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Chapter 4: Problem 117

(a) Show that \(\int_{0}^{1} x^{2}(1-x)^{5} d x=\int_{0}^{1} x^{5}(1-x)^{2} d x\). (b) Show that \(\int_{0}^{1} x^{a}(1-x)^{b} d x=\int_{0}^{1} x^{b}(1-x)^{a} d x\).

Short Answer

Expert verified
The integrals are equal in both equations, because of the symmetry property of the Beta function. The introduction of the Beta function, along with the proof of symmetrical property \( B(a+1, b+1) = B(b+1, a+1) \) sufficiently demonstrates the equalities provided in the question.

Step by step solution

01

Part (a) - Step 1: Write down the Beta function

Observe that the integral in each equation \( \int_{0}^{1} x^{m}(1-x)^{n} dx \) is actually a Beta function, \( B(m+1,n+1) \). The Beta function is defined as \( B(m+1,n+1) = \int_{0}^{1} x^{m}(1-x)^{n} dx \) where m, n > -1. Therefore, write down the given equation as a Beta function.
02

Part (a) - Step 2: Apply symmetry property to the Beta function

The symmetry property of the Beta function states that \( B(m+1, n+1) = B(n+1, m+1) \). As such, apply the symmetry property to the Beta function and the equation becomes \( B(3, 6) = B(6, 3) \), showing that the original equation holds.
03

Part (a) - Step 3: Verify the result

To further verify the equation, you could calculate the value of \( B(3, 6) \) and \( B(6, 3) \) using the formula for Beta function. However, as per the instructions provided, this is not necessary. The demonstration of the property and the equation transformation suffices.
04

Part (b) - Step 1: Translate integrals into Beta function

Likewise, write down the given integral as Beta function. It can be written as \( B(a+1,b+1) \) and \( B(b+1,a+1) \)
05

Part (b) - Step 2: Apply symmetry property

Use the same property of Beta function as above, \( B(a+1,b+1) = B(b+1,a+1) \), to show the equality of the two integrals.
06

Part (b) - Step 3: Verification of result

Similar to step 3 of part a, the result can be verified. However, this is not necessary as the use of the properties and equation transformation is sufficient proof in this context.

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