Triangular distribution mean proof

Question

Hello I am trying to find the triangular distribution mean by evaluating this integral:

$$\int_{a}^{c}x\frac{2(x-a)}{(b-a)(c-a)}dx+\int_{c}^{b}x\frac{2(b-x)}{(b-a)(b-c)}dx=\frac{ac^{3}-bc^{3}-a^{3}c+b^{3}c-ab^{3}+a^{3}b}{3(b-c)(b-a)(c-a)}$$

but as you can see that's nowhere near the expected $$\frac{a+b+c}{3}$$ Is there a way to simplify it all the way to get the correct answer? I've tried to simplify both in Symbolab and Wolfram to no avail.

Answer 1

Hint: expand $(b-c)(b-a)(c-a)(a+b+c)$ and see if you get the numerator of the result of the integral. In this case you can cancel $(b-c)(b-a)(c-a).$

Answer 2

There is a way without expanding.

The polynomial $\sum\limits_{cyc}(a^3b-a^3c)$ is equal to zero for $a=b$, for $a=c$ and for $b=c$,

which gives a factor $(a-b)(a-c)(b-c)$.

Id est, $$\sum\limits_{cyc}(a^3b-a^3c)=(a-b)(a-c)(b-c)P(a,b,c),$$ where $P$ is a cyclic polynomial with degree one, which gives $$P(a,b,c)=k(a+b+c),$$ where $k\in\mathbb R$.

Thus, $$\sum\limits_{cyc}(a^3b-a^3c)=k(a+b+c)(a-b)(a-c)(b-c).$$ Now, for $c=0$, $b=1$ and $a=2$ we obtain: $$8-2=k\cdot3\cdot2,$$ which gives $k=1$ and we are done!