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In mathematics, we usually need to find the derivative of some mathematical functions. It gives the rate of change of one variable with respect to others. Integration is the opposite process of differentiation. The fundamental use of integration is to get back the function whose derivatives are known. So, it is like an antiderivative procedure. Thus, integrals can be computed by viewing an integration as an inverse operation to differentiation. In this article we are going to discuss the concept of integration, basic integration formulas, integration formula of uv,integration formula list as well as some integration formula with examples. Let us learn it!

In Mathematics, when general addition operations cannot be performed, we use integration to add values on a large scale.

There are various different methods in mathematics that can be used to integrate functions.

Integration and differentiation are known to be a pair of inverse functions similar to addition- subtraction, and multiplication-division in Mathematics.

The process of finding functions whose derivative is given is known as integration or anti-differentiation.

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Here’s What Integration is!

Integration is the algebraic method to find the integral for a function at any point on the graph. Finding the integral of some function with respect to some variable x means finding the area to the x-axis from the curve. Therefore, the integral is also known as the anti-derivative because integrating is the reverse process of differentiating.

Not only the integral comes to determine the inverse process of taking the derivative. But it comes in the picture for solving the area problem as well. Similar to the process of differentiation for finding the slope at any point on the graph, this process of integration will be used to find the area of the curve up to any point on the graph.

The integral of the function of x from range a to b will be the sum of the rectangles to the curve at each interval of change in x as the number of rectangles goes to infinity.

We can write the integral of a function f(x) with respect to x :

\[\int\]f(x)dx

Also, integration is often considered as an inverse to the differentiation operation means that if,

\[\frac{d}{dx}\] (f(x)) = g(x)

Then,

\[\int\]g(x)dx = f(x) +C

The extra variable C is known as the constant of integration, which is really necessary. The constant of integration is important because that after all differentiation kills off constants, that is why integration and differentiation are not known as exactly inverse operations of each other.

Since the process of integration is known to be almost the inverse operation of differentiation, the recollection of formulas and processes for differentiation is often possible. So, many differentiation formulae can be used in Mathematics to provide the corresponding formula for the integration process.

Since the integral of a function isn’t definite, therefore it can be referred to as indefinite integral.

The integral of a function at a point can never be found; we always find the integral of a given function in an interval.

The Integral of a function is not unique; that is integrals of a function differ by numbers.

1.Integration by Substitution

2. Integration by Parts

3.Integration by Partial Fraction

4.Integration of Some particular fraction

5.Integration Using Trigonometric Identities

I = f(x).dx =f(g(t).g'(t).dt

The integration formula of uv : ∫f(x).g(x).dx = f(x).∫g(x).dx–∫(f′(x).∫g(x).dx).dx

Which can be further written as integral of the product of any two functions = (First function × Integral of the second function) – Integral of [ (differentiation of the first function) × Integral of the second function]. This is the integration of the uv formula.

∫ dx/ (x2 – a2) is equal to 12 a log | (x – a) / (x + a) | + c

∫ dx/ (a2 – x2) is equal to1 2 a log | (a + x) / (a – x) | + c

∫ dx / (x2 + a2) is equal to1a tan–1 (x/a) + c

∫ dx /√ (x2 – a2) is equal to log| x+√(x2 – a2) | + c

∫ dx /√ (a2 – x2) is equal to sin–1 (xa) + c

∫ dx /√ (x2 + a2) is equal to log | x + √(x2 + a2) | + c

Definite integrals are known as the special kind of integration, where both endpoints are fixed( that is both a and b are given). So, it always represents some bounded region, for computation.

\[\int_{a}^{b}\]f(x)dx= F(b)-F(a)

Standard Integrals in Integration( Basic Integration formulas) -

Q1: Solve the following definite integral.

\[\int_{-2}^{3}\] x^{3} dx

Solution:- \[\int_{-2}^{3}\] x^{3} dx

\[\int_{-2}^{3}\] x3 dx = [\[\frac{x^{4}}{4}\]]\[_{-2}^{3}\]

= \[\frac{81}{4}\] - \[\frac{16}{4}\]

= \[\frac{65}{4}\]

= 16.25

FAQ (Frequently Asked Questions)

Question 1) What are the Three Integration Methods?

Answer)Integration techniques

Integration by parts.

u-substitution.

Reverse chain rule.

Partial fraction expansion.

Integration using trigonometric identities.

Trigonometric substitution.

Question 2)What is Basic Integration?

Answer)Basic integration formulas. The fundamental use of integration can be defined as a continuous version of summing. But, paradoxically, often integrals can be computed by viewing integration as essentially an inverse operation to differentiation. (That fact is known to be the so-called Fundamental Theorem of Calculus.)

Question 3)What are Integration Techniques?

Answer)Integration Techniques. We can derive any integration formulas directly from their corresponding derivative formulas, while other integration problems do require more work. Some of the integration problems that require more work are substitution and change of variables, integration by parts, trigonometric integrals, and trigonometric substitutions.

Question 4)How do You Integrate LN?

Answer)**Strategy:** Use Integration by Parts.

Ln(x) dx. set. u = ln(x), dv = dx. then we find. du = (1/x) dx, v = x.

Then ,substitute. ln(x) dx = u dv.

And use integration by parts. = uv - v du.

Then substitute u=ln(x), v=x, and du=(1/x)dx