Cool integral property

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

SOLUTION αβ(x2(α+β)x+αβ)dx \int_\alpha^\beta (x^2 - (\alpha + \beta)x + \alpha\beta ) dx

(x33(α+β)x22+αβx)αβ \left(\frac{x^3}{3} - \frac{(\alpha + \beta)x^2}{2} + \alpha\beta x\right)\bigg|_{\alpha}^{\beta}

13(β3α3)α+β2(β2α2)+αβ(βα) \frac{1}{3}(\beta^3 - \alpha^3)-\frac{\alpha+\beta}{2}(\beta^2 - \alpha^2) + \alpha\beta(\beta-\alpha)

13(βα)(β2+αβ+α2)α+β2(βα)(β+α)+αβ(βα) \frac{1}{3}(\beta - \alpha)(\beta^2+\alpha\beta+\alpha^2)-\frac{\alpha+\beta}{2}(\beta - \alpha)(\beta + \alpha) + \alpha\beta(\beta-\alpha)

(βα)(13β2+13αβ+13α212α2αβ12β2+αβ) (\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)

(βα)(16β2+13αβ16α2) (\beta-\alpha)\left(-\frac{1}{6}\beta^2 +\frac{1}{3}\alpha\beta - \frac{1}{6}\alpha^2\right)



αβ(xα)(xβ)dx=(βα)36\boxed{\int_\alpha^\beta (x-\alpha)(x-\beta) dx = -\frac{(\beta-\alpha)^3}{6}}

Note by JohnDonnie Celestre
6 years, 5 months ago

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1 vote

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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

Soutrik Bandyopadhyay - 6 years, 5 months ago

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yeah thanks.

JohnDonnie Celestre - 5 years, 11 months ago

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