Fun with exponents

Calculus Level 4

0 1 exp ( x + e x + e e x + e e e x ) d x \Large\int_0^1 \exp({x+e^x+e^{e^x}+e^{e^{e^x}}}) \, dx

If the above integral can be expressed as

A e B e ^Ae - ^B e

for positive integers A A and B B , find A + B A+B .

Notations :

  • exp ( x ) \exp(x) denotes the exponential function, exp ( x ) = e x \exp(x) = e^x .

  • n a ^na denotes the Tetration function, n a = a a a a n number of a ’s \Large ^n a = \underbrace{a^{a^{a^{\cdot^{\cdot^a}}}}}_{n\text{ number of } a\text{'s}} .


The answer is 7.

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1 solution

Aareyan Manzoor
Feb 14, 2016

f ( n ) = 0 1 e x e e x . . . . e e e . . . n times d x f(n)=\displaystyle\int_0^1 e^x e^{e^x} .... \underbrace{e^{e^{e^{...}}}}_{\text{n times}} dx make the y-sub y = e x y=e^x and changing y to x: f ( n ) = 1 e e x e e x . . . . e e e . . . n-1 times d x f(n)=\displaystyle\int_1^e e^x e^{e^x} .... \underbrace{e^{e^{e^{...}}}}_{\text{n-1 times}} dx then by induction we would have f ( n ) = n 1 e n e d x = n e n 1 e f(n) = \int_{^{n-1}e}^{^n e} dx= ^{n}e-^{n-1}e Put n =4 and get the answer.

You got the key to solve this problem. well done!

Nihar Mahajan - 5 years, 4 months ago

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a wrong-answered level 4 question :0

Joel Yip - 5 years, 4 months ago

PS it should be 7 as the answer not 9

Aareyan Manzoor - 5 years, 4 months ago

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