Alphabetagamma!

Calculus Level 3

$\displaystyle\int_{0}^{1} x^{5} (1-x)^{5} \, dx$

If the value of the integral above can be expressed in the form $\dfrac{a}{b}$ where $a$ and $b$ are coprime positive integers, find the value of $a \times b$ .

The answer is 2772.

2 solutions

Chew-Seong Cheong
Jan 2, 2016

$\begin{aligned} I & = \int_0^1 x^5(1-x)^5 dx \\ & = \int_0^1 x^{6-1}(1-x)^{6-1} dx \\ & = \color{#3D99F6}{B (6,6)} \quad \quad \quad \quad \quad \quad \quad \small \color{#3D99F6}{B (m,n) \text{ is Beta function}} \\ & = \color{#3D99F6}{\frac{\Gamma (6)\Gamma (6)}{\Gamma (12)}} \quad \quad \quad \quad \quad \quad \small \color{#3D99F6}{\Gamma (n) \text{ is Gamma function}} \\ & = \frac{5!5!}{11!} \\ & = \frac{1}{2772} \end{aligned}$

$\Rightarrow a \times b = 1 \times 2772 = \boxed{2772}$

Using reduction formula in the third step will also do the job. BTW $\color{#D61F06} {**amazing . solution** !! }\\$

Rishabh Jain - 5 years, 5 months ago

This is incorrect. $\dfrac{5! 5!}{10!} = \dfrac1{252}$ .

You've made a mistake: it should be $B(6,6) = \dfrac{(6-1)!(6-1)!}{(6+6-1)!} = \dfrac{5!5!}{11!} = \dfrac1{2772}$ .

Plus, you don't have to use a trigonometric substitution, one of the many formulas of the Beta Function is $B(x,y) = \int_0^1 t^{x-1} (1-t)^{y-1} \, dt = \dfrac{(x-1)! \cdot (y-1)!}{(x+y-1)!} .$

Pi Han Goh - 5 years, 5 months ago

Thanks for the comments.

Chew-Seong Cheong - 5 years, 5 months ago

Jun Arro Estrella
Jan 2, 2016

Beta functions will do the job :)

Alphabetagamma!

The answer is 2772.

2 solutions

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