Question Number 94356 by ar247 last updated on 18/May/20
$$\int_{{y}} ^{\mathrm{3}} \left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)=\mathrm{40} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{y}=? \\ $$
Commented by ar247 last updated on 18/May/20
$${help} \\ $$
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](Q94368.png)
$$\mathrm{integrate}\:\mathrm{and}\:\mathrm{then}\:\mathrm{solve}\:\mathrm{for}\:{y} \\ $$$$\left[{x}^{\mathrm{3}} −{x}^{\mathrm{2}} +\mathrm{2}{x}\right]_{{y}} ^{\mathrm{3}} =−{y}^{\mathrm{3}} +{y}^{\mathrm{2}} −\mathrm{2}{y}+\mathrm{24} \\ $$$$\Rightarrow \\ $$$${y}^{\mathrm{3}} −{y}^{\mathrm{2}} +\mathrm{2}{y}+\mathrm{16}=\mathrm{0} \\ $$$$\left({y}+\mathrm{2}\right)\left({y}^{\mathrm{2}} −\mathrm{3}{y}+\mathrm{8}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{y}=−\mathrm{2} \\ $$