In mathematics, the term integral is frequently used to evaluate complex calculus problems. It is one of two main branches of calculus along with differentiation. The limit plays an important role in defining both the branches of calculus such as by taking the derivative by the first principle method and taking the integral of the function applying the upper and lower limit of the function.

**What is integral?**

Integral is the main type of calculus and is used to evaluate the new function whose original function is a differential function and area under the curve by applying the boundary values with the help of the fundamental theorem of calculus.

It is widely used to evaluate the F(x) from f’(x) (evaluate the original function from the differential function). The term F(x) is said to be the integral of the function, antiderivative of a function, Newton-Leibnitz integral, or primitive of a function f(x).

**Formulas of integral **

The formulas for integral are:

Formula with boundary values | Formula without boundary value |

ʃ f(u) du = F(u) + C | ʃ^{b}_{a} f(u) du = [F(u)]^{b}_{a} = F(b) – F(a) |

ʃ = notation of integralf(u) = integral function u = integrating variableC = constant of integration | ʃ = notation of integrala & b = boundary valuesf(u) = integral function u = integrating variableF(b) – F(a) = fundamental theorem of calculus |

**Also Check: **Basic Maths Mcqs

**Types of Integral**

There are two types of integration in calculus:

- Definite integral
- Indefinite integral

Definite integral | Indefinite integral |

It is used to evaluate the area under the curve. | It is used to evaluate the new function whose original function is a differential. |

Boundary values are involved. | Boundary values are not involved. |

The fundamental theorem of calculus is used to apply the boundary values after integrating the function. | Constant integration will be added after integrating the function. |

ʃ^{b}_{a} f(u) du = [F(u)]^{b}_{a} = F(b) – F(a) | ʃ f(u) du = F(u) + C |

An integral calculator by Meracalculator can be a useful tool for integrating functions through indefinite and definite integrals.

**Laws of integration**

Laws name | Expression |

Law of sum | ʃ [f(v) + h(v)] dv = ʃ [f(v)] dv + ʃ [h(v)] dv |

Law of difference | ʃ [f(v) – h(v)] dv = ʃ [f(v)] dv – ʃ [h(v)] dv |

Law of Power | ʃ [f(v)]^{n} dv = f(v)^{n+1} / n + 1 |

Law of constant | ʃ [C] dv = v |

Law of a constant function | ʃ [C * h(v)] dv = C ʃ [h(v)] dv |

Law of reciprocal | ʃ (1/v) dv = ln|v| + C |

Law of exponential | ʃ (e^{v}) dv = e^{v} |

Law of trigonometry | ʃ [sin(v)] dv = -cos(v) ʃ [cos(v)] dv = sin(v) |

**Also Check: Mathematics Mcqs**

**How to calculate the integral?**

Follow the below steps to evaluate the integral.

- First of all, take the given expression and put the notation of definite or indefinite integral to it.
- Use the sum, difference, and constant laws of integration.
- Evaluate the integral with the help of power, constant, and trigonometry laws of integration.
- Use the fundamental theorem of calculus in the case of the definite integral.

**Example 1: For indefinite integral**

Evaluate the indefinite integral of h(v) = 12v^{3 }– 3v^{2} + 4v^{5} + 6v + 12 with respect to v.

**Solution **

**Step 1:** First of all, take the given expression and put the notation of indefinite integral to it.

p(z) = 5z^{4 }+ 2z – 3u^{3} + sin(z) + 1

ʃ p(v) dv = ʃ [12v^{3 }– 3v^{2} + 4v^{5} + 6v + 12] dv

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

ʃ [12v^{3 }– 3v^{2} + 4v^{5} + 6v + 12] dv = ʃ [12v^{3}] dv^{ }– ʃ [3v^{2}] dv + ʃ [4v^{5}] dv + ʃ [6v] dv + ʃ [12] dv

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

ʃ [12v^{3 }– 3v^{2} + 4v^{5} + 6v + 12] dv = 12ʃ [v^{3}] dv^{ }– 3ʃ [v^{2}] dv + 4ʃ [v^{5}] dv + 6 ʃ [v] dv + ʃ [12] dv

**Step 4: **Integrate the above expression.

ʃ [12v^{3 }– 3v^{2} + 4v^{5} + 6v + 12] dv = 12 [v^{3+1} / 3 + 1]^{ }– 3 [v^{2+1} / 2 + 1] + 4 [v^{5+1} / 5 + 1] + 6 [v^{1+1} / 1 + 1] + [12v] + C

ʃ [12v^{3 }– 3v^{2} + 4v^{5} + 6v + 12] dv = 12 [v^{4} / 4]^{ }– 3 [v^{3} / 3] + 4 [v^{6} / 6] + 6 [v^{2} / 2] + [12v] + C

ʃ [12v^{3 }– 3v^{2} + 4v^{5} + 6v + 12] dv = 12/4 [v^{4}]^{ }– 3/3 [v^{3}] + 4/6 [v^{6}] + 6/2 [v^{2}] + [12v] + C

ʃ [12v^{3 }– 3v^{2} + 4v^{5} + 6v + 12] dv = 3 [v^{4}]^{ }– 1 [v^{3}] + 2/3 [v^{6}] + 3 [v^{2}] + [12v] + C

ʃ [12v^{3 }– 3v^{2} + 4v^{5} + 6v + 12] dv = 3 [v^{4}]^{ }– [v^{3}] + 2/3 [v^{6}] + 3 [v^{2}] + [12v] + C

ʃ [12v^{3 }– 3v^{2} + 4v^{5} + 6v + 12] dv = 3v^{4 }– v^{3} + 2v^{6}/3 + 3v^{2} + 12v + C

**Example 2: For indefinite integral**

Evaluate the definite integral of p(w) = 15w^{4} + 5sin(w) + 12w^{2} – 5w^{2} – 10 with respect to w & the [0, 2] is the interval.

