If $$f(x)=\binom{n}{1}(x-1)^2-\binom{n}{2}(x-2)^2+\cdots+(-1)^{n-1}\binom{n}{n}(x-n)^2$$Find the value of $$\int_0^1f(x)dx$$
I rewrote this into a compact form.$$\sum_{k=1}^n\binom{n}{k}(x-k)^2(-1)^{k-1}$$Now,$$\int_0^1\sum_{k=1}^n\binom{n}{k}(x-k)^2(-1)^{k-1}dx$$$$=\sum_{k=1}^n\binom{n}{k}\frac{(1-k)^3}{3}(-1)^{k-1}-\sum_{k=1}^n\binom{n}{k}\frac{(-k)^3}{3}(-1)^{k-1}$$$$=\sum_{k=1}^n\binom{n}{k}\frac{(1-k)^3}{3}(-1)^{k-1}+\sum_{k=1}^n\binom{n}{k}\frac{k^3}{3}(-1)^{k-1}$$After this, I took $\dfrac13$ common and did some simplifications but nothing useful came out.
Any help is greatly appreciated.
