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# Introduction to rules of integral in calculus explained with examples

In calculus, the integral is used frequently to find the new function or numerical result of that function. It integrates the function with or without the upper and lower limits of the interval. The upper and lower limits are known as boundary values of the function.

Limit is a kind of calculus used to define integrals & derivatives. In this article, we will study the definition and rules of integral along with a lot of examples.

## What is the integral in calculus?

In calculus, integral gives numbers to functions in a way that describes volume, displacement, area, & other concepts that arise by combining infinitesimal data. Integration is the process of finding integrals.

The integral calculus can be definite or indefinite. If the new function is to be calculated, then an indefinite integral is used. While if the numerical value of the function is to be calculated, then the definite integral is used.

The limits of the definite integral can be substituted by using a corollary of the fundamental theorem of calculus. Here are general equations used to denote the definite and indefinite integrals in calculus.

The general equation of indefinite integral is:

**ʃ g(t) dt = G(t) + C**

**ʃ**it is the notation of indefinite integral.- g(t) is the function to be integrated.
- dt is the integral variable.
- G(t) is the new function after integrating the given function.
- C is the integral’s constant.

The general equation of definite integral is:

**g(t) dt = G(d) – G(c) = K**

- In
**c**and**d**are the interval values and it is known as the notation of definite integral. **g(t)**is the function to be integrated.**dt**is the integral variable.**G(d) – G(c)**is the fundamental rule of calculus to find the numerical result.**K**is the final result of the definite integral.

An antiderivative calculator can be used to integrate the given function either by definite integral or indefinite integral with steps.

## Rules of integral in calculus

There are various rules of integration used to find the result of various complex problems. Let us discuss them briefly.

### Power rule of integral

When the function is given with an exponent then the power rule is used. According to this rule of integral, the power of exponential function must be increased by one and write the increased power in the denominator of the function.

ʃ g ^{n}(t) dt = g^{n+1}(t)/ n + 1

**Example: For indefinite integral**

Integrate t^{5} with respect to t.

**Solution **

**Step 1:** Identify the function, variable and apply the notation of integral.

g(t) = t^{5}

variable = t

ʃ g ^{n}(t) dt = ʃ t^{5} dt

**Step 2:** Use the power rule and integrate the above expression.

ʃ t^{5} dt = t^{5+1} / 5 + 1 + C

ʃ t^{5} dt = t^{6} / 6 + C

**Example: For definite integral**

Integrate 5t^{4} with respect to t at (2, 5).

**Solution **

**Step 1:** Identify the function, variable, boundary values, and apply the notation of integral.

g(t) = 5t^{4}

variable = t

upper limit = d = 5

lower limit = c = 2

g(t) dt = 5t^{4 }dt

**Step 2:** Use the power rule and integrate the above expression.

5t^{4 }dt = [5t^{4+1} / 4 + 1]^{5}_{2}

5t^{4 }dt = [5t^{5} / 5]^{5}_{2}

5t^{4 }dt = [t^{5}]^{5}_{2}

**Step 3:** Use the fundamental theorem of calculus to substitute the boundary values.

5t^{4 }dt = [5^{5} – 2^{5}]

5t^{4 }dt = [3125 – 32]

5t^{4 }dt = 3093

The above example can also be solved by using an integral calculator to ease up the calculations.

### Sum rule of integral

When the functions are given with a plus sign among them than the sum rule is used. According to this rule of integral, the notation of definite and indefinite integral must be applied to each function separately.

- ʃ [g(t) + h(t)] dt = ʃ [g(t)] dt + ʃ [h(t)] dt
- ʃ [g(t) + h(t) + j(t)] dt = ʃ [g(t)] dt + ʃ [h(t)] dt + ʃ [j(t)] dt

**Example: For indefinite integral**

Integrate 2t^{5} + 3t^{2} with respect to t.

**Solution **

**Step 1:** Identify the function, variable and apply the notation of integral.

Function = 2t^{5} + 3t^{2}

variable = t

ʃ [g(t) + h(t)] dt = ʃ [2t^{5} + 3t^{2}] dt

**Step 2:** Use the sum rule and write the notation of integral separately.

ʃ [2t^{5} + 3t^{2}] dt = ʃ [2t^{5}] dt + ʃ [3t^{2}] dt

**Step 3:** Use the power rule and integrate the above expression.

ʃ [2t^{5} + 3t^{2}] dt = [2 t^{5+1} / 5 + 1] + [3 t^{2+1} / 2 + 1] + C

ʃ [2t^{5} + 3t^{2}] dt = [2 t^{6 }/ 6] + [3 t^{3} / 3] + C

ʃ [2t^{5} + 3t^{2}] dt = t^{6 }/ 3 + t^{3} + C

**Example: For definite integral**

Integrate 5t + 3t^{2} with respect to t at (1, 2).

