Function integral

Calculus Level 3

A continuous real function f f satisfies f ( 2 x ) = 3 f ( x ) f(2x)=3f(x) for all real x x . If ∫ 0 1 f ( x ) d x = 1 \displaystyle\int _{ 0 }^{ 1 }{ f\left( x \right) \, dx=1 } , then compute the value of definite integral ∫ 1 2 f ( x ) d x \displaystyle \int _{ 1 }^{ 2 }{ f\left( x \right) \, dx } .


The answer is 5.

Shubham Singh by shubham singh , 23, India

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1 solution

Observe that

∫ 1 2 f ( x ) d x = 2 ∫ 1 2 2 2 f ( 2 u ) d u = 2 ⋅ 3 ∫ 1 2 2 2 f ( x ) d x = ( 2 ⋅ 3 ) 2 ∫ 1 2 2 2 2 2 f ( x ) d x = ⋯ , \int_1^2 f(x) dx = 2 \int_{\frac{1}{2}}^{\frac{2}{2}} f(2u) du = 2 \cdot 3 \int_{\frac{1}{2}}^{\frac{2}{2}} f(x) dx = (2 \cdot 3)^2 \int_{\frac{1}{2^2}}^{\frac{2}{2^2}} f(x) dx = \cdots,

and one could generalize this result to

∫ 1 2 f ( x ) d x = 6 n ∫ 1 2 n 1 2 n − 1 f ( x ) d x . \int_1^2 f(x) dx = 6^n \int_{\frac{1}{2^n}}^{\frac{1}{2^{n-1}}} f(x) dx.

Therefore

∫ 1 2 f ( x ) d x ∑ n = 1 ∞ 1 6 n = 1 5 ∫ 1 2 f ( x ) d x = ∑ n = 1 ∞ ∫ 1 2 n 1 2 n − 1 f ( x ) d x = ∫ 0 1 f ( x ) d x = 1. \int_1^2 f(x) dx \sum_{n=1}^\infty \frac{1}{6^n} = \frac{1}{5} \int_1^2 f(x) dx = \sum_{n=1}^\infty \int_{\frac{1}{2^n}}^{\frac{1}{2^{n-1}}} f(x) dx = \int_0^1 f(x)dx = 1.

We conclude:

∫ 1 2 f ( x ) d x = 5. \int_1^2 f(x) dx = 5.

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