In mathematics, many functions are used, a beta function is one of them. Functions are very essential in mathematics for different purposes. Each function contains inputs and required the unique out for each by applying different laws of mathematics.

The first kind of Euler integral is another name of the beta function. But in that integral gamma functions are also used as beta and gamma functions are related to each other. The concept of integration is used in these functions.

## What is Beta Function?

The beta function is that function in mathematics that is unique and is organized by the First kind of Euler integral. It is mostly stated as the domain of real numbers. The beta function is generally denoted by the Greek letter **β. The **beta function** is written as B(a, b), and a, b are the real numbers.**

The beta function is symmetrical. So, B(a, b) must be equal to B(b, a). The connotation among the set of the inputs and the outputs in mathematics is usually explained by a function known as the beta function.

You can also find **factorial of 100 **also by checking above examples.

In the beta function, for each input value, there must be at least one output that is associated with the input. The real values of the beta function must be greater than zero. This function follows the definite integral as it has the upper limit one and the lower limit zero.

### The formula of Beta Function

The formula for this function can be written in the form of.

**B (a, b) = ∫ ^{0}_{1} t^{a-1} (1 – t)^{b-1} dt**

In this equation, ** a **and

**are real numbers and must be greater than zero and**

*b***is the variable whose integral is to be calculated.**

*t*There is another formula known as the factorial form of the beta function and is written as.

**B (a, b) = (a – 1)! (b – 1)! / (a + b – 1)!**

## Solving problems of beta function by using formula

For the calculation of beta function, the knowledge of integral and exponent is very important. Without the knowledge of these concepts, the problems of the beta function are not solved. The integral in beta function is used for definite integral.

And exponents of the functions are calculated by multiplying the base by itself in power times. By using the integral and exponent, the problems of this function can be solved easily.

**Example 1:**

Calculate the beta function, when the real values are a = 7 and b = 6.

**Solution**

**Step 1:** Identify the real values of the given function.

a = 7

b = 6

**Step 2:** Take the general equation of the beta function.

B (a, b) = ∫^{0}_{1} t^{a-1} (1 – t)^{b-1} dt

**Step 3:** Put the values of the real number in the above equation.

B (7, 6) = ∫^{0}_{1} t^{7-1} (1 – t)^{6-1} dt

B (7, 6) = ∫^{0}_{1} t^{6} (1 – t)^{5} dt

**Step 4:** Now open the powers of the function to make it simple for computing.

B (7, 6) = ∫^{0}_{1} t^{6} (-t^{5} + 5t^{4} – 10t^{3} + 10t^{2} – 5t + 1) dt

B (7, 6) = ∫^{0}_{1} (-t^{11} + 5t^{10} – 10t^{9} + 10t^{8} – 5t^{7} + t^{6}) dt

**Step 5:** Now apply the sum and difference rule of integration.

B (7, 6) = ∫^{0}_{1} (-t^{11}) dt + ∫^{0}_{1} (5t^{10}) dt – ∫^{0}_{1} (10t^{9}) dt + ∫^{0}_{1} (10t^{8}) dt – ∫^{0}_{1} (5t^{7}) dt + ∫^{0}_{1} (t^{6}) dt

**Step 6:** Solve the integral.

B (7, 6) = ∫^{0}_{1} (-t^{11}) dt + ∫^{0}_{1} (5t^{10}) dt – ∫^{0}_{1} (10t^{9}) dt + ∫^{0}_{1} (10t^{8}) dt – ∫^{0}_{1} (5t^{7}) dt + ∫^{0}_{1} (t^{6}) dt

B (7, 6) = (-t^{11+1}/11 + 1)^{1}_{0} + (5t^{10+1}/10 + 1)^{1}_{0} – (10t^{9+1}/9 + 1)^{1}_{0} + (10t^{8+1}/8 + 1)^{1}_{0} – (5t^{7+1}/7 + 1)^{1}_{0 }+ (t^{6+1}/6 + 1)^{1}_{0}

B (7, 6) = (-t^{12}/12)^{1}_{0} + (5t^{11}/11)^{1}_{0} – (10t^{10}/10)^{1}_{0} + (10t^{9}/9)^{1}_{0} – (5t^{8}/8)^{1}_{0 }+ (t^{7}/7)^{1}_{0}

B (7, 6) = -1/12 (t^{12})^{1}_{0} + 5/11 (t^{11})^{1}_{0} – 10/10 (t^{10})^{1}_{0} + 10/9 (t^{9})^{1}_{0} – 5/8 (t^{8})^{1}_{0 }+ 1/7 (t^{7})^{1}_{0}

**Step 7:** Now apply the fundamental theorem of calculus for solving the limits of the integral.

∫^{b}_{a} f(t) dt = F(a) – F(b)

B (7, 6) = -1/12 (1^{12} – 0^{12}) + 5/11 (1^{12} – 0^{12}) – 10/10 (1^{12} – 0^{12}) + 10/9 (1^{12} – 0^{12}) – 5/8 (1^{12} – 0^{12}) + 1/7 (1^{12} – 0^{12})

B (7, 6) = -1/12 (1 – 0) + 5/11 (1 – 0) – 10/10 (1 – 0) + 10/9 (1 – 0) – 5/8 (1 – 0) + 1/7 (1 – 0)

B (7, 6) = -1/12 (1) + 5/11 (1) – 10/10 (1) + 10/9 (1) – 5/8 (1) + 1/7 (1)

B (7, 6) = -1/12 + 5/11– 10/10 + 10/9 – 5/8 + 1/7

B (7, 6) = -1/12 + 5/11– 1 + 10/9 – 5/8 + 1/7

**Step 8:** Now solve the fractions.

B (7, 6) = -1/12 + 5/11– 1 + 10/9 – 5/8 + 1/7

B (7, 6) = 0.00018038

These calculations are too large and very time-consuming. To avoid this, you can simply use the beta function calculator which gives you the required output in few seconds.

**Example 2: **

Find the beta function of, B (7, 6) = ∫^{0}_{1} t^{6} (1 – t)^{5} dt.

**Solution**

**Step 1:** Write the given function.

B (7, 6) = ∫^{0}_{1} t^{6} (1 – t)^{5} dt

**Step 2:** Now write the general form of the beta function.

B (a, b) = ∫^{0}_{1} t^{a-1} (1 – t)^{b-1} dt

**Step 3:** Compare the given function with the general function.

a – 1 = 6

a = 6 + 1

a = 7

b – 1 = 5

b = 5 + 1

b = 6

**Step 4:** Now take the factorial form of the beta function.

B (a, b) = (a – 1)! (b – 1)! / (a + b – 1)!

**Step 5:** Put the value of a and b in the above formula.

B (a, b) = (a – 1)! (b – 1)! / (a + b – 1)!

B (7, 6) = (7 – 1)! (6 – 1)! / (7 + 6 – 1)!

B (7, 6) = (6)! (5)! / (13 – 1)!

B (7, 6) = (6)! (5)! / (12)!

B (7, 6) = (720) (120) / (479001600)

B (7, 6) = 86400 / 479001600

B (7, 6) = 0.00018038

## Summary

The beta function is a function used widely in mathematics for the calculation of the Euler integral. You can solve this function by two formulas one is by using exponent and integral and the other is by using the factorial form. Both the formulas are explained above, you can perform any problem by following these two formulas.