**Solution **

**Step 1:** Identify the function, variable, boundary values, and apply the notation of integral.

function = 5t + 3t^{2}

variable = t

upper limit = d = 2

lower limit = c = 1

g(t) dt = [5t + 3t^{2}] dt

**Step 2:** Use the sum rule and write the notation of integral separately.

[5t + 3t^{2}] dt = [5t] dt + [3t^{2}] dt

**Step 3:** Use the power rule and integrate the above expression.

[5t + 3t^{2}] dt = [5 t^{1+1} / 1 + 1]^{2}_{1 }+ [3 t^{2+1} / 2 + 1]^{2}_{1}

[5t + 3t^{2}] dt = [5 t^{2} / 2]^{2}_{1 }+ [3 t^{3} / 3]^{2}_{1}

[5t + 3t^{2}] dt = [5 t^{2} / 2]^{2}_{1 }+ [t^{3}]^{2}_{1}

[5t + 3t^{2}] dt = 5/2[t^{2}]^{2}_{1 }+ [t^{3}]^{2}_{1}

**Step 3:** Use the fundamental theorem of calculus to substitute the boundary values.

[5t + 3t^{2}] dt = 5/2[2^{2} – 1^{2}] + [2^{3} – 1^{3}]

[5t + 3t^{2}] dt = 5/2[4 – 1] + [8 – 1]

[5t + 3t^{2}] dt = 5/2[3] + [7]

[5t + 3t^{2}] dt = 15/2 + 7

[5t + 3t^{2}] dt = 7.5 + 7

[5t + 3t^{2}] dt = 14.5

### Difference rule of integral

When the functions are given with a minus sign among them than the difference rule is used. According to this rule of integral, the notation of definite and indefinite integral must be applied to each function separately.

- ʃ [g(t) – h(t)] dt = ʃ [g(t)] dt – ʃ [h(t)] dt
- ʃ [g(t) – h(t) – j(t)] dt = ʃ [g(t)] dt – ʃ [h(t)] dt – ʃ [j(t)] dt

**Example: For indefinite integral**

Integrate 3t^{2} – 4t^{3} with respect to t.

**Solution **

**Step 1:** Identify the function, variable and apply the notation of integral.

Function = 3t^{2} – 4t^{3}

variable = t

ʃ [g(t) – h(t)] dt = ʃ [3t^{2} – 4t^{3}] dt

**Step 2:** Use the difference rule and write the notation of integral separately.

ʃ [3t^{2} – 4t^{3}] dt = ʃ [3t^{2}] dt – ʃ [4t^{3}] dt

**Step 3:** Use the power rule and integrate the above expression.

ʃ [3t^{2} – 4t^{3}] dt = [3 t^{2+1} / 2 + 1] – [4 t^{3+1} / 3 + 1] + C

ʃ [3t^{2} – 4t^{3}] dt = [3 t^{3} / 3] – [4 t^{4} / 4] + C

ʃ [3t^{2} – 4t^{3}] dt = t^{3} – t^{4} + C

**Example: For definite integral**

Integrate 5 – t^{2} with respect to t at (1, 3).

**Solution **

**Step 1:** Identify the function, variable, boundary values, and apply the notation of integral.

function = 5 – t^{2}

variable = t

upper limit = d = 3

lower limit = c = 1

g(t) dt = [5 – t^{2}] dt

**Step 2:** Use the difference rule and write the notation of integral separately.

[5 – t^{2}] dt = [5] dt – [t^{2}] dt

**Step 3:** Use the power rule and integrate the above expression.

[5 – t^{2}] dt = [5t]^{3}_{1} – [t^{2+1} / 2 + 1]^{3}_{1 }

[5 – t^{2}] dt = [5t]^{3}_{1} – [t^{3} / 3]^{3}_{1 }

[5 – t^{2}] dt = 5[t]^{3}_{1} – 1/3[t^{3}]^{3}_{1 }

**Step 3:** Use the fundamental theorem of calculus to substitute the boundary values.

[5 – t^{2}] dt = 5[3 – 1] – 1/3[3^{3} – 1^{3}]

[5 – t^{2}]dt = 5[2] – 1/3[27 – 1]

[5 – t^{2}]dt = 5[2] – 1/3[26]

[5 – t^{2}]dt = 10 – 26/3

[5 – t^{2}] dt = 10 – 8.6667

[5 – t^{2}] dt = 1.334

## Summary

Now you can grab all the basics of integral from this post. After reading the above post, you can solve any problem of definite or indefinite integral easily. You can learn the rules of integral to make the calculations easier.