A problem asked by Himanshu Sharma

Given that $f(2x) = 3f(x)$ and $\int_0^1 f(x)\; dx = 1$, find the value of $\int_1^2 f(x) \; dx $

Answer:

Substitute $x = 2u$ and $dx = 2\; du$, then we have

$ \int_0^1 f(x)\; dx$

$= 2 \int_0^\frac{1}{2} f(2u)\; du $

$= 6 \int_0^\frac{1}{2} f(u)\; du $

So,  $ \int_0^\frac{1}{2} f(u)\; du = \frac{1}{6}  \int_0^1 f(x)\; dx = \frac{1}{6} $

And, $ \int_{\frac{1}{2}}^1 f(u)\; du = \frac{5}{6} $

Also,

$ \int_1^2 f(x)\; dx$

$= 2 \int_{\frac{1}{2}}^1 f(2u)\; du $

$= 6 \int_{\frac{1}{2}}^1 f(u)\; du $

$= 6 * \frac{5}{6} = 5$