**Solution **

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

ʃ^{b}_{a} [p(w)] dw = ʃ^{2}_{0} [15w^{4} + 5sin(w) + 12w^{2} – 5w^{2} – 10] dw

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

ʃ^{2}_{0} [15w^{4} + 5sin(w) + 12w^{2} – 5w^{2} – 10] dw = ʃ^{2}_{0} [15w^{4}] dw + ʃ^{2}_{0} [5sin(w)] dw + ʃ^{2}_{0} [12w^{2}] dw – ʃ^{2}_{0} [5w^{2}] dw – ʃ^{2}_{0} [10] dw

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

ʃ^{2}_{0} [15w^{4} + 5sin(w) + 12w^{2} – 5w^{2} – 10] dw = 15ʃ^{2}_{0} [w^{4}] dw + 5ʃ^{2}_{0} [sin(w)] dw + 12ʃ^{2}_{0} [w^{2}] dw – 5ʃ^{2}_{0} [w^{2}] dw – ʃ^{2}_{0} [10] dw

**Step 3:** Now integrate the above expression.

ʃ^{2}_{0} [15w^{4} + 5sin(w) + 12w^{2} – 5w^{2} – 10] dw = 15 [w^{4+1} / 4 + 1]^{2}_{0} + 5 [-cos(w)]^{2}_{0} + 12 [w^{2+1} / 2 + 1]^{2}_{0} – 5 [w^{2+1} / 2 + 1]^{2}_{0} – [10w]^{2}_{0}

ʃ^{2}_{0} [15w^{4} + 5sin(w) + 12w^{2} – 5w^{2} – 10] dw = 15 [w^{5} / 5]^{2}_{0} + 5 [-cos(w)]^{2}_{0} + 12 [w^{3} / 3]^{2}_{0} – 5 [w^{3} / 3]^{2}_{0} – 10 [w]^{2}_{0}

ʃ^{2}_{0} [15w^{4} + 5sin(w) + 12w^{2} – 5w^{2} – 10] dw = 15/5 [w^{5}]^{2}_{0} + 5 [-cos(w)]^{2}_{0} + 12/3 [w^{3}]^{2}_{0} – 5/3 [w^{3}]^{2}_{0} – 10 [w]^{2}_{0}

ʃ^{2}_{0} [15w^{4} + 5sin(w) + 12w^{2} – 5w^{2} – 10] dw = 3 [w^{5}]^{2}_{0} – 5 [cos(w)]^{2}_{0} + 4 [w^{3}]^{2}_{0} – 5/3 [w^{3}]^{2}_{0} – 10 [w]^{2}_{0}

**Step 4:** Apply the boundary values.

ʃ^{2}_{0} [15w^{4} + 5sin(w) + 12w^{2} – 5w^{2} – 10] dw = 3 [2^{5} – 0^{5}] – 5 [cos(2) – cos(0)] + 4 [2^{3} – 0^{3}] – 5/3 [2^{3} – 0^{3}] – 10 [2 – 0]

ʃ^{2}_{0} [15w^{4} + 5sin(w) + 12w^{2} – 5w^{2} – 10] dw = 3 [32 – 0] – 5 [cos(2) – cos(0)] + 4 [8 – 0] – 5/3 [8 – 0] – 10 [2 – 0]

ʃ^{2}_{0} [15w^{4} + 5sin(w) + 12w^{2} – 5w^{2} – 10] dw = 3 [32] – 5 [cos(2) – 1] + 4 [8] – 5/3 [8] – 10 [2]

ʃ^{2}_{0} [15w^{4} + 5sin(w) + 12w^{2} – 5w^{2} – 10] dw = 69 – 5 [cos(2) – 1] + 32 – 40/3 – 20

ʃ^{2}_{0} [15w^{4} + 5sin(w) + 12w^{2} – 5w^{2} – 10] dw = 69 – 5cos(2) + 5 + 32 – 13.34 – 20

ʃ^{2}_{0} [15w^{4} + 5sin(w) + 12w^{2} – 5w^{2} – 10] dw = 106 – 5cos(2) – 33.34

ʃ^{2}_{0} [15w^{4} + 5sin(w) + 12w^{2} – 5w^{2} – 10] dw = 72.66 – 5cos(2)

**Sum Up**

Integral is a main branch of calculus that is further divided into two categories i.e., definite integral and indefinite integral. Both types are used to evaluate the area under the curve and the new function whose original function is the differential function.