What is integral calculus and how to calculate it?

Integral is one of the main branches of calculus used to determine the volume under the curve with or without taking the boundary values.  The integral calculus involves two ways one is the definite integral and the other is indefinite integral.

It is the inverse process of differential calculus. The limit calculus is frequently used to define integral calculus. While the derivative is another branch of calculus used to differentiate the functions.

In this article, we will learn the definition, rules, and types of integral in calculus along with solved examples.

What is integral in calculus?

The integral in calculus is a newly formed function whose differential is the actual function or a numerical result of the function to find the area under the curve or graph by using the boundary values.

The integral in calculus is divided into two types. One is the definite integral while the other is the indefinite integral. A definite integral is used to determine the numerical value of the function with the help of upper and lower limits.

While the indefinite integral is used to find the new function by integrating the function with respect to the integrating variable. The indefinite integral does not have any boundary values. Both kinds of integral are widely used in calculus to solve various kinds of problems.

The definite integral uses the fundamental theorem of calculus for substituting the upper and lower limit values. Below is the general formula for the definite integral.

p(s) ds = P(c) – P(b) = M

  • Where c & b are the upper and lower limits of the function respectively.
  • p(s) is the function.
  • “s” is the integrating variable.
  • P(c) – P(b)is the substitution of limits with the help of the fundamental theorem of calculus.
  • After simplification, M is the final result.

The indefinite integral is used to find the new function with respect to the integrating variable. Below is the general expression for the indefinite integral.

ʃ p(s) ds = P(s) + C

  • p(s) is the function.
  • “s” is the integrating variable.
  • P(s) is the result of indefinite integral.
  • C is the constant of integration.

Rules of integral in calculus

Below are some commonly used rules of integral calculus.

Rule nameRules
Difference ruleʃ [p(s) – q(s)] ds =ʃ [p(s)]ds – ʃ [q(s)]ds
constant ruleʃ k ds = k(s) + C
Sum ruleʃ [p(s) + q(s)] ds =ʃ [p(s)]ds + ʃ [q(s)]ds
constant function ruleʃ (k * p(s) ds = k ʃ (p(s) ds
power ruleʃ [(p(s)]n = [p(s)]n+1 / n + 1]
Exponential ruleʃ es ds = es +C
Reciprocal ruleʃ (1/s) ds = ln|s| + C

How to calculate the problems of integral?

The problems of the integral can be solved easily by using its types and rules.

For indefinite integral

Example 1:

Evaluate the indefinite integral of p(s) = 24s3 + 4s – 3s4 + 9sin(s) + 7s + 5 with respect to s.

Solution

Step 1: First of all, take the function p(s) and apply the notation of integration on it.

p(s) = 24s3 + 4s – 3s4 + 9sin(s) + 7s + 5

integrating variable = s

ʃ p(s) ds = ʃ [24s3 + 4s – 3s4 + 9sin(s) + 7s + 5] ds

Step 2: Now apply the notation of integration to each function separately by using the sum and difference rules of integral calculus.

ʃ p(s) ds = ʃ [24s3 + 4s – 3s4 + 9sin(s) + 7s + 5] ds = ʃ [24s3] ds+ ʃ [4s] ds – ʃ [3s4] ds + ʃ [9sin(s)] ds + ʃ [7s] ds + ʃ [5] ds

Step 3: Now write the constant coefficients outside the notation of integration with the help of constant function rule of integral calculus.

= 24 ʃ [s3] ds+ 4 ʃ [s] ds – 3 ʃ [s4] ds + 9 ʃ [sin(s)] ds + 7 ʃ [s] ds + ʃ [5] ds

Step 4: Integrate the above expression by using the trigonometric and power rules of integral.

= 24 [s3+1 / 3 + 1]+ 4 [s1+1 / 1 + 1] – 3 [s4+1 / 4 + 1] + 9 [-cos(s)] + 7 [s1+1 / 1 + 1] + [5s] + C

= 24 [s4 / 4]+ 4 [s2 / 2] – 3 [s5 / 5] + 9 [-cos(s)] + 7 [s2 / 2] + [5s] + C

= 24/4 [s4]+ 4/2 [s2] – 3/5 [s5] + 9 [-cos(s)] + 7/2 [s2] + [5s] + C

= 6 [s4]+ 2 [s2] – 3/5 [s5] + 9 [-cos(s)] + 7/2 [s2] + [5s] + C

= 6s4 + 2s2 – 3 s5/5 – 9cos(s) + 3.5s2 + 5s + C

= 6s4 + 5.5s2 – 3 s5/5 – 9cos(s) + 5s + C

Use an antiderivative calculator to avoid such a large number of steps for finding the integral of the function.

For definite integral

Example 2:

Evaluate the definite integral of p(s) = 7s4 – 5s3 + 6s2 + 2s + 1 with respect to s and the boundary values are (2, 3)

Solution

Step 1: First of all, take the function p(s) and apply the notation of integration on it.

p(s) = 7s4 – 5s3 + 6s2 + 2s + 1

integrating variable = s

 [p(s)] ds = [7s4 – 5s3 + 6s2 + 2s + 1] ds

Step 2: Now apply the notation of integration to each function separately by using the sum and difference rules of integral calculus.

 [p(s)] ds = [7s4 – 5s3 + 6s2 + 2s + 1] ds = [7s4] ds – [5s3] ds + [6s2] ds + [2s] ds + [1] ds

Step 3: Now write the constant coefficients outside the notation of integration with the help of the constant function rule of integral calculus.

 = 7[s4] ds – 5[s3] ds + 6[s2] ds + 2[s] ds + [1] ds

Step 4: Integrate the above expression by using the power rule of integral.

= 7 [s4+1 / 4 + 1]32 – 5 [s3+1 / 3 + 1]32 + 6 [s2+1 / 2 + 1]32 + 2 [s1+1 / 1 + 1]32 + [s]32

= 7 [s5 / 5]32 – 5 [s4 / 4]32 + 6 [s3 / 3]32 + 2 [s2 / 2]32 + [s]32

= 7/5 [s5]32 – 5/4 [s4]32 + 6/3 [s3]32 + 2/2 [s2]32 + [s]32

= 7/5 [s5]32 – 5/4 [s4]32 + 2 [s3]32 + 1 [s2]32 + [s]32

= 7/5 [s5]32 – 5/4 [s4]32 + 2 [s3]32 + [s2]32 + [s]32

Step 5: Now apply the upper and lower limit with the help of the fundamental theorem of calculus.

= 7/5 [35 – 25] – 5/4 [34 – 24] + 2 [33 – 23] + [32 – 22] + [3 – 2]

= 7/5 [243 – 32] – 5/4 [81 – 16] + 2 [27 – 8] + [9 – 4] + [3 – 2]

= 7/5 [211] – 5/4 [65] + 2 [19] + [4] + [1]

= 1477/5 – 325/4 + 38 + 4 + 1

= 295.4 – 81.25 + 38 + 4 + 1

= 257.15

Summary

In this post, we have covered all the basics of integral along with its definition, kinds, rules, and solved examples. You can grab all the basics of integration just by following the above post. Now you are witnessed that this topic is not much difficult.

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