Question

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

Answer 1

Explanation:

I can't understand this question

Answer 2

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
`\frac{\mathrm dy}k=\mathrm dx` and the integral becomes


`\displaystyle\int_(y=0)^(y=2k)ke^y\,\frac{\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=\frac{\ln 2}2=\ln\sqrt2`