2x)dx = ∫(A/x + B/(x – √2) + C/(x + √2))dx = A ln|x| + B ln|x – √2| + C ln|x + √2| + D, where A = -1/4, B = 1/(4√2), C = -1/(4√2), and D is an arbitrary constant.

Integration of Irrational Functions

An irrational function is a function that involves a radical or a fractional exponent, such as f(x) = √(x² + 1). Integrating irrational functions can be tricky, but there are some methods that can help us transform them into simpler functions. One of these methods is called trigonometric substitution, which allows us to replace a radical expression with a trigonometric function.

Trigonometric substitution is based on the fact that some trigonometric identities can be used to eliminate radicals. For example, if we have a radical of the form √(a² – x²), where a is a constant, then we can use the identity sin²θ + cos²θ = 1 to write x = a sin θ and dx = a cos θ dθ. Then we can substitute these expressions and rewrite the radical as √(a² – x²) = √(a² – a² sin²θ) = √(a² cos²θ) = a cos θ. The new expression may be easier to integrate than the original one.

For example, suppose we want to find an antiderivative of f(x) = 1/√(4 – x²). This function looks difficult because it involves a radical in the denominator. However, we can notice that the radical has the form √(a² – x²), where a = 2. Then we can use the trigonometric substitution x = 2 sin θ and dx = 2 cos θ dθ, and rewrite the integral as ∫1/√(4 – x²)dx = ∫1/√(4 – 4 sin²θ)2 cos θ dθ = (1/2)∫1/cos θ dθ. The new integral is much simpler to evaluate: (1/2)∫1/cos θ dθ = (1/2)ln|sec θ + tan θ| + C. To get back to the original variable x, we can use the right triangle with hypotenuse 2, opposite side x, and angle θ, and write sec θ = 2/√(4 – x²) and tan θ = x/√(4 – x²). Therefore, an antiderivative of f(x) = 1/√(4 – x²) is F(x) = (1/2)ln|(2 + x)/√(4 – x²)| + C.

Integration of Trigonometric Functions

Trigonometric functions are functions that involve the ratios of the sides of a right triangle, such as sin x, cos x, tan x, etc. Integrating trigonometric functions can be easy or hard, depending on the form of the function. Some basic trigonometric integrals are:

∫sin x dx = -cos x + C
∫cos x dx = sin x + C
∫tan x dx = ln|sec x| + C
∫cot x dx = ln|sin x| + C
∫sec x dx = ln|sec x + tan x| + C
∫csc x dx = -ln|csc x + cot x| + C

However, some trigonometric functions may require some manipulation or transformation before they can be integrated. For example, if we have a product of two trigonometric functions, such as sin x cos x, then we can use a trigonometric identity to rewrite it as a single function, such as (1/2)sin 2x. Then we can integrate it easily: ∫sin x cos x dx = (1/2)∫sin 2x dx = -(1/4)cos 2x + C.

Another example is if we have a power of a trigonometric function, such as sin²x. In this case, we can use another trigonometric identity to rewrite it as a function of double angle, such as (1/2)(1 – cos 2x). Then we can integrate it easily: ∫sin²x dx = (1/2)∫(1 – cos 2x)dx = (1/2)x – (1/4)sin 2x + C.

Reduction Formulae

Sometimes, we may encounter integrals that involve powers or products of trigonometric functions that are not easy to simplify or transform. In such cases, we can use a technique called reduction formulae to reduce the complexity of the integral. Reduction formulae are recursive relations that express an integral of a higher degree or order in terms of an integral of a lower degree or order.

For example, one reduction formula for ∫sinⁿx dx is ∫sinⁿx dx = -(1/n)sinⁿ⁻¹xcos x + (n – 1)/n ∫sinⁿ⁻²x dx. This formula allows us to reduce the power of sin x by one and obtain another integral that is simpler to evaluate. For example, if n = 3, then ∫sin³x dx = -(1/3)sin²xcos x + (2/3)∫sin x dx = -(1/3)sin²xcos x – (2/3)cos x + C.

Another example of a reduction formula is for ∫cosⁿx sin mx dx, where n and m are positive integers. The formula is ∫cosⁿx sin mx dx = -(1/m)cosⁿx sin(m – 1)x + (n/m)∫cosⁿ⁻¹x sin(m – 2)x dx. This formula allows us to reduce the order of the product of cos x and sin x by two and obtain another integral that is simpler to evaluate. For example, if n = 4 and m = 3, then ∫cos⁴x sin³x dx = -(1/3)cos⁴x sin²x + (4/3)∫cos³x sin x dx.

Quadrature

One of the most important applications of integration is to find the area of a region bounded by curves. This process is also called quadrature, because it involves finding a square (or a rectangle) that has the same area as the region. The idea is to divide the region into thin vertical (or horizontal) strips, and approximate the area of each strip by a rectangle whose height (or width) is given by a function. Then we can add up the areas of all the rectangles and take the limit as the width (or height) of the rectangles approaches zero. This gives us the exact area of the region as an integral of the function.

For example, suppose we want to find the area of the region bounded by y = x² and y = 4 – x². We can sketch the region and see that it is symmetric about the y-axis and has endpoints at x = -2 and x = 2. We can divide the region into n equal vertical strips, each with width ∆x = 4/n. The height of each strip is given by the difference between the upper and lower functions: y = 4 – x² – x² = 4 – 2x². Therefore, the area of each strip is approximately Aₖ = yₖ∆x = (4 – 2xₖ²)∆x, where xₖ is any point in the kth strip. The total area of the region is approximately A = A₁ + A₂ + … + Aₙ = ∑ₖ=₁ⁿAₖ = ∑ₖ=₁ⁿ(4 – 2xₖ²)∆x.

To find the exact area, we need to take the limit as n → ∞ and ∆x → 0. This gives us A = limₙ→∞∑ₖ=₁ⁿ(4 – 2xₖ²)∆x = ∫₋₂²(4 – 2x²)dx = [4x – (2/3)x³]₋₂² = (16/3) – (-16/3) = 32/3.

Therefore, the area of the region bounded by y = x² and y = 4 – x² is 32/3.

Volumes and Surfaces of Solids of Revolution

Another application of integration is to find the volume or surface area of a solid obtained by rotating a region around an axis. This process is also called solids of revolution, because it involves finding a solid that has a circular cross-section. The idea is to divide the region into thin slices perpendicular to the axis of rotation, and approximate the volume or surface area of each slice by a disk or a ring whose radius is given by a function. Then we can add up the volumes or surface areas of all the disks or rings and take the limit as the thickness of the slices approaches zero. This gives us the exact volume or surface area of the solid as an integral of a function.

For example, suppose we want to find the volume of the solid obtained by rotating

Conclusion

In this article, we have explored some of the topics covered in Golden Integral Calculus by N P Bali, a comprehensive book for students of mathematics and engineering. We have seen how to perform integration by various methods, such as substitution, parts, partial fractions, and trigonometric substitution. We have also learned how to use reduction formulae to simplify complex integrals involving powers or products of trigonometric functions. Finally, we have applied integration to find the areas, volumes, and surface areas of regions and solids of revolution. We hope that this article has given you a glimpse of the beauty and power of integral calculus, and has motivated you to read the book and learn more.

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