Prove $\int_\alpha ^\beta (x-\alpha)(x-\beta)dx = -\frac{1}{6}(\beta -\alpha )^3$

Question

In addition, does anyone know what that equation is called?

I know that it’s a property of definite integral

Answer 1

As $\alpha\le x\le\beta$

$-(\beta-\alpha)\le2x-(\alpha+\beta)\le\beta-\alpha$

WLOG $2x-(\alpha+\beta)=(\beta-\alpha)\cos2t\iff x=\alpha\sin^2t+\beta\cos^2t$

to reach at Beta Function

Answer 2

Let $\xi = (x-\alpha)/(\beta-\alpha)$.

Observe that \begin{equation} \int_\alpha ^\beta (x-\alpha)(x-\beta)dx = (\beta-\alpha)^3\int_0^1\xi(\xi-1) d\xi = -\frac{1}{6}(\beta-\alpha)^3 \end{equation}

Answer 3

You can employ symmetry to solve the problem. Let $\gamma = \frac{\beta-\alpha}{2}$ and $s = x - \frac{\alpha+\beta}{2}$ then

\begin{align} \int_\alpha^\beta (x-\alpha)(x-\beta)\ dx &= \int_{-\gamma}^\gamma (s+\gamma)(s-\gamma)\ ds \\ &= \int_{-\gamma}^{\gamma} (s^2 - \gamma^2) ds \\ &= -\frac{4\gamma^3}{3} \\ & = -\frac{(\beta-\alpha)^3}{6} \end{align}

Answer 4

I guess you can also integrate by parts:

$$\int_\alpha^\beta (x-\alpha)(x-\beta) \; dx = \left.\frac{(x-\alpha)^2(x-\beta)}{2} \right|_\alpha^\beta - \int_\alpha^\beta \frac{(x-\alpha)^2}{2}\; dx $$

$$= 0 -\left.\frac{(x-\alpha)^3}{6}\right|_\alpha^\beta =-\frac{(\beta-\alpha)^3}{6}. $$

Answer 5

Denote: $x-\alpha=t \Rightarrow x-\beta=t+\alpha-\beta \Rightarrow dx=dt$. Then: $$\begin{align}\int_\alpha^\beta (x-\alpha)(x-\beta) \; dx &=\int_0^{\beta-\alpha} t(t+\alpha-\beta) \; dx =\\ &=\int_0^{\beta-\alpha} t^2+t(\alpha-\beta) \; dx =\\ &=\left(\frac{t^3}{3}+\frac{t^2}{2}(\alpha-\beta)\right)\big{|}_0^{\beta-\alpha}=\\ &=\frac{(\beta-\alpha)^3}{3}-\frac{(\beta-\alpha)^3}{2}=\\ &=-\frac{(\beta-\alpha)^3}{6}.\end{align}$$