# Types of Integral Calculus: Explained with examples

In calculus, the integral is the reverse process of differentiation. The integral and differential are the two main branches of calculus used to find the solution to various calculus terms such as a slope of the tangent line, the area under the curve, etc.

Integrals revert the process that a differential does. Both well-known kinds of calculus can be defined by using the limit calculus. Limit is also a branch of calculus. In this post, we will learn the basics of integration such as its definition, types, and examples.

## What is integral calculus?

In calculus, the integral is a term used to find the numerical value of the function or a new function whose original function is the differential. It is generally used to find the area under the curve or a graph. It uses the limit values of the function.

The limit values are the upper and lower limits also known as the boundary values used to find the numerical value of the function through integration. It integrates the functions by taking the integrating variables.

## Types of integral calculus

Integral calculus is of two types.

• Definite integral
• Indefinite integral

Let us discuss the kinds of integral briefly.

### Definite integral

In calculus, a definite integral is used to find the numerical values of the function by taking the upper and lower limit values of the interval. The upper and lower limit values are to be a substitute in the integral function with the help of the fundamental theorem of calculus.

According to the fundamental theorem, the upper limit must be applied to the function first and then the lower limit value along with the negative sign among them. below is the general expression of the definite integral.

• In the definite integral equation, u & v are the lower and upper limits values respectively
• p(z) is the integrand function.
• “z” is the integrating variable.
• P(v) – P(u) is the applied upper & lower limits with the help of the fundamental theorem of calculus
• After solving the “P(v) – P(u)” the final result will be L.

### Indefinite integral

In calculus, the indefinite integral is used to find the new function by integrating whose original function is differential. The indefinite integral does not have any boundary values. It simply integrates the function with respect to integrating variables.

Below is the general expression of this type of integral.

ʃ p(z) dz = P(z) + C

• p(z) is the integrand function.
• “z” is the integrating variable.
• P(z) is the new function after integrating the given function.
• C is the integral constant.

## Rules of integral calculus

The basics rules of integral are:

## How to solve problems of integral calculus?

The rules and types of integral calculus are used to solve its problems easily. Below are a few solved examples.

### For indefinite integral

Example

Find the definite integral of p(z) = 5z4 + 2z – 3u3 + sin(z) + 1 with respect to z.

Solution

Step 1: First of all, take the given integrand and apply the integration notation to it.

p(z) = 5z4 + 2z – 3u3 + sin(z) + 1

ʃ p(z) dz = ʃ [5z4 + 2z – 3u3 + sin(z) + 1] dz

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

ʃ [5z4 + 2z – 3u3 + sin(z) + 1] dz = ʃ [5z4] dz + ʃ [2z] dz – ʃ [3u3] dz + ʃ [sin(z)] dz + ʃ [1] dz

Step 3: Now take the constant coefficients outside the integral notation by using the constant function rule of integral.

= 5ʃ [z4] dz + 2ʃ [z] dz – 3ʃ [u3] dz + ʃ [sin(z)] dz + ʃ [1] dz

Step 4: Integrate the above expression.

= 5 [z4+1 / 4 + 1] + 2 [z1+1 / 1 + 1] – 3 [u3+1 / 3 + 1] + [-cos(z)] + [z] + C

= 5 [z5 / 5] + 2 [z2 / 2] – 3 [u4 / 4] + [-cos(z)] + [z] + C

= 5/5 [z5] + 2/2 [z2] – 3/4 [u4] + [-cos(z)] + [z] + C

= [z5] + [z2] – 3/4 [u4] + [-cos(z)] + [z] + C

= z5 + z2 – 3u4/4 – cos(z) + z + C

Use an integration calculator to solve the problems of integral calculus with steps in a couple of seconds.

### For definite integral

Example

Find the definite integral of p(z) = 3z3 – 4z3 + 6z4 + 5 with respect to z & the [1, 3] is the interval.

Solution

Step 1: First of all, take the given integrand and apply the integration notation to it.

p(z) = 3z3 – 4z3 + 6z4 + 5

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

Step 3: Now take the constant coefficients outside the integral notation by using the constant function rule of integral.

Step 3: Now integrate the above expression.

= 3 [z3+1 / 3 + 1]31 – 4 [z3+1 / 3 + 1]31 + 6 [z5+1 / 5 + 1]31 + [5z]31

= 3 [z4 / 4]31 – 4 [z4 / 4]31 + 6 [z6 / 6]31 + [5z]31

= 3/4 [z4]31 – 4/4 [z4]31 + 6/6 [z6]31 + 5 [z]31

= 3/4 [z4]31 – [z4]31 + [z6]31 + 5 [z]31

Step 4: Apply the boundary values.

= 3/4 [34 – 14] – [34 – 14] + [36 – 16] + 5 [3 – 1]

= 3/4 [81 – 1] – [81 – 1] + [729 – 1] + 5 [3 – 1]

= 3/4 [80] – [80] + [728] + 5 [2]

= 240/4 – 80 + 728 + 10

= 60 – 80 + 728 + 10

= 718

## Conclusion

In this post, we have covered all the basics of integration such as the definition, types, formulas, and examples. Now you can solve any problem of kinds of integral easily by following the above post.