**Integral calculus: what is it and how to calculate it?**

Table of Contents

Integral calculus is a widely used concept in mathematics that inverses the process of differential. In mathematics, calculus has two main branches one is integral calculus and the other is differential calculus.

The differential calculus is used to calculate the slope of the tangent line while the integral calculus is used to find the area under the curve. In this post, we will discuss integral calculus along with an explanation and calculation.

**Integral calculus**

A term used to determine the new function or the numerical value of the function whose original function is derivative is known as the integral calculus. The area under the curve can be calculated with the help of this type of calculus.

For determining the numerical result of the function, it takes the upper and lower limit value of the function i.e., the boundary values of the function. The integral is solved with the help of taking the integrating variable of the function.

**Laws of integral calculus **

The basics rules of integral are:

Law name | Laws |

Sum Law | ʃ [m(x) + n(x)] dx = ʃ [m(x)] dx + ʃ [n(x)] dx |

constant Law | ʃ k dx = k * x + C |

constant function Law | ʃ (k * m(x) dx = k ʃ (m(x) dx |

power Law | ʃ [(m(x)]^{n} = [m(x)]^{n+1} / n + 1] |

Exponential Law | ʃ e^{x} dx = e^{x} + C |

Reciprocal Law | ʃ (1/x) dx = ln|x| + C |

Difference Law | ʃ [m(x) – n(x)] dx = ʃ [m(x)] dx – ʃ [n(x)] dx |

**Types of integral calculus explained with examples**

Below are two commonly used types of integral calculus.

- Definite integral
- Indefinite integral

These types of integral calculus can be found quite easily by using an integral calculator that provides steps to make it more understanding.

Let us discuss the types of integral calculus along with examples.

**Definite integral **

In integral calculus, to find the numerical values of the function by taking the boundary values a definite integral is used. First of all, you have to integrate the function with respect to the integrating variable.

After that apply the boundary values of the function with the help of the fundamental theorem of calculus. This theorem states that the upper limit of the function must be applied first to the function then the lower limit with a negative sign among them.

The general formula for the definite integral is:

ab **m(x) dx = M(b) – M(a) = N**

Where, a & b are the lower and the upper limit of the function respectively, m(x) is the integrand function, x is the variable of integration, M(b) – M(a) is the fundamental theorem of the calculus, and N is the final result of the function.

**Example**

Evaluate the definite integral of m(x) = 12x^{3} – 14x^{3} – 10x^{4} + 15cos(x) + 5 with respect to x & the [2, 3] is the interval.

**Solution **

**Step I:** Take the given integrand and apply the integral notation to it.

m(x) = 12x^{3} – 14x^{3} – 10x^{4} + 15cos(x) + 5

23 [m(x)] dx = 23 [12x^{3} – 14x^{3} – 10x^{4} + 15cos(x) + 5] dx

**Step II:** Use the sum and difference laws of integral calculus and apply the integral notation to each function separately.

23 [12x^{3} – 14x^{3} – 10x^{4} + 15cos(x) + 5] dx = 23 [12x^{3}] dx – 23 [14x^{3}] dx – 23 [10x^{4}] dx + 23 [15cos(x)] dx + 23 [5] dx

**Step III:** Now use the constant function law of integral and take the constant coefficients outside the integral notation.

= 1223 [x^{3}] dx – 1423 [x^{3}] dx – 1023 [x^{4}] dx + 1523 [cos(x)] dx + 23 [5] dx

**Step IV: **Use the power and trigonometry law of integral calculus and solve the above expression.

= 12 [x^{3+1} / 3 + 1]^{3}_{2} – 14 [x^{3+1} / 3 + 1]^{3}_{2} – 10 [x^{4+1} / 4 + 1]^{3}_{2} + 15 [sin(x)]^{3}_{2} + [5x]^{3}_{2}

