# Cool integral property

Evaluate the integral: $\int_\alpha^\beta (x-\alpha)(x-\beta) dx$

SOLUTION $\int_\alpha^\beta (x^2 - (\alpha + \beta)x + \alpha\beta ) dx$

$\left(\frac{x^3}{3} - \frac{(\alpha + \beta)x^2}{2} + \alpha\beta x\right)\bigg|_{\alpha}^{\beta}$

$\frac{1}{3}(\beta^3 - \alpha^3)-\frac{\alpha+\beta}{2}(\beta^2 - \alpha^2) + \alpha\beta(\beta-\alpha)$

$\frac{1}{3}(\beta - \alpha)(\beta^2+\alpha\beta+\alpha^2)-\frac{\alpha+\beta}{2}(\beta - \alpha)(\beta + \alpha) + \alpha\beta(\beta-\alpha)$

$(\beta-\alpha)\left(\frac{1}{3}\beta^2 + \frac{1}{3}\alpha\beta + \frac{1}{3}\alpha^2 - \frac{1}{2}\alpha^2 - \alpha\beta - \frac{1}{2}\beta^2 + \alpha\beta\right)$

$(\beta-\alpha)\left(-\frac{1}{6}\beta^2 +\frac{1}{3}\alpha\beta - \frac{1}{6}\alpha^2\right)$

$(\beta-\alpha)\left[\frac{(\beta-\alpha)^2}{-6}\right]$

$\underline{\underline{-\frac{(\beta-\alpha)^3}{6}}}$

$\boxed{\int_\alpha^\beta (x-\alpha)(x-\beta) dx = -\frac{(\beta-\alpha)^3}{6}}$ Note by JohnDonnie Celestre
5 years ago

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Couldn't we substitute x - average(alpha,beta) = u ?

That would have significantly reduced the computational time. Please think over it

yeah thanks.

- 4 years, 6 months ago