Question Number 135544 by 0731619177 last updated on 13/Mar/21
Answered by Olaf last updated on 13/Mar/21
![f(x) = sinx+cosx = (√2)cos((π/4)−x) x∈[0,(π/4)] ⇒ f(x)∈[1,(√2)] ⇒ ∀x∈[0,(π/4)] ⌊f(x)⌋ = 1 Ω = ∫_0 ^(π/4) cos(2x)×2^1 dx Ω = [sin(2x)]_0 ^(π/4) = 1](https://www.tinkutara.com/question/Q135545.png)
$${f}\left({x}\right)\:=\:\mathrm{sin}{x}+\mathrm{cos}{x}\:=\:\sqrt{\mathrm{2}}\mathrm{cos}\left(\frac{\pi}{\mathrm{4}}−{x}\right) \\ $$$${x}\in\left[\mathrm{0},\frac{\pi}{\mathrm{4}}\right]\:\Rightarrow\:{f}\left({x}\right)\in\left[\mathrm{1},\sqrt{\mathrm{2}}\right] \\ $$$$\Rightarrow\:\forall{x}\in\left[\mathrm{0},\frac{\pi}{\mathrm{4}}\right]\:\lfloor{f}\left({x}\right)\rfloor\:=\:\mathrm{1} \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{cos}\left(\mathrm{2}{x}\right)×\mathrm{2}^{\mathrm{1}} {dx} \\ $$$$\Omega\:=\:\left[\mathrm{sin}\left(\mathrm{2}{x}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:=\:\mathrm{1} \\ $$
Commented by 0731619177 last updated on 13/Mar/21
$${thanks} \\ $$