Integration is one of the most important branches of mathematics that plays a crucial role in the study of mathematical core concepts especially the core dimensional concepts of calculus. Integration primarily deals with the computation of the integral of a function of areas under the curve, the concept of accumulation, and is the reverse of differentiation.
With its powerful theorems, versatile techniques, and broad applications, integral calculus continues to shape our understanding of the universe. Integral calculus plays a key role in several different fields i.e. physics, engineering, economics, and biology, where quantities (dimensions) are to be computed or measured.
In this article, we will address the core concept of integration. We will explore its definition, a significant theorem that will help to solve integral problems, and important methods that are very useful for the computation of integral problems with some examples.
Definition of integration
At its core, integration involves finding the antiderivative of a function. An antiderivative, also known as an indefinite integral, is the reverse of differentiation.
Let us have a function f(x), the integral of f(x) w.r.t x is represented as ∫ f(x) dx and represents the family of the functions whose derivative is equal to f(x). The integral of a function illustrates the area under the curve of the function. It allows us to determine a function whose derivative matches a given function.
Here it is important to elaborate on the concept of definite & indefinite integrals.
- Indefinite Integrals: Indefinite integrals are expressed using the symbol ∫, which represents the integral. For example, ∫(2x) dx is the indefinite integral of 2x with respect to x. The process of finding the antiderivative involves adding a constant term called the constant of integration (C).
- Definite Integrals: Definite integrals are used to calculate the accumulated quantity over a specific interval. Moreover, a definite integral denotes the area under a curve within a given specific range (interval). It is represented as ∫[a, b] f(x) dx, where “a” and “b” are the interval limits, and f(x) is the function.
The Fundamental Theorem of Calculus:
At the heart of integral calculus lies the Fundamental Theorem of Calculus. This theorem establishes a crucial connection between differentiation and integration. In essence, it states that if you have a continuous function f(x) defined on a closed interval [a, b], its definite integral can be calculated by finding an antiderivative F of the function f(x) and evaluating it at the endpoints of the interval.
Techniques of Integration:
Mastering integral calculus involves various techniques to solve different types of integrals.
- Substitution Method: The substitution method is a powerful technique for solving integrals by substituting a variable with a new one to simplify the integral’s form.
- Integration by Parts: Integration by parts is a method that transforms complicated integrals into simpler ones by using the product rule for differentiation.
∫ u.v dx = u ∫ v dx – ∫ (∫ v dx). u’ dx
- Partial Fractions: Partial fraction decomposition breaks down a complex rational function into simpler fractions, making integration more manageable.
- Trigonometric integrals: Trigonometric integrals deal with integrals involving trigonometric functions.
- Improper integrals: Improper integrals handle functions with infinite bounds.
- Solving Differential Equations with Integrals: Integral calculus is integral in solving first-order differential equations by finding an integrating factor. Second-order differential equations can be solved using integration techniques, offering solutions for various physical phenomena.
How to integrate a function?
Integrating a function is a fundamental concept in calculus, representing the reverse process of differentiation. The integral of a function computes the area under its curve. Below are a few solved examples of integrating functions.
Compute the integral of the function 3t4 + 2t3 + 4t – 9.
Step 1: Given data
∫(3t4 + 2t3 + 4t – 9) dt
Step 2: Applying integral rules and simplifying.
∫(3t4 + 2t3 + 4t – 9) dt = ∫(3t4)dt + ∫(2t3)dt + ∫(4t) dt – ∫(9) dt (Apply the sum & difference rule)
∫(3t4 + 2t3 + 4t – 9) dt = 3∫t4 dt + 2∫t3 dt + 4∫t dt – 9∫ dt (Apply constant multiple rule)
∫(3t4 + 2t3 + 4t – 9) dt = 3 (1/4+1) t4+1 + 2 (1/3+1) t3+1 + 4 (1/1+1) t1+1 – 9t + c (Power & constant rule)
∫(3t4 + 2t3 + 4t – 9) dt = (3/5) t4 + (1/2) t4 + 2t2 – 9t + c Ans.
Evaluate the integral by parts for the functions u(x) = x2 and v(x) = sin x.
Step 1: Given defined functions
u = x2 and v = sin x
Step 2: Let
I = ∫ x2 sin x dx
Step 3: Integrating by parts and simplifying.
I = x2 (-cos x) – ∫ (-cos x) . 2x dx
I = – x2 cos x + 2 ∫ x cos x dx
Again integrating parts
I = – x2 cos x + 2 (x sin x – ∫ sin x (1) dx)
I = – x2 cos x + 2x sin x – 2 (-cos x) + c
I = – x2 cos x + 2x sin x + 2 cos x + c
You can use online tools like the integral calculator by MeraCalculator to integrate the given function to get rid of lengthy calculations.
Evaluate the integral for the given function [(x + b) / (x2 + 2bx + c)1/2]
Step 1: Given data
Given function = [(x + b) / (x2 + 2bx + c)1/2]
Step 2: Let
I = ∫ [(x + b) / (x2 + 2bx + c)1/2]
Step 3: Simplifying
Put t = x2 + 2bx + c
=> dt = (2x + 2b) dx
=> dt = 2(x + b) dx
=> ½ dt = (x +b) dx
I = ∫ (½ dt / t1/2) = 1/2 ∫ t-1/2 dt
I = 1/2 . (t1/2 + 1) / (- 1/2 + 1) + c1
I = ½ . (t1/2/ ½ ) + c1
I = x2 + 2bx + c)1/2 + c1
I = √(x2 + 2bx + c) + c1 Ans.
We can summarize that integral calculus is a captivating mathematical discipline that enables us to understand accumulation, area, and a myriad of real-world phenomena. In this article, we have addressed the concept of integration. We have described its definition, briefed definite, and indefinite integrals as well as techniques to tackle the integral problems.
In the last section, we solved some examples in order to comprehend the concept of integration. Hopefully, apprehending this article you will be able to address the problems related to finding the integral of a function.