Reading this as
[tex]f_X(x)=\begin{cases}ke^{kx}&\text{for }x\in[0,2]\\0&\text{otherwise}\end{cases}[/tex]
For [tex]f_X(x)[/tex] to be a valid PDF, the integral over its support must equal 1:
[tex]\displaystyle\int_{-\infty}^\infty f_X(x)\,\mathrm dx=\int_{x=0}^{x=2} ke^{kx}\,\mathrm dx=1[/tex]
Let [tex]y=kx[/tex], so that [tex]\dfrac{\mathrm dy}k=\mathrm dx[/tex] and the integral becomes
[tex]\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[/tex]
[tex]e^{2k}-e^0=1[/tex]
[tex]e^{2k}-1=1[/tex]
[tex]e^{2k}=2[/tex]
[tex]\ln e^{2k}=\ln2[/tex]
[tex]2k=\ln 2[/tex]
[tex]k=\dfrac{\ln 2}2=\ln\sqrt2[/tex]
