A roaring paper tiger (My sixth integral problem)

Calculus Level 5

∫ − 1 1 x 3 + 3 x 2 + 27 x 3 + 27 x 2 + 729 x 3 + 243 x 2 + 19683 x 3 + 2182 x 2 + ⋯ 3 3 3 3 d x \displaystyle \int_{-1}^{1} \sqrt[3]{x^3 + 3x^2 + \sqrt[3]{27x^3 + 27x^2 + \sqrt[3]{729x^3 + 243x^2 +\sqrt[3]{19683x^3 + 2182 x^2 +\cdots} }}}dx

Evaluate the integral above.

Clarification:

  • The coefficient of x 3 x^3 is given by 3 3 n − 3 3^{3n-3} , where n n is a positive integer.

  • The coefficient of x 2 x^2 is given by 3 2 n − 1 3^{2n-1} , where n n is a positive integer.


The answer is 2.

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Hobart Pao by Hobart Pao , 23, USA

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1 solution

Rishabh Jain
Feb 6, 2016

x + 1 = ( x + 1 ) 3 3 x+1=\sqrt[3]{(x+1)^3} = x 3 + 3 x 2 + 3 x + 1 3 =\sqrt[3]{x^3+3x^2+\color{#D61F06}{3x+1}} = x 3 + 3 x 2 + ( 3 x + 1 ) 3 3 3 =\sqrt[3]{x^3+3x^2+\sqrt[3]{\color{#D61F06}{(3x+1)^3}}} = x 3 + 3 x 2 + 27 x 3 + 27 x 2 + 9 x + 1 3 3 =\sqrt[3]{x^3+3x^2+\sqrt[3]{27x^3+27x^2+\color{#20A900}{9x+1}}} = x 3 + 3 x 2 + 27 x 3 + 27 x 2 + ( 9 x + 1 ) 3 3 3 3 =\sqrt[3]{x^3+3x^2+\sqrt[3]{27x^3+27x^2+\sqrt[3]{\color{#20A900}{(9x+1)^3}}}} = x 3 + 3 x 2 + 27 x 3 + 27 x 2 + 729 x 3 + 243 x 2 + 27 x + 1 3 3 3 =\sqrt[3]{x^3+3x^2+\sqrt[3]{27x^3+27x^2+\sqrt[3]{729x^3+243x^2+\color{#3D99F6}{27x+1}}}} = ⋯ =\cdots which is what required in the question.
Hence, ∫ − 1 1 ( x + 1 ) d x = 0 + 2 = 2 \Large \int_{-1}^1( x +1)dx= 0+2=\boxed{\color{#007fff}{2}}

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Lovely solution! =D =D =D

Pi Han Goh - 5 years, 4 months ago

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T h a n k s ! ! \Large\color{#007fff}{\mathcal{Thanks!!}}

Rishabh Jain - 5 years, 4 months ago

correct me if i'm wrong but in line 5 did you mean

x 3 + 3 x 2 + 27 x 3 + 27 x 2 + ( 9 x + 1 ) 3 3 3 3 \sqrt [ 3 ]{ x^{ 3 }+3x^{ 2 }+\sqrt [ 3 ]{ 27x^{ 3 }+27x^{ 2 }+\sqrt [ 3 ]{ (9x+1)^3 } } }

Hamza A - 5 years, 4 months ago

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Right...edited... :)

Rishabh Jain - 5 years, 4 months ago

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