Hey "e" why are you lying on the floor ??? (Simple integration)

Level 1

Integrate the following:

$\int_0^\infty \! \left\lfloor 2e^{-x} \right\rfloor dx,$

where $\lfloor \cdot \rfloor$ denotes the greatest integer function.

\ln 2

\frac2e

e^{2}

2 solutions

Jack Lam
Feb 25, 2016

For $0<x<\log{2}$ , $1<e^{-x}<\frac{1}{2}$ and so $f(x) = 1$

For $\log{2}<x<\infty$ , $\frac{1}{2}<e^{-x}<0$ , and so $f(x)=0$

Thus, the integral is simply a rectangle with height $1$ and length $\log{2}$

And the integral is therefore $\log{2}$

these identities are wrong since there is no positive number which satisfies 1 < e^(-x)<1/2

Ankit Singh - 4 years ago

No th identity is always right...e^(-x) is always a positive no. Irrespective of x. Suppose e^(-2) =1/e^2=0.135. Simar is the case with positive x.

Koulick Sadhu - 3 years, 5 months ago

Tootie Frootie
Feb 20, 2014

it,quite interesting