Improper Integral 2

Calculus Level 3

∫ 64 ∞ 3 x 2 d x ( x 3 − 4 ) 3 = 1 H \large \int _{64}^\infty \frac {3x^2 \ dx}{ \left (\sqrt {x^3-4}\right)^3} =\frac 1{\sqrt H }

The equation above holds true for positive integer H H . Let H = 2 a − 1 H = 2^a-1 ; find a a .


The answer is 16.

Liam P by Liam P , 22, Canada

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1 solution

Chew-Seong Cheong
Mar 15, 2017

I = ∫ 64 ∞ 3 x 2 d x ( x 3 − 4 ) 3 Let u = x 3 − 4 , d u = 3 x 2 d x = ∫ 6 4 3 − 4 ∞ d u u 3 2 = − 2 u ∣ 6 4 3 − 4 ∞ = 0 + 2 6 4 3 − 4 = 2 2 18 − 4 = 1 2 16 − 1 \begin{aligned} I & = \int_{64}^\infty \frac {3x^2 \ dx }{(\sqrt{x^3-4})^3} & \small \color{#3D99F6} \text{Let }u = x^3-4, \ du = 3x^2 \ dx \\ & = \int_{64^3-4}^\infty \frac {du}{u^\frac 32} \\ & = \frac {-2}{\sqrt u} \bigg|_{64^3-4}^\infty \\ & = 0 + \frac 2{\sqrt{64^3-4}} \\ & = \frac 2{\sqrt{2^{18}-4}} \\ & = \frac 1{\sqrt{2^{16}-1}} \end{aligned}

⟹ a = 16 \implies a = \boxed{16}

3 Helpful 0 Interesting 1 Brilliant 0 Confused

2 18 − 4 = 262144 − 4 = 262140 → 2 262140 → 1 65535 \sqrt { 2 ^ { 18 } - 4 } = \sqrt { 262144 - 4 } = \sqrt { 262140 } \rightarrow \frac { 2 } { \sqrt { 262140 } } \rightarrow \frac { 1 } { \sqrt { 65535 } }

1 2 16 − 1 = 1 65536 − 1 = 1 65535 \frac { 1 } { \sqrt { 2 ^ { 16 } - 1 } } = \frac { 1 } { \sqrt { 65536 - 1 } } = \frac { 1 } { \sqrt { 65535 } } .

The values are same?

. . - 3 months, 3 weeks ago

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