# Introduction to Integral: Its Formulas, Types and Calculation

Integral plays a crucial role in mathematics as it helps us to determine valuable quantities, including areas, volumes, and displacement. When discussing integrals, our typical reference is toward the concept of definite integrals. The indefinite integrals help find antiderivatives. The process of calculating the integral is known as integration.

Integration considers one of the main topics in calculus. Let’s discuss integral, its formulas, and the types of the integral. We will solve some examples at the last of this article.

**What is an Integral?**

The integral, represented by the symbol ∫ is a mathematical concept in calculus. It has two parts: the integrand and the differential of the variable. The integrand is the function we want to integrate, and the differential represents the variable with respect to which we perform the integration.

The integral is commonly defined as the opposite process of differentiation. The process of differentiation determines the slope or rate of change of a function, whereas integration recovers the original function from its derivatives.

**Types of Integral**

There are two main types of Integral:

· Definite integral

· Indefinite Integral

**Definite Integral**

The definite integral gives us the area under the curve between the limit of the integration. The definite integral is represented by ∫_{a}^{b} f(x) dx, where ‘a’ and ‘b’ are the lower and upper limits of the interval respectively.

**Indefinite Integral**

The indefinite integral, also known as the antiderivative, is the reverse operation of differentiation. It finds the original function when given its derivative. The indefinite integral has no specific limits and includes a constant of integration (c). The indefinite integral is represented by ∫ f(x) dx.

**Integral formulas**

To evaluate the integral of the function, you should remember the following basic formulas:

**1.** **Power Rule**

∫ x^{n} dx = ( x^{n+1} / (n + 1)) + C (where ‘n’ is any real number except -1)

**2.** **Integral for Constant**

∫ a dx = ax + C (where ‘a’ is any constant)

**3.** **Sum and Difference Rule**

∫ (f(x) ± g(x)) dx = ∫ f(x) dx ± ∫ g(x)) dx

**4.** **Integral formula of Exponential function**

∫ e^{x} dx = e^{x} + C

**5.** **Integral formula for Natural Logarithm**

∫ (1/x) dx = ln|x| + C

∫ a^{x} dx = a^{x}/ln(a) + C

**6.** **Integral formulas for Trigonometric function**

∫ Sin(x) dx = – cos(x) + C

∫ Cos(x) dx = Sin (x) + C

∫ Tan(x) dx = ln|sec (x)| + C = – ln|cos(x)| + C

∫ Sec(x) dx = ln|tan(x) + sec (x)| + C

∫ Csc(x) dx = ln|csc(x) – cot (x)| + C

∫ Cot(x) dx = ln|sin(x)| + C

∫ Sec^{2}(x) dx = tan (x) + C

∫ Csc^{2}(x) dx = – cot (x) + C

∫ Sec(x) tan(x) dx = sec (x) + C

∫ Csc(x) cot(x) dx = – csc (x) + C

**7.** **Integral formulas for inverse Trigonometric function**

∫ (1 / (√1 – x^{2})) dx = arcsin (x) + C

∫ (-1 / (√1 – x^{2})) dx = arcos (x) + C

∫ (1 / (1 + x^{2})) dx = arctan (x) + C

∫ (-1 / (1 + x^{2})) dx = arccot (x) + C

∫ (1 / (x√ x^{2} – 1)) dx = arcsec (x) + C

∫ (-1 / (x√ x^{2} – 1)) dx = arccsc (x) + C

An integration calculator could be used to solve the problems of integral with the help of above formulas and properties.

**Different Methods of Finding the Integration**

There are several methods used to find the integration of function in calculus. Some commonly used techniques are here:

· Integration by Substitution method

· Integration by part

· Finding Integration by Partial fraction

**Integration by Substation method**

Substitution involves replacing variables or expressions in the integrand with new variables or expressions. This technique is useful for simplifying the integrand and making it easier to integrate. It involves choosing an appropriate substitution and applying the chain rule in reverse to find the integral. Some useful trigonometric substitutions are given below:

Expression involving | Suitable Substitution |

√(a^{2} – x^{2}) | x = a sin θ |

√(x^{2} – a^{2}) | x = a sec θ |

√(a^{2} + x^{2}) | x = a tan θ |

√(a ± x) | √(a ± x) = t |

√(2ax – x^{2}) | x – a = a sin θ |

√(2ax + x^{2}) | x + a = a sec θ |

**Integration by part**

We may integrate the product of two functions using integration by part. Consider two unique functions, f(x) and g(x). The order will follow the ILATE (inverse function, logarithm, algebraic expression, Trigonometric function, Exponential function) rule. If f(x) is first and g(x) 2^{nd} function then the formula of integration by part is:

∫ (f(x) * g(x)) dx = f(x) ∫ g(x) dx – ∫ (f’(x) *∫ g(x) dx) dx

**Finding Integration by Partial Fraction**

This method is especially useful for integrating rational functions that cannot be easily integrated directly. By decomposing the rational function into partial fractions, we can integrate each term separately.

**Fundamental Theorem and Properties of Definite Integrals**

If f is continuous on [a, b], then

· ∫_{a}^{b} f(x) dx = f(b) – f(a)

· ∫_{a}^{b} f(x) dx = – ∫_{b}^{a} f(x) dx

· ∫_{a}^{b} f(x) dx = ∫_{a}^{c} f(x) dx + ∫_{c}^{b} f(x) dx

**Solved Examples of Integral**

Let’s solve some examples to gain more knowledge about integral.

**Example 1:**

Evaluate ∫ (x^{4} + 4x^{3} + 9x +4) dx

**Solution**

∫ (x^{4} + 4x^{3} + 9x +4) dx

= ∫ x^{4} dx + ∫ 4x^{3} dx + ∫ 9x dx + ∫ 4 dx

= ∫ x^{4} dx + 4∫ x^{3} dx + 9∫ x dx + 4∫1dx

Applying the power and constant rule of integral

= (x^{4 +1} / 4 + 1) + 4(x^{3+1} / 3 + 1) + 9(x^{1+1} / 1 + 1) + 4x + C

= (x^{4 +1} / 4 + 1) + 4(x^{3+1} / 3 + 1) + 9(x^{1+1} / 1 + 1) + 4x + C

= (x^{5 }/ 5) + 4(x^{4} / 4) + 9(x^{2} / 2) + 4x + C

= (x^{5 }/ 5) + x^{4} + 9(x^{2} / 2) + 4x + C

Hence, ∫ (x^{4} + 4x^{3} + 9x +4) dx = (x^{5 }/ 5) + x^{4} + 9(x^{2} / 2) + 4x + C

**Example 2:**

Evaluate ∫_{0}^{𝜋}^{ }cos (5x) dx

**Solution**

∫_{0}^{𝜋}^{ }cos (5x) dx

∴ ∫ cos (kx) = sin kx / k

= [sin (5x) / 5]_{0}^{𝜋}^{ }= (1/5) [sin (5x)]_{ 0}^{𝜋}^{ }

∴ ∫_{a}^{b} f(x) dx = f (b) – f (a)

= (1/5) [sin (5^{ }𝜋) – sin (5*0)]

= (1/5) [0 – 0]_{ }= 0

Hence, ∫_{0}^{𝜋}^{ }cos (5x) dx = 0

**Conclusion**

Integration is considered one of the major topics in calculus. In this article, we have discussed the integral with its types. We have comprehensively dealt with every essential formula. During our discussion, we deliberated on various methods for obtaining integrals where the formula cannot be directly applied.