# What are the indefinite and definite integrals and how to evaluate them?

Definite and indefinite integrals are two types of integral calculus. Integral is one of the main branches of calculus that is used to find the area under the curve. The main purpose of integral calculus is to find a new function or a numerical value of the function.

In this article, we’ll learn the basics of the integral, its types, and how to calculate the advanced problems of definite and indefinite integrals.

## What is the definite and indefinite integral?

Integral calculus is a well-known technique used to evaluate the area under the curve. The integral is a process that is used to find the anti-derivative of the function. In simple words, the process of finding the original function f(y) from the differential function f’(y) is said to be integral.

The integral is divided into two types.

• Definite integral
• Indefinite integral

Both types of integral are very essential for integrating the functions with respect to integrating variables.

### 1.  Definite Integral

The definite integral is a type of integral used to find the numerical value of the function by using the boundary points as upper and lower limit values. The upper and lower limit values are to be applied with the help of a fundamental theorem of calculus.

The fundamental theorem of calculus is a well-known method for finding the numerical value of the function by applying the boundary points.

ba f(x) dx = F(b) – F(a)

### 2.  Indefinite Integral

The other type of integral calculus is indefinite integral. In this type of integral, the original function is to be evaluated whose function is differential. It is widely used to find a new function. The boundary values are not used in this type of integral.

∫ f(x) dx = F(x) + C

## Rules of integral calculus

Here are some well-known rules of integral calculus.

## What is the difference between definite and indefinite integral?

Here are some basics differences between the definite and indefinite integral.

## How to find the problems of definite and indefinite integral?

The rules and formulas of the integral are used to find the definite and indefinite integral of the function. Let us take a few examples of definite and indefinite integrals.

### Example 1: For indefinite integral

Integrate the given function with respect to “z”.

f(z) = 2z5– 3cos(z) – 4z + 12z5 + 20z

Solution

Step 1:First of all, write the given expression according to the general expression of the indefinite integral.

f(z) = 2z5 – 3cos(z) – 4z + 12z5 + 20z

ʃf(z) dz = ʃ [2z5 – 3cos(z) – 4z + 12z5 + 20z] dz

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

ʃ [2z5 – 3cos(z) – 4z + 12z5 + 20z] dz = ʃ [2z5] dz– ʃ [3cos(z)] dz – ʃ [4z] dz + ʃ [12z5] dz + ʃ [20z] dz

Step 3:Now use the constant function rule of integral calculus.

ʃ [2z5 – 3cos(z) – 4z + 12z5 + 20z] dz = 2ʃ [z5] dz– 3ʃ [cos(z)] dz – 4ʃ [z] dz + 12ʃ [z5] dz + 20ʃ [z] dz

Step 4:Now integrate the above expression by applying the power rule and trigonometric formulas.

ʃ [2z5 – 3cos(z) – 4z + 12z5 + 20z] dz = 2[z5+1 / 5 + 1]– 3 [sin(z)] – 4 [z1+1 / 1 + 1] + 12 [z5+1 / 5 + 1] + 20 [z1+1 / 1 + 1] + C

ʃ [2z5 – 3cos(z) – 4z + 12z5 + 20z] dz = 2[z6 / 6]– 3 [sin(z)] – 4 [z2 / 2] + 12 [z6 / 6] + 20 [z2 / 2] + C

ʃ [2z5 – 3cos(z) – 4z + 12z5 + 20z] dz = 2/6[z6]– 3 [sin(z)] – 4/2 [z2] + 12/6 [z6] + 20/2 [z2] + C

ʃ [2z5 – 3cos(z) – 4z + 12z5 + 20z] dz = 1/3[z6]– 3 [sin(z)] – 2 [z2] + 2 [z6] + 10 [z2] + C

ʃ [2z5 – 3cos(z) – 4z + 12z5 + 20z] dz = z6/3– 3sin(z) – 2z2 + 2z6 + 10z2 + C

ʃ [2z5 – 3cos(z) – 4z + 12z5 + 20z] dz = (1/3 + 2)z6 – 3sin(z) + [-2 + 10]z2+ C

ʃ [2z5 – 3cos(z) – 4z + 12z5 + 20z] dz = (7/3)z6 – 3sin(z) + 8z2 + C

You can check the result of the above calculated integral problem with the help of an online integral calculator by MeraCalculator to confirm the accuracy.

### Example 2: For definite integral

Integrate the given function with respect to u& the [1, 3] is the interval.

f(u) = 4u5 +2u – 6u2 + 5u

Solution

Step 1:First of all, write the given expression according to the general expression of the definite integral.

f(u) = 4u5 + 2u – 6u2 + 5u

ʃ32 [f(u)] du = ʃ32[4u5 + 2u – 6u2 + 5u] du

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

ʃ32[4u5 + 2u – 6u2 + 5u] du = ʃ32[4u5] du + ʃ32[2u] du – ʃ32[6u2] du + ʃ32[5u] du

Step 3:Now use the constant function rule of integral calculus.

ʃ32[4u5 + 2u – 6u2 + 5u] du = 4ʃ32[u5] du + 2ʃ32[u] du – 6ʃ32[u2] du + 5ʃ32[u] du

Step 4:Now apply the power rule to integrate the above expression.

ʃ32[4u5 + 2u – 6u2 + 5u] du = 4[u5+1 / 5 + 1]31 + 2 [u1+1 / 1 + 1]31 – 6 [u2+1 / 2 + 1]31 + 5 [u1+1 / 1 + 1]31

ʃ32[4u5 + 2u – 6u2 + 5u] du = 4 [u6/ 6]31 + 2 [u2 / 2]31 – 6 [u3 / 3]31 + 5 [u2 / 2]31

ʃ32[4u5 + 2u – 6u2 + 5u] du = 4/6 [u6]31 + 2/2 [u2]31 – 6/3 [u3]31 + 5/2 [u2]31

ʃ32[4u5 + 2u – 6u2 + 5u] du = 2/3 [u6]31 + 1 [u2]31 – 2 [u3]31 + 5/2 [u2]31

ʃ32[4u5 + 2u – 6u2 + 5u] du = 2/3 [u6]31 + [u2]31 – 2 [u3]31 + 5/2 [u2]31

Step 5:Apply the boundary values.

ʃ32[4u5 + 2u – 6u2 + 5u] du = 2/3 [36 – 16] + [32 – 12] – 2 [33 – 13] + 5/2 [32 – 12]

ʃ32[4u5 + 2u – 6u2 + 5u] du = 2/3 [729 – 1] + [9 – 1] – 2 [27 – 1] + 5/2 [9 – 1]

ʃ32[4u5 + 2u – 6u2 + 5u] du = 2/3  +  – 2  + 5/2 

ʃ32[4u5 + 2u – 6u2 + 5u] du = 1456/3 + 8 – 52 + 40/2

ʃ32[4u5 + 2u – 6u2 + 5u] du = 485.33 + 8 – 52 + 20

ʃ32[4u5 + 2u – 6u2 + 5u] du = 493.33 – 52 + 20

ʃ32[4u5 + 2u – 6u2 + 5u] du = 441.33 + 20

ʃ32[4u5 + 2u – 6u2 + 5u] du = 461.33

## Conclusion

The definite and indefinite integrals are discussed with examples in this post. Now you can grab the basics of these types of integration from this post. They are easy to find and evaluating their problems is not a difficult task.