That's a great mapping

Calculus Level 4

Define a function $f(x)$ such that $f(x)=x^{2}$ for $x \in [-1,1]$ and $f(x)=f(\frac{1}{x})$ for all non-zero $x$ .

Find $\displaystyle \int_{-\infty}^{\infty} f(x)\, dx$ .

1 solution

First Last
Jul 12, 2017

Because $x$ on $(1,\infty)$ maps to $\frac1{x}$ on $(0,1)$ then f can be written as piecewise:

$\text{f}(x)=\begin{cases} x^2 & [-1,1] \\ \frac1{x^2} & (-\infty,-1)U(1,\infty) \\ \end{cases}$

$\displaystyle\int_{-\infty}^\infty\text{f}(x)dx=\quad\int_{-\infty}^{-1}\frac{dx}{x^2}+\int_{-1}^1 x^2dx+\int_1^\infty\frac{dx}{x^2} = \frac8{3}$

How come it's level 5???

saptarshi dasgupta - 3 years, 2 months ago

@saptarshi dasgupta Srsly!!! I agree!!!

Aaghaz Mahajan - 3 years ago

Lol now,it is on lvl4..

A Former Brilliant Member - 11 months, 4 weeks ago