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

$${\int }_y^3(3x^2-2x+2)=40$$$$\frac{1}{2}y=?$$

Commented by ar247 last updated on 18/May/20

$$help$$

Commented by MJS last updated on 18/May/20

$$\mathrm{integrate}\mathrm{and}\mathrm{then}\mathrm{solve}\mathrm{for}y$$$${[x^3-x^2+2x]}_y^3=-y^3+y^2-2y+24$$$$\Rightarrow$$$$y^3-y^2+2y+16=0$$$$(y+2)(y^2-3y+8)=0$$$$\Rightarrow y=-2$$

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