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# Integration in calculus: Defined and Explained with examples

Calculus is a branch of mathematics that has further sub-branches such as continuity, differentiation, integration, etc. The sub-branches of calculus are helpful in finding the slope of the tangent line, the nature of the function, numerical value, etc.

Integration and differentiation are the main branches of calculus because it deals with the properties and calculations of derivative and integral. In this article, we will learn the definition, explanation, examples, and solutions of integration in calculus.

## Integration in calculus

In calculus, integration is a process of finding the integral of the function with respect to integrating variables with or without limit values. It is usually used to calculate the area under the curve, a new function, and the numerical value of the function.

There are two further sub-types of integration in calculus.

1. Definite integral
2. Indefinite integral

The indefinite integral is used to calculate the new function whose original function is the derivative of the function. In this sub-type of integral, there are o boundary values are involved. It is also known as the antiderivative of the function as it reverses the process that a derivative does.

The definite integral is used to determine the numerical value of the function with the help of boundary values (limit values) of the function. The upper and lower limit points are involved in this sub-type of integration.

The upper and lower limiting values are substituted for the function after integrating the function with the help of the fundamental theorem of calculus. According to this theorem, the upper limit value is applied first to the integrated function and then the lower limit value with the minus sign among them.

### Formulas of definite and indefinite integrals

The formula for the definite integral is:

p(z) dz = P(u) – P(v) = N

The formula for the indefinite integral is:

ʃ p(z) dz = P(z) + C

## How to calculate the problems of integration?

The following methods are widely used to calculate the problems of integration.

• By using an integral calculator
• Manual method

Here we are going to learn how to calculate the problems of integration either by using an integral calculator or a manual method.

### By using an integral calculator

An integral calculator is a helpful tool that can be used to solve the problems of integration with steps to save the time of performing larger calculations.

How to use this integration calculator?

Here are a few steps to solve the problems of integration by using this calculator.

• Select the type of integral i.e., definite or indefinite integral.
• Input the function into the required input field.
• Select the integrating variable “x” is selected by default.
• Enter the number of upper and lower limits in the case of the definite integral.
• Press the calculate button.
• The step by step solution will come below the calculate button. Integration in calculus

### Manual Method

To calculate the problems of integration manually, follow the below examples.

Example 1: For indefinite integral

Integrate the given function with respect to “z”.

p(z) = 5.5z2 + 12/3z3 + 10cos(z) + 15z + 12

Solution

Step I: First of all, apply the integration notation and integrating variable to the function.

ʃ p(z) dz = ʃ [5.5z2 + 12/3z3 + 10cos(z) + 15z + 12] dz

Step II: Now before calculating the integral of the function, use the sum and difference rules of integration and apply the notation of integral to each function separately.

ʃ [5.5z2 + 12/3z3 + 10cos(z) + 15z + 12] dz = ʃ [5.5z2] dz + ʃ [12/3z3] dz + ʃ [10cos(z)] dz + ʃ [15z] dz + ʃ  dz

Step III: Now take out the constant coefficients outside the integral notation.

ʃ [5.5z2 + 12/3z3 + 10cos(z) + 15z + 12] dz = 5.5ʃ [z2] dz + 12/3ʃ [z3] dz + 10ʃ [cos(z)] dz + 15ʃ [z] dz + ʃ  dz

Step IV: Integrate the above expression with the help of power and trigonometry laws of integration.

ʃ [5.5z2 + 12/3z3 + 10cos(z) + 15z + 12] dz = 5.5 [z2+1 / 2 + 1] + 12/3 [z3+1 / 3 + 1] + 10 [sin(z)] + 15 [z1+1 / 1 + 1] + [12z] + C

ʃ [5.5z2 + 12/3z3 + 10cos(z) + 15z + 12] dz = 5.5 [z3 / 3] + 12/3 [z4 / 4] + 10 [sin(z)] + 15 [z2 / 2] + [12z] + C

ʃ [5.5z2 + 12/3z3 + 10cos(z) + 15z + 12] dz = 5.5/3 [z3] + 12/3 * 4 [z4] + 10 [sin(z)] + 15/2 [z2] + [12z] + C

ʃ [5.5z2 + 12/3z3 + 10cos(z) + 15z + 12] dz = 5.5/3 [z3] + 12/12 [z4] + 10 [sin(z)] + 15/2 [z2] + [12z] + C

ʃ [5.5z2 + 12/3z3 + 10cos(z) + 15z + 12] dz = 1.834 [z3] + 1 [z4] + 10 [sin(z)] + 7.5 [z2] + [12z] + C

ʃ [5.5z2 + 12/3z3 + 10cos(z) + 15z + 12] dz = 1.834z3 + z4 + 10sin(z) + 7.5z2 + 12z + C

Example 2: For definite integral

Integrate the given function with respect to “t”.

p(t) = 2t3 – 21t2 + 12t5 + 3t in the interval of [1, 2].

Solution

Step I: First of all, apply the integration notation and integrating variable to the function.

p(t) dt = [2t3 – 21t2 + 12t5 + 3t] dt

Step II: Now before calculating the integral of the function, use the sum and difference rules of integration and apply the notation of integral to each function separately.

[2t3 – 21t2 + 12t5 + 3t] dt = [2t3] dt – [21t2] dt + [12t5] dt + [3t] dt

Step III: Now take out the constant coefficients outside the integral notation.

[2t3 – 21t2 + 12t5 + 3t] dt = 2[t3] dt – 21[t2] dt + 12[t5] dt + 3[t] dt

Step IV: Integrate the above expression with the help of the power law of integration.

[2t3 – 21t2 + 12t5 + 3t] dt = 2 [t3+1 / 3 + 1]21 – 21 [t2+1 / 2 + 1]21 + 12 [t5+1 / 5 + 1]21 + 3 [t1+1 / 1 + 1]21

[2t3 – 21t2 + 12t5 + 3t] dt = 2 [t4 / 4]21 – 21 [t3 / 3]21 + 12 [t6 / 6]21 + 3 [t2 / 2]21

[2t3 – 21t2 + 12t5 + 3t] dt = 2/4 [t4]21 – 21/3 [t3]21 + 12/6 [t6]21 + 3/2 [t2]21

[2t3 – 21t2 + 12t5 + 3t] dt = 1/2 [t4]21 – 7/1 [t3]21 + 2/1 [t6]21 + 3/2 [t2]21

[2t3 – 21t2 + 12t5 + 3t] dt = 0.5 [t4]21 – 7 [t3]21 + 2 [t6]21 + 1.5 [t2]21

Step V: Apply the boundary values.

[2t3 – 21t2 + 12t5 + 3t] dt = 0.5 [24 – 14]– 7 [23 – 13] + 2 [26 – 16] + 1.5 [22 – 12]

[2t3 – 21t2 + 12t5 + 3t] dt = 0.5 [16 – 1]– 7 [8 – 1] + 2 [64 – 1] + 1.5 [4 – 1]

[2t3 – 21t2 + 12t5 + 3t] dt = 0.5 – 7  + 2  + 1.5 

[2t3 – 21t2 + 12t5 + 3t] dt = 7.5 – 49 + 126 + 4.5

[2t3 – 21t2 + 12t5 + 3t] dt = -41.5 + 126 + 4.5

[2t3 – 21t2 + 12t5 + 3t] dt = 84.5 + 4.5

[2t3 – 21t2 + 12t5 + 3t] dt = 89

## Wrap up

Now you can solve any problem of integration either by using an integral calculator or manually as we have discussed both the methods in the above post. In this article, we have covered almost each basic of the integration along with examples.