Question Number 209550 by Tony6400 last updated on 14/Jul/24
Answered by Berbere last updated on 15/Jul/24
![x∈]−(π/2),(π/2)[=I;∀x∈I cos(x)≥0 1+e^(−x) ≥1⇒∀x∈I cos(x)≥((cos(x))/(1+e^(−x) )) ∫_(−(π/2)) ^(π/2) (f(x)−g(x))dx=A=∫_(−(π/2)) ^(π/2) cos(x)−∫_(−(π/2)) ^(π/2) ((cos(x))/(1+e^(−x) ))dx= 2−∫_(−(π/2)) ^(π/2) ((cos(x))/(1+e^(−x) ))dx_(x→−x) =2−∫_(−(π/2)) ^(π/2) ((cos(x))/(1+e^x ))dx =2−∫_(−(π/2)) ^(π/2) e^(−x) .((cos(x))/(1+e^(−x) ))dx⇒2A=4−∫_(−(π/2)) ^(π/2) cos(x)=2 A=1 u.a](https://www.tinkutara.com/question/Q209569.png)
$$\left.{x}\in\right]−\frac{\pi}{\mathrm{2}},\frac{\pi}{\mathrm{2}}\left[={I};\forall{x}\in{I}\:{cos}\left({x}\right)\geqslant\mathrm{0}\right. \\ $$$$\mathrm{1}+{e}^{−{x}} \geqslant\mathrm{1}\Rightarrow\forall{x}\in{I}\:\:{cos}\left({x}\right)\geqslant\frac{{cos}\left({x}\right)}{\mathrm{1}+{e}^{−{x}} } \\ $$$$\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \left({f}\left({x}\right)−{g}\left({x}\right)\right){dx}=\mathscr{A}=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} {cos}\left({x}\right)−\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}\left({x}\right)}{\mathrm{1}+{e}^{−{x}} }{dx}= \\ $$$$\mathrm{2}−\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}\left({x}\right)}{\mathrm{1}+{e}^{−{x}} }{dx}_{{x}\rightarrow−{x}} =\mathrm{2}−\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}\left({x}\right)}{\mathrm{1}+{e}^{{x}} }{dx} \\ $$$$=\mathrm{2}−\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} {e}^{−{x}} .\frac{{cos}\left({x}\right)}{\mathrm{1}+{e}^{−{x}} }{dx}\Rightarrow\mathrm{2}\mathscr{A}=\mathrm{4}−\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} {cos}\left({x}\right)=\mathrm{2} \\ $$$$\mathscr{A}=\mathrm{1}\:{u}.{a} \\ $$
Answered by mr W last updated on 15/Jul/24
![for x≥0: h(x)=f(x)−g(x) =cos x−((cos x)/(1+e^(−x) )) =cos (−x)−((cos (−x))/(1+e^(−x) )) =((cos (−x)+e^(−x) cos (−x)−cos (−x))/(1+e^(−x) )) =((e^(−x) cos (−x))/(1+e^(−x) )) =((cos (−x))/(1+e^(−(−x)) )) =cos (−x)−[cos (−x)−((cos (−x))/(1+e^(−(−x)) ))] =cos x−h(−x) area=∫_(−(π/2)) ^(π/2) h(x)dx =∫_(−(π/2)) ^0 h(x)dx+∫_0 ^(π/2) h(x)dx =∫_(−(π/2)) ^0 h(x)dx+∫_0 ^(π/2) (cos x−h(−x))dx =∫_(−(π/2)) ^0 h(x)dx+∫_0 ^(π/2) cos x dx−∫_0 ^(π/2) h(−x))dx =∫_(−(π/2)) ^0 h(x)dx+∫_0 ^(π/2) cos x dx+∫_0 ^(−(π/2)) h(x)dx =∫_(−(π/2)) ^0 h(x)dx+∫_0 ^(π/2) cos x dx−∫_(−(π/2)) ^0 h(x)dx =∫_0 ^(π/2) cos x dx =[sin x]_0 ^(π/2) =1 ✓](https://www.tinkutara.com/question/Q209575.png)
$${for}\:{x}\geqslant\mathrm{0}: \\ $$$${h}\left({x}\right)={f}\left({x}\right)−{g}\left({x}\right) \\ $$$$=\mathrm{cos}\:{x}−\frac{\mathrm{cos}\:{x}}{\mathrm{1}+{e}^{−{x}} } \\ $$$$=\mathrm{cos}\:\left(−{x}\right)−\frac{\mathrm{cos}\:\left(−{x}\right)}{\mathrm{1}+{e}^{−{x}} } \\ $$$$=\frac{\mathrm{cos}\:\left(−{x}\right)+{e}^{−{x}} \mathrm{cos}\:\left(−{x}\right)−\mathrm{cos}\:\left(−{x}\right)}{\mathrm{1}+{e}^{−{x}} } \\ $$$$=\frac{{e}^{−{x}} \mathrm{cos}\:\left(−{x}\right)}{\mathrm{1}+{e}^{−{x}} } \\ $$$$=\frac{\mathrm{cos}\:\left(−{x}\right)}{\mathrm{1}+{e}^{−\left(−{x}\right)} } \\ $$$$=\mathrm{cos}\:\left(−{x}\right)−\left[\mathrm{cos}\:\left(−{x}\right)−\frac{\mathrm{cos}\:\left(−{x}\right)}{\mathrm{1}+{e}^{−\left(−{x}\right)} }\right] \\ $$$$=\mathrm{cos}\:{x}−{h}\left(−{x}\right) \\ $$$${area}=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} {h}\left({x}\right){dx} \\ $$$$=\int_{−\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} {h}\left({x}\right){dx}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {h}\left({x}\right){dx} \\ $$$$=\int_{−\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} {h}\left({x}\right){dx}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{cos}\:{x}−{h}\left(−{x}\right)\right){dx} \\ $$$$\left.=\int_{−\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} {h}\left({x}\right){dx}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}\:{x}\:{dx}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {h}\left(−{x}\right)\right){dx} \\ $$$$=\int_{−\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} {h}\left({x}\right){dx}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}\:{x}\:{dx}+\int_{\mathrm{0}} ^{−\frac{\pi}{\mathrm{2}}} {h}\left({x}\right){dx} \\ $$$$=\int_{−\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} {h}\left({x}\right){dx}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}\:{x}\:{dx}−\int_{−\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} {h}\left({x}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}\:{x}\:{dx} \\ $$$$=\left[\mathrm{sin}\:{x}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=\mathrm{1}\:\checkmark \\ $$
Commented by mr W last updated on 15/Jul/24
