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!
 A: While expanding seems the most straightforward solution, we can also integrate by parts
$\begin{align}
I&=\int_{a}^{b} (x-\alpha)(x-\beta) \, dx
\\&=\frac{1}{2}(x-\alpha)^2(x-\beta)\bigg|_{x=a}^{x=b}-\frac{1}{2}\int_{a}^{b} (x-\alpha)^2 \, dx
\\&=\frac{1}{2}(x-\alpha)^2(x-\beta)-\frac{1}{6}(x-\alpha)^3\bigg|_{x=a}^{x=b} 
\end{align}$
A: Using the identity
$$ab=\frac14\left((a+b)^2-(a-b)^2\right)$$
we get
\begin{align}\int_a^b(x-\alpha)(x-\beta)dx&=\frac14\int_a^b(2x-(\alpha+\beta))^2dx-\frac14\int_a^b(\alpha-\beta)^2dx\\&=\frac1{24}(2x-(\alpha-\beta))^3\Bigg|_a^b-\frac14(\alpha-\beta)^2(b-a)\end{align}
A: $$\int_a^b (x-\alpha)(x - \beta) \ dx = \int_a^b \left(x -\frac{\alpha + \beta}{2}\right)^2 - \left(\frac{\alpha - \beta}{2}\right)^2 \ dx \\= \left[\frac{1}{3}\left(x -\frac{\alpha + \beta}{2}\right)^3  - \left(\frac{\alpha - \beta}{2}\right)^2 x\right]_a^b$$
A: Since you have a second order degree polynomial, Simpson's rule becomes exact. Thus
$$
\begin{align}
\int_a^b (x-\alpha)(x-\beta)\,dx &= \frac{1}{6(b-a)}\Bigl[(a-\alpha)(a-\beta)\\
&\quad +4\bigl((a+b)/2-\alpha\bigr)\bigl((a+b)/2-\beta\bigr)+(b-\alpha)(b-\beta)\Bigr].
\end{align}
$$
A: $$\int (x-\alpha)(x-\beta)dx = \int (x^2-(\alpha+\beta)x+\alpha\beta )dx = \frac{x^3}{3}-(\alpha+\beta)\frac{x^2}{2}+\alpha\beta x+k$$
where $k$ is constant.
