Answer with explanation:
The Normalization Principle states that
[tex]\int_{-\infty }^{+\infty }f(x)dx=1[/tex]
Given
[tex]f(x)=xe^{-kx}(x>0\\\\0(x<0)[/tex]
Thus solving the integral we get
[tex]\int_{0 }^{+\infty }A\cdot xe^{-kx}dx=1\\\\A\int_{0 }^{+\infty }\cdot xe^{-kx}dx=1[/tex]
The integral shall be solved using chain rule initially and finally we shall apply the limits as shown below
[tex]I=\int xe^{-kx}dx\\\\x\int e^{-kx}dx-\int \frac{d(x)}{dx}\int e^{-kx}dx\\\\-\frac{xe^{-kx}}{k}-\int 1\cdot \frac{-e^{-kx}}{k}\\\\\therefore I=\frac{e^{-kx}}{k}-\frac{xe^{-kx}}{k}[/tex]
Applying the limits and solving for A we get
[tex]I=\frac{1}{k}[\frac{1}{e^{kx}}-\frac{x}{e^{kx}}]_{0}^{+\infty }\\\\I=-\frac{1}{k}\\\\\therefore A=-k[/tex]