Recursions in Real Analysis.

Calculus Level 4

Let f : R → R f : \mathbb{R \rightarrow R} be a continuous function which satisfies the identity f ( 2 x ) = 3 f ( x ) ∀ x ∈ R f(2x) = 3f(x)\ \forall x \in \mathbb R .

If ∫ 0 1 f ( x ) d x = 1 \displaystyle \int _{ 0 }^{ 1 } f(x) \ dx = 1 , evaluate 1 2 ∫ 1 2 f ( x ) d x \displaystyle \frac { 1 }{ 2 } \int _{ 1 }^{ 2 } f( x) \ dx .


The answer is 2.5.

Priyanshu Mishra by Priyanshu Mishra , 20, India

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1 solution

∫ 0 1 f ( x ) d x = 1 Given ∫ 0 1 3 f ( x ) d x = 3 also that f ( 2 x ) = 3 f ( x ) ∫ 0 1 f ( 2 x ) d x = 3 Let u = 2 x ⟹ d u = 2 d x 1 2 ∫ 0 2 f ( u ) d u = 3 1 2 ∫ 1 2 f ( u ) d u + 1 2 ∫ 0 1 f ( u ) d u = 3 Replace u with x 1 2 ∫ 1 2 f ( x ) d x + 1 2 ∫ 0 1 f ( x ) d x = 3 1 2 ∫ 1 2 f ( x ) d x + 1 2 = 3 ⟹ 1 2 ∫ 1 2 f ( x ) d x = 3 − 1 2 = 2.5 \begin{aligned} \int_0^1 f(x) \ dx & = 1 & \small \color{#3D99F6} \text{Given} \\ \int_0^1 {\color{#3D99F6}3f(x)} \ dx & = 3 & \small \color{#3D99F6} \text{also that }f(2x) = 3f(x) \\ \int_0^1 {\color{#3D99F6}f(2x)} \ dx & = 3 & \small \color{#3D99F6} \text{Let } u = 2x \implies du = 2\ dx \\ \frac 12 \int_0^2 f(u) \ du & = 3 \\ \frac 12 \int_1^2 f(u) \ du + \frac 12 \int_0^1 f(u) \ du & = 3 & \small \color{#3D99F6} \text{Replace } u \text{ with }x \\ \frac 12 \int_1^2 f(x) \ dx + \color{#3D99F6} \frac 12 \int_0^1 f(x) \ dx & = 3 \\ \frac 12 \int_1^2 f(x) \ dx + \color{#3D99F6} \frac 12 & = 3 \\ \implies \frac 12 \int_1^2 f(x) \ dx & = 3 - \frac 12 = \boxed {2.5} \end{aligned}

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As a supplement (in order to know the solution is self-consistent) we can note that f ( x ) = log ⁡ 2 6 ⋅ ∣ x ∣ log ⁡ 2 3 f(x) = \log_26 \cdot |x|^{\log_23} is one such function. Therefore the solution makes sense (are there any other such functions?)

Brian Moehring - 2 years, 7 months ago

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Thanks, good suggestion.

Chew-Seong Cheong - 2 years, 7 months ago

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