Question-223461

Question Number 223461 by MathematicalUser2357 last updated on 26/Jul/25

Answered by mr W last updated on 26/Jul/25

Commented by mr W last updated on 26/Jul/25

I=((2(β−α))/3)×(((α+β)/2)−α)(((α+β)/2)−β) =−(((β−α)^3 )/6)

$${I}=\frac{\mathrm{2}\left(\beta−\alpha\right)}{\mathrm{3}}×\left(\frac{\alpha+\beta}{\mathrm{2}}−\alpha\right)\left(\frac{\alpha+\beta}{\mathrm{2}}−\beta\right) \\ $$$$\:\:=−\frac{\left(\beta−\alpha\right)^{\mathrm{3}} }{\mathrm{6}} \\ $$

Commented by MathematicalUser2357 last updated on 26/Jul/25

$$\:\underline{\underbrace{\:}} \\ $$

Commented by Ghisom last updated on 26/Jul/25

yes but ((sin (π/6))/( ((2187))^(1/7) ))×[(α−β)(α^2 +αβ+β^2 )+3α(β−α)β] is correct, too plus it′s very pleasant

$$\mathrm{yes}\:\mathrm{but} \\ $$$$\frac{\mathrm{sin}\:\frac{\pi}{\mathrm{6}}}{\:\sqrt[{\mathrm{7}}]{\mathrm{2187}}}×\left[\left(\alpha−\beta\right)\left(\alpha^{\mathrm{2}} +\alpha\beta+\beta^{\mathrm{2}} \right)+\mathrm{3}\alpha\left(\beta−\alpha\right)\beta\right] \\ $$$$\mathrm{is}\:\mathrm{correct},\:\mathrm{too}\:\mathrm{plus}\:\mathrm{it}'\mathrm{s}\:\mathrm{very}\:\mathrm{pleasant} \\ $$

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Question-223461

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