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Question Number 211008 by mnjuly1970 last updated on 26/Aug/24

Commented by Frix last updated on 26/Aug/24

$$y=\alpha x+p?$$

Commented by mnjuly1970 last updated on 26/Aug/24

$$y=\alpha x+\beta \underset{―}{\underbrace{⋚}}$$

Commented by Frix last updated on 26/Aug/24

$$I=\underset{-\frac{\pi }{2}}{\stackrel{\frac{\pi }{2}}{\int }}({\sin }^{2}x-2\beta \sin x-2\alpha x\sin x+{(\alpha x+\beta )}^2)dx=$$$$=[\frac{x-\cos x\sin x}{2}+2\beta \cos x+2\alpha {(x\cos x-\sin x+\frac{{(\alpha x+\beta )}^3}{3\alpha }]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}=$$$$=\frac{{\alpha }^2{\pi }^3}{12}-4\alpha +{\beta }^2\pi +\frac{\pi }{2}$$$$\mathrm{The}\mathrm{minimum}\mathrm{occurs}\mathrm{at}\alpha =\frac{24}{{\pi }^3}\wedge \beta =0$$$$\min I=\frac{{\pi }^4-96}{2{\pi }^3}$$$$y=\alpha x+\beta =\frac{24x}{{\pi }^3}$$

Commented by mnjuly1970 last updated on 26/Aug/24

$$gratefulsirFrix$$

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