To Integrate or not to Integrate

Calculus Level 4

∫ 1 2 x x x ⋯ 5 4 3 d x = A B e − C − D F e − G \int_1^2 \sqrt{ x\sqrt[3]{x\sqrt[4]{x\sqrt[5]{\cdots}}}}\text{d}x=\frac{A^{Be-C}-D}{Fe-G}

The equation above holds true for positive integers A , B , C , D , F , G A,B,C,D,F,G . Find A + B + C + D + F + G A+B+C+D+F+G .

Notation: e ≈ 2.718 e\approx 2.718 denotes the Euler's number .


The answer is 7.

Hamza A by Hamza A , 19, USA

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2 solutions

Chew-Seong Cheong
Sep 10, 2019

I = ∫ 1 2 x x x ⋯ 5 4 3 d x = ∫ 1 2 ( x ( x ( x ( ⋯ ) 1 5 ) 1 4 ) 1 3 ) 1 2 d x = ∫ 1 2 x 1 2 + 1 3 ! + 1 4 ! + ⋯ d x = ∫ 1 2 x e − 2 d x = x e − 1 e − 1 ∣ 1 2 = 2 e − 1 − 1 e − 1 \begin{aligned} I & = \int_1^2 \sqrt{x\sqrt[3]{x\sqrt[4]{x\sqrt[5]{\cdots}}}} dx \\ & = \int_1^2 \left(x\left(x\left(x\left(\cdots \right)^\frac 15\right)^\frac 14\right)^\frac 13\right)^\frac 12 dx \\ & = \int_1^2 x^{\frac 12 + \frac 1{3!} + \frac 1{4!}+\cdots}\ dx \\ & = \int_1^2 x^{e-2}\ dx = \frac {x^{e-1}}{e-1}\ \bigg|_1^2 = \frac {2^{e-1}-1}{e-1} \end{aligned}

Therefore, A + B + C + D + F + G = 2 + 1 + 1 + 1 + 1 + 1 = 7 A+B+C+D+F+G = 2+1+1+1+1+1 = \boxed 7 .

1 Helpful 0 Interesting 1 Brilliant 0 Confused

The integrand is x e − 2 x^{e-2} . The value of the integral is 2 e − 1 − 1 e − 1 \dfrac{2^{e-1}-1}{e-1} .

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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