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Question Number 94356 by ar247 last updated on 18/May/20

∫_y ^3 (3x^2 −2x+2)=40  (1/2)y=?

$$\int_{{y}} ^{\mathrm{3}} \left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)=\mathrm{40} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{y}=? \\ $$

Commented by ar247 last updated on 18/May/20

help

$${help} \\ $$

Commented by MJS last updated on 18/May/20

integrate and then solve for y  [x^3 −x^2 +2x]_y ^3 =−y^3 +y^2 −2y+24  ⇒  y^3 −y^2 +2y+16=0  (y+2)(y^2 −3y+8)=0  ⇒ y=−2

$$\mathrm{integrate}\:\mathrm{and}\:\mathrm{then}\:\mathrm{solve}\:\mathrm{for}\:{y} \\ $$$$\left[{x}^{\mathrm{3}} −{x}^{\mathrm{2}} +\mathrm{2}{x}\right]_{{y}} ^{\mathrm{3}} =−{y}^{\mathrm{3}} +{y}^{\mathrm{2}} −\mathrm{2}{y}+\mathrm{24} \\ $$$$\Rightarrow \\ $$$${y}^{\mathrm{3}} −{y}^{\mathrm{2}} +\mathrm{2}{y}+\mathrm{16}=\mathrm{0} \\ $$$$\left({y}+\mathrm{2}\right)\left({y}^{\mathrm{2}} −\mathrm{3}{y}+\mathrm{8}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{y}=−\mathrm{2} \\ $$

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