Existence of function $f$ with integral properties $\int_{-1}^1 f(x) (1 - |x|) = r, \int_{-1}^1 f^k(x) (1 - |x|) = r^{k + 1}$?

Question

Let $r > 1$ and fix an integer $k>1$. I am wondering if there exists a function $f \colon [-1, 1] \to \mathbb{R}_+$, such that $$ \int_{-1}^1 f(x) \big(1 - |x|\big)\, \mathrm{d} x = r, \quad \mbox{and} \quad \int_{-1}^1 \big(f(x)\big)^k \big(1 - |x|\big)\, \mathrm{d} x = r^{k+1}. $$ Comment: Note that the right hand side has an exponent $k+1$ in the second equation; for this reason $f(x) \equiv r$ sadly does not work. I have tried playing with simple things (like polynomials) and was unable to get anything useful.

Answer 1

There is one of the form $f(x)=a(1-|x|)^b$; specifically, let $\lambda>0$ be the solution of $$\frac{(1+\lambda)^k}{1+k\lambda}=r$$ (exists since $\lambda\mapsto(1+\lambda)^k/(1+k\lambda)$ is increasing), and take $f(x)=r(1+\lambda)(1-|x|)^{2\lambda}$.