Question

Find the value of k for which the given function is a probability density function. f(x = kekx on [0, 2]

Answer 1

Reading this as

$f_X(x)=\begin{cases}ke^{kx}&\text{for }x\in[0,2]\\0&\text{otherwise}\end{cases}$

For $f_X(x)$ to be a valid PDF, the integral over its support must equal 1:

$\displaystyle\int_{-\infty}^\infty f_X(x)\,\mathrm dx=\int_{x=0}^{x=2} ke^{kx}\,\mathrm dx=1$

Let $y=kx$, so that $\dfrac{\mathrm dy}k=\mathrm dx$ and the integral becomes

$\displaystyle\int_{y=0}^{y=2k}ke^y\,\dfrac{\mathrm dy}k=\int_0^{2k}e^y\,\mathrm dy=e^y\bigg|_{y=0}^{y=2k}=1$
$e^{2k}-e^0=1$
$e^{2k}-1=1$
$e^{2k}=2$
$\ln e^{2k}=\ln2$
$2k=\ln 2$
$k=\dfrac{\ln 2}2=\ln\sqrt2$