Integral calculus: Its Types, Formulas, and Solved examples

Integral calculus is the branch of calculus in which we study the integral and its application. In the 17^th century, the first time the term “integration” was used by Isaac Newton and Leibniz. In this article, we will discuss the basic concept of integration, its types, formulas, and methods that how to solve integral in detail.

What is integration?

Mainly the integration is used when we want to calculate the area under a curve. The process of finding the Anti-derivative of the function is known as integration. It is represented as . If f is a differentiable function and f^’ is the derivative of the function f. Then integration of f^’ returns the function f.

For example:

Let f(x) = x³ + 4x² + 3x differentiate f(x) with respect to x is denoted by f^’(x) is 3x² + 8x + 3, if integrate f^’(x) w.r.t x then it is given by:          

∫f(x) d(x) = x³ + 4x² + 3x

Formulas of integration:  

If we want to calculate the integration of any function, first of all, we should remember some basic integral formulas. So basic formulas of the integral are given below

Types of Integral:

There are two types of integral:

1. Definite Integral
2. Indefinite Integral 

Indefinite Integral:

In which types of integral, integral no bounded in limit. The upper or lower limit is not given. It is denoted as ∫f(x) dx= F(x) +C Where f(x) is known as the integrand, dx is the integrating agent and C is any arbitrary constant.

Definite Integral:

If the integral is bounded in limit such that both a lower limit and upper limit are given. Such that ∫^b_a f(x) dx. Where Limit bound in [a, b] and a is the lower limit and b is an upper limit.  If a>b then ∫^b_a f(x) dx = f(a) - f(b) but if a = b then ∫^b_af(x)dx = 0. Remember that in Definite integral we do not use the arbitrary constant.

An integral calculator is a handy tool to evalaute the problems of definite and indefinte integral according to the formulas and rules of integration.

Examples of integration:

In this section, we’ll discuss the method of calculating definite and indefinite integrals.

Example 1:

f(x) = x³ + 4x² + 5x + 9

Solution:

Step 1: Apply the integral on both sides.

∫f(x) dx= (x³ + 4x² + 5x + 9) dx

Step 2: Separate the integrals using the appropriate rule

∫f(x) dx= ∫ (x³) dx + ∫ (4x²) dx + ∫ (5x) dx + ∫ (9) dx

Step 3: Write the constants outside the integral using the constant function rule. 

∫f(x) dx= ∫ (x³) dx + 4∫ (x²) dx + 5∫ (x) dx + ∫ (9) dx

Step 4: Integrate each term by using formulas (in this example we use the power rule of integration i.e.

∫ (xⁿ) dx = xⁿ⁺¹/n+1 and for constant we introduce variable ∫a dx = ax +c where “a” and “c” is constant) 

∫f(x) dx = (x³⁺¹ / 3+1) + 4(x²⁺¹ / 2+1) + (5x¹⁺¹ / 1+1) + 9(x) + C

(Where “C” is any arbitrary constant) 

Step 5: Simplify: 

∫f(x)dx = (x⁴ / 4) + (4x³ / 3) + (5x² / 2) + 9(x) + C

∫ (x³ + 4x² + 5x + 9) dx = (x⁴ / 4) + (4x³ / 3) + (5 x² / 2) + 9(x) + C

Example 2:

f(x) = cos(5x)

Solution:

Step 1: Take integration on both sides concerning x. i.e.

∫ f(x) dx = ∫ (cos(5x)) dx

Step 2: As we know, ∫cos(x) dx = sin(x) and derivative of the angle write in denominator

∫ f(x) dx = [sin(5x) / 5] + C

Example 3:

Evaluate the integral of the function f(x) = x²+4x+5 bounded in the region [1,2]

Solution:

Step 1: Apply the integral on both sides with its limit.

∫²₁ f(x) dx = ∫²₁ (x²  + 4x + 5) dx

Step 2: Separate the integrals using the appropriate rule

∫²₁ f(x) dx = ∫²₁ (x²) dx +  ∫²₁ 4(x) dx +  ∫²₁ 5dx

Step 3: Write the constants outside the integral using the constant function rule. 

∫²₁ f(x) dx= ∫²₁ (x²) dx + 4∫²₁ (x) dx + 5∫²₁ 1dx

Step 4: Integrate each term by using formulas (in this example we use the power rule of integration i.e.

 ∫ (xⁿ) dx = xⁿ⁺¹/n+1 and for constant we introduce variable ∫ a dx = ax where a is constant) 

∫²₁ f(x) dx = [x²⁺¹ / 2+1]₁² +  4 [x¹⁺¹ / 1+1]₁²   +  5[x]₁²

Step 5: Simplify

∫²₁ f(x) dx = [x³ / 3]₁² +  4 [x² / 2]₁²   +  5[x]₁²

Step 6: Use f(x) |_a^b = f(b) – f(a)

∫²₁ f(x) dx = [(2³ / 3)  -  (1³ / 3)] +  4 [(2²  / 2 ) - (1²  / 2 )]  +  5 (2 – 1)

Step 7: Evaluate the expression

∫²₁ f(x) dx = [(8 / 3) - (1 / 3)] + 4 [(4 / 2) - (1  / 2)] + 5 (2 – 1)

∫²₁ f(x) dx = (7 / 3)  +  (12 / 2) + 5 (1)

∫²₁ f(x) dx = (7 / 3)  +  (6)    +  5 (1)

Step 8: By taking L.C.M, we get

∫²₁ f(x) dx = (7 + 18 + 15) / 3

Step 9: Now add the numerator expression

∫²₁ f(x) dx = 40 / 3

∫²₁ (x² + 4x + 5) dx = 40 / 3

Conclusion:

In this article, we have discussed the basic definition of the term “Integral” and covered some basic history. This term was first introduced by the English mathematician Sir Isaac Newton. Further in this article, we have discussed the types of integrals and some basic rules and formulas.

In the example section, we have covered the step-by-step solution of both definite and indefinite integrals in detail. You have witnessed that it is not a difficult topic after going through this article, you’ll able to solve all the problems related to integrals.