Evaluating $\int_a^b (x-\alpha)(x-\beta) \ dx$ without expanding the integrand Is there a property for definite integrals of the form:

$$\int_{a}^{b} (x-\alpha)(x-\beta) \ dx$$

that would allow one to evaluate it without expanding the integrand? 

Thanks!
While expanding seems the most straightforward solution, we can also integrate by parts

$\begin{align}
I&amp;=\int_{a}^{b} (x-\alpha)(x-\beta) \, dx
\\&amp;=\frac{1}{2}(x-\alpha)^2(x-\beta)\bigg|_{x=a}^{x=b}-\frac{1}{2}\int_{a}^{b} (x-\alpha)^2 \, dx
\\&amp;=\frac{1}{2}(x-\alpha)^2(x-\beta)-\frac{1}{6}(x-\alpha)^3\bigg|_{x=a}^{x=b} 
\end{align}$
$$\int (x-\alpha)(x-\beta)dx = \int (x^2-(\alpha+\beta)x+\alpha\beta )dx = \frac{x^3}{3}-(\alpha+\beta)\frac{x^2}{2}+\alpha\beta x+k$$
where $k$ is constant.