A calculus problem by A Former Brilliant Member

Calculus Level 4

5050 × ∫ 0 1 ( 1 − x 50 ) 100 d x ∫ 0 1 ( 1 − x 50 ) 101 d x = ? \large 5050 \times \dfrac{\displaystyle \int_0^1 (1-x^{50})^{100} \, dx }{\displaystyle \int_0^1 (1-x^{50})^{101} \, dx } = \, ?


The answer is 5051.

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A Former Brilliant Member by A Former Brilliant Member

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2 solutions

Chew-Seong Cheong
Aug 16, 2016

I 0 = ∫ 0 1 ( 1 − x 50 ) 100 d x Let u = x 50 ⟹ d x = u − 49 50 50 d u = ∫ 0 1 u − 49 50 ( 1 − u ) 100 50 d u = B ( 1 − 49 50 , 1 + 100 ) 50 B ( m , n ) is beta function = B ( 1 50 , 101 ) 50 = Γ ( 1 50 ) Γ ( 101 ) 50 Γ ( 5051 50 ) Γ ( x ) is gamma function \begin{aligned} I_0 & = \int_0^1 \left(1-\color{#3D99F6}{x^{50}} \right)^{100} dx & \small \color{#3D99F6}{\text{Let }u=x^{50} \implies dx = \frac {u^{-\frac{49}{50}}}{50}du} \\ & = \int_0^1 \frac {u^{-\frac{49}{50}} (1-u)^{100}}{50} du \\ & = \frac {\color{#3D99F6}{B \left(1-\frac {49}{50}, 1+100 \right)}}{50} & \small \color{#3D99F6}{B(m,n) \text{ is beta function}} \\ & = \frac {B \left(\frac 1{50}, 101 \right)}{50} \\ & = \frac {\Gamma \left(\frac 1{50} \right) \Gamma \left(101 \right)}{50 \Gamma \left(\frac {5051}{50} \right)} & \small \color{#3D99F6}{\Gamma (x) \text{ is gamma function}} \end{aligned}

Similarly, I 1 = ∫ 0 1 ( 1 − x 50 ) 101 d x = Γ ( 1 50 ) Γ ( 102 ) 50 Γ ( 5101 50 ) \begin{aligned} I_1 & = \int_0^1 \left(1-x^{50} \right)^{101} dx = \frac {\Gamma \left(\frac 1{50} \right) \Gamma \left(102 \right)}{50 \Gamma \left(\frac {5101}{50} \right)} \end{aligned}

⟹ 5050 × I 0 I 1 = 5050 ⋅ Γ ( 1 50 ) Γ ( 101 ) 50 Γ ( 5051 50 ) ⋅ 50 Γ ( 5101 50 ) Γ ( 1 50 ) Γ ( 102 ) = 5050 ⋅ Γ ( 101 ) Γ ( 5101 50 ) Γ ( 102 ) Γ ( 5051 50 ) = 5050 ⋅ 100 ! ⋅ 5051 ! ! 5 0 2526 ⋅ Γ ( 1 50 ) 101 ! ⋅ 5049 ! ! 5 0 2525 ⋅ Γ ( 1 50 ) = 5050 ⋅ 5051 101 ⋅ 50 = 5051 \begin{aligned} \implies 5050 \times \frac {I_0}{I_1} & = 5050 \cdot \frac {\Gamma \left(\frac 1{50} \right) \Gamma \left(101 \right)}{50 \Gamma \left(\frac {5051}{50} \right)} \cdot \frac {50 \Gamma \left(\frac {5101}{50} \right)}{\Gamma \left(\frac 1{50} \right) \Gamma \left(102 \right)} \\ & = 5050 \cdot \frac {\Gamma \left(101 \right) \Gamma \left(\frac {5101}{50} \right)}{\Gamma \left(102 \right) \Gamma \left(\frac {5051}{50} \right)} \\ & = 5050 \cdot \frac {100! \cdot \frac {5051!!}{50^{2526}} \cdot \Gamma \left(\frac 1{50} \right)}{101! \cdot \frac {5049!!}{50^{2525}} \cdot \Gamma \left(\frac 1{50} \right)} \\ & = 5050 \cdot \frac {5051}{101 \cdot 50} \\ & = \boxed{5051} \end{aligned}

References:

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If the lower integral is I 2 = ∫ 0 1 ( 1 − x 50 ) 101 d x \displaystyle I_2=\int_{0}^{1} (1-x^{50})^{101} dx & upper integral I 1 = ∫ 0 1 ( 1 − x 50 ) 100 d x \displaystyle I_1=\int_{0}^{1} (1-x^{50})^{100} dx

then we want to evaluate S = 5050 I 1 I 2 \displaystyle S=\frac{5050I_1}{I_2}

Apply IBP on I 2 I_2 to get, I 2 = x ( 1 − x 50 ) ] 0 1 + 5050 ∫ 0 1 x 50 ( 1 − x 50 ) 100 = 0 + 5050 ∫ 0 1 ( 1 − ( 1 − x 50 ) ) ( 1 − x 50 ) 100 d x \displaystyle I_2 = x(1-x^{50})]_{0}^{1}+5050\int_{0}^{1} x^{50}(1-x^{50})^{100} = 0+5050\int_{0}^{1}(1-(1-x^{50}))(1-x^{50})^{100} dx

So, I 2 = 5050 ( ∫ 0 1 ( 1 − x 50 ) 1 00 ) d x − 5050 ∫ 0 1 ( 1 − x 50 ) 101 d x ⟹ I 2 = 5050 ( I 1 − I 2 ) \displaystyle I_2=5050(\int_{0}^{1} (1-x^{50})^100)dx - 5050\int_{0}^{1} (1-x^{50})^{101} dx \implies I_2=5050(I_1-I_2)

Thus , 5050 I 1 I 2 = 5051 \displaystyle \frac{5050I_1}{I_2}=5051

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