Finding $k$ when $f(x)=kx^{-(k+1)}$.

Question

I am trying to do the following question:

How do I find the constant k? Do I use $\int_{1}^{\infty}f(x)dx = 1$? If so, then the integral is just equal to $1$...

Answer 1

Yes, you are supposed to use that the integral is equal to $1$, so $$\int_1^\infty kx^{-(k+1)}dx=1\\\left.\frac k{-(k)}x^{-k}\right|_1^\infty=1\\ \frac {k}{k}=1$$ and I find any $k \gt 0$ is a solution because you need the integral to converge.

Answer 2

Yes for the integral to exist as x goes to $\infty$ , k shall be bigger than 0 $$\int_1^\infty f(x) \, dx = -\frac{1}{x^k}+1$$ you can clearly see if k is negative or zero the integral diverges