23 [12x^{3} – 14x^{3} – 10x^{4} + 15cos(x) + 5] dx = 12 [x^{4} / 4]^{3}_{2} – 14 [x^{4} / 4]^{3}_{2} – 10 [x^{5} / 5]^{3}_{2} + 15 [sin(x)]^{3}_{2} + [5x]^{3}_{2}

= 12/4 [x^{4}]^{3}_{2} – 14/4 [x^{4}]^{3}_{2} – 10/5 [x^{5}]^{3}_{2} + 15 [sin(x)]^{3}_{2} + 5 [x]^{3}_{2}

= 3 [x^{4}]^{3}_{2} – 7/2 [x^{4}]^{3}_{2} – 2 [x^{5}]^{3}_{2} + 15 [sin(x)]^{3}_{2} + 5 [x]^{3}_{2}

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

= 3 [3^{4} – 2^{4}] – 7/2 [3^{4} – 2^{4}] – 2 [3^{5} – 2^{5}] + 15 [sin(3) – sin(2)] + 5 [3 – 2]

= 3 [81 – 16] – 7/2 [81 – 16] – 2 [243 – 64] + 15 [sin(3) – sin(2)] + 5 [3 – 2]

= 3 [65] – 7/2 [65] – 2 [179] + 15 [sin(3) – sin(2)] + 5 [1]

= 195 – 455/2 – 358 + 15 [sin(3) – sin(2)] + 5

= 195 – 227.5 – 358 + 15 [sin(3) – sin(2)] + 5

= –395.5 + 15sin(3) – 15sin(2)

**Indefinite integral **

In integral calculus, to find the new function whose original function is derivative an indefinite integral is used. It does not have any boundary values as limit values are used to find the numerical result not for finding the new function.

It integrates the function with respect to the integrating variable. The general formula for the indefinite integral is:

**ʃ m(x) dx = M(x) + C**

Where m(x) is the integrand function, x is the variable of integration, M(x) is a new function, and C is the constant of integration.

**Example **

Evaluate the indefinite integral of m(x) = 15x^{4 }– 20x + 6x^{3} – cos(x) + 21 with respect to x.

**Solution **

**Step I:** Take the given integrand and apply the integral notation to it.

m(x) = 15x^{4 }– 20x + 6x^{3} – cos(x) + 21

ʃ m(x) dx = ʃ [15x^{4 }– 20x + 6x^{3} – cos(x) + 21] dx

**Step II:** Use the sum and difference laws of integral calculus and apply the integral notation to each function separately.

ʃ [15x^{4 }– 20x + 6x^{3} – cos(x) + 21] dx = ʃ [15x^{4}] dx^{ }– ʃ [20x] dx + ʃ [6x^{3}] dx – ʃ [cos(x)] dx + ʃ [21] dx

**Step III:** Now use the constant function law of integral and take the constant coefficients outside the integral notation.

= 15ʃ [x^{4}] dx^{ }– 20ʃ [x] dx + 6ʃ [x^{3}] dx – ʃ [cos(x)] dx + ʃ [21] dx

**Step IV: **Use the power and trigonometry law of integral calculus and solve the above expression.

= 15 [x^{4+1} / 4 + 1]^{ }– 20 [x^{1+1} / 1 + 1] + 6 [x^{3+1} / 3 + 1] – [sin(x)] + [21x]

= 15 [x^{5} / 5]^{ }– 20 [x^{2} / 2] + 6 [x^{4} / 4] – [sin(x)] + [21x]

= 15/5 [x^{5}]^{ }– 20/2 [x^{2}] + 6/4 [x^{4}] – [sin(x)] + [21x]

= 3 [x^{5}]^{ }– 10 [x^{2}] + 3/2 [x^{4}] – [sin(x)] + [21x]

= 3x^{5 }– 10x^{2} + 1.5x^{4} – sin(x) + 21x

Use an antiderivative calculator to check the result of the above example.

**Conclusion **

Now the confusion about solving the complex problems of integral calculus must be cleared. As all the basics of integral calculus along with explanations and calculations are covered in this post.