Question

The value of \(\int_{0}^{1} x^{3 / 2}(1-x)^{3 / 2} d x\) is (a) \(\pi=32\) (b) \(-\pi / 32\) \(\begin{array}{ll}\text { (c) } 3 \pi / 128 & (d)-3 \pi / 128\end{array}\)

Short answer

(c) \(3\pi/128\).

Step-by-step solution

Recognize the Integral as a Beta Function

The integral \( \int_{0}^{1} x^{3/2}(1-x)^{3/2} \, dx \) can be recognized as a Beta function. The Beta function is defined as \( B(m+1, n+1) = \int_0^1 x^m (1-x)^n \, dx \). In this case, we identify \( m = 3/2 \) and \( n = 3/2 \). So, the integral corresponds to \( B(5/2, 5/2) \).

Use the Beta and Gamma Function Relation

The Beta function can be related to the Gamma function as follows: \[ B(m, n) = \frac{\Gamma(m) \Gamma(n)}{\Gamma(m+n)} \] Using this, we calculate \( B(5/2, 5/2) = \frac{\Gamma(5/2) \Gamma(5/2)}{\Gamma(5)} \).

Calculate the Gamma Functions

The Gamma function for \( \Gamma(5/2) \) can be calculated using the property \( \Gamma(n+1) = n\Gamma(n) \) and \( \Gamma(1/2) = \sqrt{\pi} \). So, we have:- \( \Gamma(5/2) = \frac{3}{2} \times \Gamma(3/2) \)- \( \Gamma(3/2) = \frac{1}{2} \times \Gamma(1/2) = \frac{1}{2} \times \sqrt{\pi} \)Combining these, \( \Gamma(5/2) = \frac{3}{2} \times \frac{1}{2} \times \sqrt{\pi} = \frac{3\sqrt{\pi}}{4} \).

Substitute Gamma Values and Simplify

Substitute the calculated values into the Beta function relation: \[ B(5/2, 5/2) = \frac{\left(\frac{3\sqrt{\pi}}{4}\right)^2}{\Gamma(5)} \]Calculate \( \Gamma(5) = 4! = 24 \).Substitute and simplify the expression:\[ B(5/2, 5/2) = \frac{9\pi/16}{24} = \frac{9\pi}{384} = \frac{3\pi}{128} \] This corresponds to option (c).

Key concepts

Gamma Function

The Gamma function, denoted as \( \Gamma(n) \), is an extension of the factorial function to complex numbers and real numbers. For non-negative integers, \( n \), it is defined as \( \Gamma(n) = (n-1)! \). It is extensively used in various fields such as probability, statistics, and complex analysis. The function is defined for all complex numbers except the negative integers. One of the key functional equations for the Gamma function is \( \Gamma(n+1) = n\Gamma(n) \), which mirrors the factorial property \( n! = n \times (n-1)! \). This property allows the calculation of gamma values iteratively. For instance, \( \Gamma(1/2) \) is a special case and equals \( \sqrt{\pi} \), which helps derive results where the argument is a half-integer.Understanding these properties simplifies dealing with complex integrals and connects directly to calculating Beta functions.

Integral Calculus

Integral calculus is the branch of mathematics that deals with integrals, or the accumulation of quantities. It includes finding areas under curves, volumes of solids, and solving differential equations. There are two main types of integrals: definite and indefinite.

• Definite integrals compute a number representing the signed area under a curve, between two limits, while
• Indefinite integrals provide a function that describes the accumulation of a quantity over time.

The integration technique used in our exercise involves recognizing a specific form inside the integrand, suitable for substitution into a known function, such as the Beta function. The transformation of the integral into a Beta function enabled the use of properties from special functions to solve it. This showcases how integral calculus intersects with special functions to simplify complex integrals.

Special Functions

Special functions are particular mathematical functions that have established names and importance due to their widespread utility in mathematics and its applications. Examples of special functions include the Gamma function, Beta function, Bessel functions, and others.The Beta function, \( B(m, n) \), is one such special function, defined as an integral from 0 to 1 that involves powers of \( x \) and \( (1-x) \). It is closely related to the Gamma function through the relation \( B(m, n) = \frac{\Gamma(m) \Gamma(n)}{\Gamma(m+n)} \). This relationship is particularly useful for evaluating complex integrals that match the form of a Beta function.In applied mathematics and physics, these functions provide solutions to differential equations, help in approximating series, and simplify complicated expressions. Their extensive properties and inter-relations can save time in problem-solving, when recognized and applied appropriately.