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?

You can first get rid of the integration bounds by the linear transform $a+(b-a)t$:
$$\int_a^b (x-a)(x-b)\,dx=(b-a)^3\int_0^1t(t-1)\,dt.$$
Mentally expanding the polynomial, the integral is $\frac13-\frac12=-\frac16$.
For the other case, $a+(b-a)t/2$:
$$ \int_a^{(a+b)/2} (x-a)(x-\frac{a+b}{2})(x-b)\, dx=\frac{(b-a)^4}{16}\int_0^1t\left(t-1\right)(t-2)\,dt.$$
Also mentally, $(\frac14-1+1)$$\frac1{16}=\frac1{64}$.

Also, using a remark by @mickep, Newton's formula applies and
$$\int_0^1t(t-1)\,dt=\frac16\left(0-4\frac12\frac12+0\right)=-1,$$
$$\int_0^1t(t-1)(t-2)\,dt=\frac16\left(0+4\frac12\frac12\frac32+0\right)=\frac14.$$
