Introduction to definite & indefinite integral explained with examples

In calculus, the definite & indefinite integrals are the basics kinds of integration. Integral is a branch of calculus used frequently to find the area under the curve. It takes the function & respective integrating variables to solve the complex calculus problems.

It is widely used to find the numerical values of the function & a new function (reverse of differentiation). In this post, we will study all the basics of the definite and indefinite integrals along with a lot of examples.

What are the definite and indefinite integrals?

The definite and indefinite integrals are the basic types of integrals used frequently to find the numerical value or a new function with respect to integrating variables. Let us discuss both types of integrals briefly.

Definite integral in calculus

In calculus, the definite integral is a type of integration used to find the numerical value of the function by taking the upper & lower limits according to the fundamental rule of calculus. The interval of initial & final values is used in this type of integral.

Such as, (s, t) is the interval of integration also known as boundaries values. The equation used to denote the definite type of integral is:

∫_s^tp(w) dw = P(t) - P(s) = K

• In the above equation of definite integral, s&t are the initial & final terms of the interval.
• p(w) is the given function.
• dw is the integrating variable of the function.
• P(t) - P(s) is the fundamental theorem of calculus used to find the numerical result of the function.
• K is the numerical value of the function.

Indefinite integral in calculus

The indefinite integral is the other type of integral used to find the new function after integrating the function by using its rules & methods. In this type of integral, the upper & lower limits are not involved.

This type of integration is also known as the antiderivative. The word antiderivative is used because it reverses the process that a derivative does. The equation used to denote the indefinite types of integral is:

∫p(w) dw = P(w) + C

• In the formula of indefinite integral, ∫ is the integration notation.
• p(w) is the given function.
• dw is the integrating variable of the function.
• P(w) is the new function after integrating the function with respect to the integrating variable.
• "C" is the constant of integration.

Rules of integration

Here are some commonly used rules of integration.

How to solve the problems of the definite & indefinite integrals?

The problems of the definite & indefinite integrals can be solved easily either by using the rules of integration or an integral calculator. You can get the solution with steps by using the calculator. Follow the below steps to solve the problems of integrals.

1. Identify the function and integrating variable of the function.
2. Apply the definite or indefinite notation of integral.
3. Apply the rules of the integral.
4. Integrate the functions by using the power, constant, & trigonometry rules of integral.
5. In the case of definite integral, use the fundamental theorem of calculus to find the numerical result.

Example

Evaluate p(w) = 6w⁵ + 2sin(w) - 6w³ + 11w² + 12 with respect to "w" having (2, 3) as an interval,

Solution

Step I: Identify the function, variable, interval, and apply the integration notation.

∫_s^tp(w) dw =∫₂³[6w⁵ + 2sin(w) - 6w³ + 11w² + 12] dw

Step II: Apply the integral notation separately to each function by using the sum & difference rule of integration. And write the constant coefficient outside the integral notation.

∫₂³[6w⁵ + 2sin(w) - 6w³ + 11w² + 12] dw =∫₂³[6w⁵] dw + ∫₂³[2sin(w)] dw - ∫₂³[6w³] dw + ∫₂³[11w²] dw + ∫₂³[12] dw

∫₂³[6w⁵ + 2sin(w) - 6w³ + 11w² + 12] dw = 6∫₂³[w⁵] dw + 2∫₂³[sin(w)] dw - 6∫₂³[w³] dw + 11∫₂³[w²] dw + ∫₂³[12] dw

Step III: Apply the constant, trigonometry, and power rule of integration to integrate into the above expression.

∫₂³[6w⁵ + 2sin(w) - 6w³ + 11w² + 12] dw = 6 [w⁵⁺¹ / 5 + 1]₃² + 2[-cos(w)]₃² - 6[w³⁺¹ / 3 + 1]₃² + 11 [w²⁺¹ / 2 + 1]₃² + [12w]₃²

∫₂³[6w⁵ + 2sin(w) - 6w³ + 11w² + 12] dw = 6 [w⁵⁺¹ / 6]₃² + 2 [-cos(w)]₃² - 6[w³⁺¹ / 4]₃² + 11 [w²⁺¹ / 3]₃² + [12w]₃²

