How to integrate $\int_a^b (x-a)(x-b)\,dx=-\frac{1}{6}(b-a)^3$ in a faster way? $\displaystyle \int_a^b (x-a)(x-b)\,dx=-\frac{1}{6}(b-a)^3$
$\displaystyle \int_a^{(a+b)/2} (x-a)(x-\frac{a+b}{2})(x-b)\, dx=\frac{1}{64}(b-a)^4$ 
Instead of expanding the integrand, or doing integration by part, is there any faster way to compute this kind of integral?

Hint. One may use the fact that
$$
4uv=(u+v)^2-(u-v)^2 \tag1
$$ giving
$$
\begin{align}
4\int_a^b (x-a)(x-b)\:dx=4\int_a^b \left(x-\frac{a+b}2\right)^2\:dx-\int_a^b (b-a)^2\:dx \tag2
\end{align}
$$ which is easier to evaluate.
Then, one has $$
\begin{align}
&4\int_a^b (x-a)(x-b)\left(x-\frac{a+b}2\right)\:dx
\\\\&=\int_a^b \left[4\left(x-\frac{a+b}2\right)^2- (b-a)^2\right]\left(x-\frac{a+b}2\right)\:dx
\\\\&=4\int_a^b \left(x-\frac{a+b}2\right)^3\:dx- (b-a)^2\int_a^b\left(x-\frac{a+b}2\right)\:dx\tag3
\end{align}
$$ which is easier to evaluate.
