Show the integral equality $\int_0^{k} x(k-x) \,\mathrm{d}x =1/2 \int_{0}^k x^2 \,\mathrm{d}x$ without direct computation

Question

Is it possible to prove that

$$\int_0^{k} x(k-x) \,\mathrm{d}x =\frac{1}{2} \int_{0}^k x^2 \,\mathrm{d}x$$

without directly computing the right hand side, left hand side and see that they are equal? I can simply evaluate it as follows, but that is not what I am asking =)

$$ \int_0^{k} x(k-x) \mathrm{d}x = \Bigl[\frac{k}{2}\cdot x^2 - \frac{1}{3}\cdot x^3\Bigr]_0^k = \frac{k}{2}k^2 - \frac{1}{3}k^3 = \frac{k^3}{2} - \frac{k^3}{3} = \frac{k^3}{6} = \frac{1}{2} \int_0^k x^2 \mathrm{d}x $$

Is there some symmetry argument or substitution that could be made? I already know that

$$ \int_0^{k} x(k-x) \,\mathrm{d}x = \frac{1}{2} \int_0^{k/2} x(k-x)\,\mathrm{d}x = \frac{1}{2} \int_{k/2}^k x(k-x)\,\mathrm{d}x $$

I also attempted a few substitutions such as $x=k/2-y$ without much success.

Answer 1

Integration by parts: $$ \int_0^{k} x(k-x) \, dx = \underbrace{ \left[ (\frac 12 x^2)(k-x)\right]_{x=0}^{x=k}}_{= 0}- \int_0^{k} (\frac 12 x^2) (-1) \, dx $$

As a small generalization you can prove (using repeated integration by parts) that $$ \int_0^{k} x^p(k-x)^q \, dx = \frac{p! q!}{(p+q)!}\int_0^{k} x^{p+q} \, dx $$ for positive integers $p$ and $q$.

Answer 2

Take derivatives; you'll get $x^2/2=x^2/2$, hence both integrals differ only by a constant. Now substitute $k=0$.