Astro? No! Asteroid!

Calculus Level 2

$\large \frac{ \displaystyle \int_0^1 (1-x^{9995})^{1/1999} \, dx } { \displaystyle \int_0^1 (1-x^{1999})^{1/9995} \, dx } = \ ?$

The answer is 1.

1 solution

Aditya Kumar
Aug 8, 2015

${ I }_{ 1 }=\int _{ 0 }^{ 1 } (1-x^{ 9995 })^{ 1/1999 }\, dx\\ \quad \quad =\int _{ 0 }^{ 1 }{ (1-t)^{ 1/1999 } } .\frac { 1 }{ 9995 } .{ t }^{ \frac { -9994 }{ 9995 } }dt\\ \quad \quad =\frac { \Gamma \left( \frac { 2000 }{ 1999 } \right) \Gamma \left( \frac { 1 }{ 9995 } \right) }{ \Gamma \left( \frac { 10000+1-9995 }{ 9995 } \right) } =1\\ { I }_{ 2 }=\int _{ 0 }^{ 1 } (1-x^{ 1999 })^{ 1/9995 }\, dx\\ \quad \quad =\int _{ 0 }^{ 1 }{ (1-t)^{ 1/9995 } } .\frac { 1 }{ 1999 } .{ t }^{ \frac { -1998 }{ 1999 } }dt\\ \quad \quad =\frac { \Gamma \left( \frac { 1 }{ 1999 } \right) \Gamma \left( \frac { 9996 }{ 9995 } \right) }{ \Gamma \left( \frac { 10000+1-9995 }{ 9995 } \right) } =1\\ \therefore \frac { { I }_{ 1 } }{ { I }_{ 2 } }=1$

Or you can interpret it as the area swept for $x^{9995} + y^{1999} = 1$ . This works too!

Pi Han Goh - 5 years, 10 months ago

Oh wow! I didn't think of it. Thanks!

Aditya Kumar - 5 years, 10 months ago

No. Thank you. It didn't occur to me to use Beta Functions.

Pi Han Goh - 5 years, 10 months ago

@Pi Han Goh – Haha thanks!

Aditya Kumar - 5 years, 10 months ago

Astro? No! Asteroid!

The answer is 1.

1 solution

0 pending reports