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(β3−α3)−α+β2(β2−α2)+αβ(β−α) \frac{1}{3}(\beta^3 - \alpha^3)-\frac{\alpha+\beta}{2}(\beta^2 - \alpha^2) + \alpha\beta(\beta-\alpha)

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

(β−α)(13β2+13αβ+13α2−12α2−αβ−12β2+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)

(β−α)(−16β2+13αβ−16α2) (\beta-\alpha)\left(-\frac{1}{6}\beta^2 +\frac{1}{3}\alpha\beta - \frac{1}{6}\alpha^2\right)

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

−(β−α)36‾‾\underline{\underline{-\frac{(\beta-\alpha)^3}{6}}}

∫αβ(x−α)(x−β)dx=−(β−α)36\boxed{\int_\alpha^\beta (x-\alpha)(x-\beta) dx = -\frac{(\beta-\alpha)^3}{6}}

#Calculus #Integrals #Properties

Note by JohnDonnie Celestre
6 years, 5 months ago

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Comments

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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