Null Function/Examples/Example 3

From ProofWiki
Jump to navigation Jump to search

Example of Null Function

Let $f: \R \to \R$ be the real function defined as:

$\forall x \in \R: \map f x = \begin {cases} 1 & : t = 1 \le t \le 2 \\ 0 & : \text {otherwise} \end {cases}$


Then $f$ is not a null function.


Proof

Let $x > 2$.

\(\ds \int_0^x \map f u \rd u\) \(=\) \(\ds \int_0^1 \map f u \rd u + \int_1^2 \map f u \rd u + \int_2^x \map f u \rd u\)
\(\ds \) \(=\) \(\ds \int_0^1 0 \rd u + \int_1^2 1 \rd u + \int_2^x 0 \rd u\) Definition of $\map f x$
\(\ds \) \(=\) \(\ds 0 + 1 + 0\) Definite Integral of Constant

Hence the result by definition of null function.

$\blacksquare$


Sources