A useful approximation!

Calculus Level 5

Find the approximate value of:

1 e e x 1 x d x \Large \int_{\frac{1}{e}}^{e}x^{\frac{1}{x}}dx

Inspiration


The answer is 2.65861.

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1 solution

Aditya Kumar
Aug 18, 2015

a p p r o x i m a t i o n o f 1 e e x 1 x d x 1 e e x 1 x d x = 1 e e e l o g x x d x = I S i n c e w e w a n t t o a p p r o x i m a t e t h e v a l u e o f I , w e c a n u s e t a y l o r s e r i e s o f e x p a n s i o n o f e x I = 1 e e ( 1 + l o g x x 1 ! + ( l o g x x ) 2 2 ! + ( l o g x x ) 3 3 ! + . . . ) d x o n s p l i t t i n g t h e i n t e g r a l s a n d s o l v i n g t h e m , w e g e t I 2.65861 approximation\quad of\quad \int _{ \frac { 1 }{ e } }^{ e }{ { x }^{ \frac { 1 }{ x } } } dx\\ \int _{ \frac { 1 }{ e } }^{ e }{ { x }^{ \frac { 1 }{ x } } } dx\quad =\quad \int _{ \frac { 1 }{ e } }^{ e }{ { e }^{ \frac { logx }{ x } } } dx\quad =\quad I\\ Since\quad we\quad want\quad to\quad approximate\quad the\quad value\quad of\quad I,\\ we\quad can\quad use\quad taylor\quad series\quad of\quad expansion\quad of\quad { e }^{ x }\\ I\quad =\quad \int _{ \frac { 1 }{ e } }^{ e }{ \left( 1+\frac { \frac { logx }{ x } }{ 1! } +\frac { { \left( \frac { logx }{ x } \right) }^{ 2 } }{ 2! } +\frac { { \left( \frac { logx }{ x } \right) }^{ 3 } }{ 3! } +... \right) } dx\\ on\quad splitting\quad the\quad integrals\quad and\quad solving\quad them,\quad we\quad get\\ I\quad \approx \quad 2.65861

@Pi Han Goh solve this in 10 seconds!

Aditya Kumar - 5 years, 9 months ago

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Nope. Impossible. Too hard. You can't do that yourself in a hour. (No calculator/Wolframalpha).

Pi Han Goh - 5 years, 9 months ago

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But u had read my solution in the inspiration problem. U should've entered that.

Aditya Kumar - 5 years, 9 months ago

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@Aditya Kumar Oh wait... no wonder this looks so familiar. ahha!!! me dumb

Pi Han Goh - 5 years, 9 months ago

@Abhishek Bakshi see this and comment.

Aditya Kumar - 5 years, 8 months ago

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