∫₂³[6w⁵ + 2sin(w) - 6w³ + 11w² + 12] dw = 6/6 [w⁵⁺¹]₃² + 2 [-cos(w)]₃² - 6/4 [w³⁺¹]₃² + 11/3 [w²⁺¹]₃² + [12w]₃²

∫₂³[6w⁵ + 2sin(w) - 6w³ + 11w² + 12] dw = [w⁶]₃² + 2 [-cos(w)]₃² - 3/2 [w⁴]₃² + 11/3 [w³]₃² + [12w]₃²

∫₂³[6w⁵ + 2sin(w) - 6w³ + 11w² + 12] dw = [w⁶]₃² - 2 [cos(w)]₃² - 3/2 [w⁴]₃² + 11/3 [w³]₃² + [12w]₃²

Step IV: Use "∫_s^tp(w) dw = P(t) - P(s)" on the above expression.

∫₂³[6w⁵ + 2sin(w) - 6w³ + 11w² + 12] dw = [3⁶ - 2⁶] - 2 [cos(3) - cos(2)] - 3/2 [34 - 24] + 11/3 [33 - 23] + 12[3 - 2]

∫₂³[6w⁵ + 2sin(w) - 6w³ + 11w² + 12] dw = [729 - 64] - 2 [cos (3) - cos (2)] - 3/2 [81 - 16] + 11/3 [27 - 8] + 12 [3 - 2]

∫₂³[6w⁵ + 2sin(w) - 6w³ + 11w² + 12] dw = [665] - 2 [cos (3) - cos (2)] - 3/2 [65] + 11/3 [19] + 12 [1]

∫₂³[6w⁵ + 2sin(w) - 6w³ + 11w² + 12] dw =665 - 2 [cos (3) - cos (2)] - 195/2 + 209/3 + 12

∫₂³[6w⁵ + 2sin(w) - 6w³ + 11w² + 12] dw =665 - 2 [cos (3) - cos (2)] - 97.5 + 69.6667 + 12

∫₂³[6w⁵ + 2sin(w) - 6w³ + 11w² + 12] dw =567.5 - 2 [cos (3) - cos (2)] + 69.6667 + 12

∫₂³[6w⁵ + 2sin(w) - 6w³ + 11w² + 12] dw =649.1667 - 2 [cos (3) - cos (2)]

For indefinite integral

Example

Integrate p(w) = 2w³ - 6w⁵ + 3w + 10 with respect to "w".

Solution

Step I: Identify the function, variable, and apply the integration notation.

p(w) = 2w³ - 6w⁵ + 3w + 10

integrating variable = w

∫ p(w) dw = ∫ [2w³ - 6w⁵ + 3w + 10] dw

Step II: Apply the integral notation separately to each function by using the sum & difference rule of integration. And write the constant coefficient outside the integral notation.

∫ p(w) dw = ∫ [2w³ - 6w⁵ + 3w + 10] dw = ∫ [2w³] dw - ∫ [6w⁵] dw + ∫ [3w] dw + ∫ [10] dw

∫ p(w) dw = ∫ [2w³ - 6w⁵ + 3w + 10] dw = 2 ∫ [w³] dw - 6 ∫ [w⁵] dw + 3 ∫ [w] dw + ∫ [10] dw

Step III: Apply the constant and power rule of integration to integrate into the above expression.

∫ [2w³ - 6w⁵ + 3w + 10] dw = 2[w³⁺¹ / 3 + 1] - 6 [w⁵⁺¹ / 5 + 1] + 3[w¹⁺¹ / 1 + 1] + [10w] + C

∫ [2w³ - 6w⁵ + 3w + 10] dw = 2[w4 / 4] - 6 [w6 / 6] + 3[w2 / 2] + [10w] + C

∫ [2w³ - 6w⁵ + 3w + 10] dw = 2/4[w⁴] - 6/6[w⁶] + 3/2[w²] + [10w] + C

∫ [2w³ - 6w⁵ + 3w + 10] dw = 1/2[w⁴] - [w⁶] + 3/2[w²] + [10w] + C

∫ [2w³ - 6w⁵ + 3w + 10] dw = w⁴/2 - w⁶ + 3w²/2 + 10w + C

Summary

After reading the above post, you can easily solve the problems of integration. In this post, we have discussed all the basics of the definite & indefinite integrals. You can grab all the basics of integration from this post.