Question Number 204409 by mr W last updated on 16/Feb/24
$${find}\:\lfloor\int_{\mathrm{0}} ^{\mathrm{2023}} \frac{\mathrm{2}}{{x}+{e}^{{x}} }{dx}\rfloor=? \\ $$
Commented by witcher3 last updated on 16/Feb/24
$$\mathrm{nice}\:\mathrm{problems}\:\mathrm{sir}\:\:\mathrm{Thanx}\:\mathrm{for}\:\mathrm{share}\:\mathrm{it} \\ $$
Answered by witcher3 last updated on 16/Feb/24
$$\lfloor\:\mathrm{x}\rfloor\:\mathrm{floor}\:\mathrm{function}? \\ $$
Commented by mr W last updated on 16/Feb/24
$${yes}. \\ $$
Answered by mr W last updated on 18/Feb/24
![x∈[0, 2023] ⇒x≥0 (2/(x+e^x ))<(2/e^x ) ⇒∫_0 ^(2023) (2/(x+e^x ))dx<∫_0 ^(2023) (2/e^x )dx=2[−(1/e^x )]_0 ^(2023) =2(1−(1/e^(2023) ))<2](https://www.tinkutara.com/question/Q204463.png)
$${x}\in\left[\mathrm{0},\:\mathrm{2023}\right]\:\Rightarrow{x}\geqslant\mathrm{0} \\ $$$$\frac{\mathrm{2}}{{x}+{e}^{{x}} }<\frac{\mathrm{2}}{{e}^{{x}} } \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{2023}} \frac{\mathrm{2}}{{x}+{e}^{{x}} }{dx}<\int_{\mathrm{0}} ^{\mathrm{2023}} \frac{\mathrm{2}}{{e}^{{x}} }{dx}=\mathrm{2}\left[−\frac{\mathrm{1}}{{e}^{{x}} }\right]_{\mathrm{0}} ^{\mathrm{2023}} =\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{1}}{{e}^{\mathrm{2023}} }\right)<\mathrm{2} \\ $$
Commented by mr W last updated on 18/Feb/24
Commented by mr W last updated on 18/Feb/24
![y=(2/(x+e^x )) y′=−((2(1+e^x ))/((x+e^x )^2 ))<0 ⇒y is decreasing y′′=−((2e^x )/((x+e^x )^2 ))+((4(1+e^x )^2 )/((x+e^x )^3 ))=((2[2+4e^x +e^x (e^x −x)])/((x+e^x )^3 ))>0 ⇒y is ⌣ ∫_0 ^(2023) ydx=red area > blue area ...](https://www.tinkutara.com/question/Q204467.png)
$${y}=\frac{\mathrm{2}}{{x}+{e}^{{x}} } \\ $$$${y}'=−\frac{\mathrm{2}\left(\mathrm{1}+{e}^{{x}} \right)}{\left({x}+{e}^{{x}} \right)^{\mathrm{2}} }<\mathrm{0}\: \\ $$$$\Rightarrow{y}\:{is}\:{decreasing} \\ $$$${y}''=−\frac{\mathrm{2}{e}^{{x}} }{\left({x}+{e}^{{x}} \right)^{\mathrm{2}} }+\frac{\mathrm{4}\left(\mathrm{1}+{e}^{{x}} \right)^{\mathrm{2}} }{\left({x}+{e}^{{x}} \right)^{\mathrm{3}} }=\frac{\mathrm{2}\left[\mathrm{2}+\mathrm{4}{e}^{{x}} +{e}^{{x}} \left({e}^{{x}} −{x}\right)\right]}{\left({x}+{e}^{{x}} \right)^{\mathrm{3}} }>\mathrm{0} \\ $$$$\Rightarrow{y}\:{is}\:\smile \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2023}} {ydx}={red}\:{area}\:>\:{blue}\:{area} \\ $$$$… \\ $$
Answered by witcher3 last updated on 16/Feb/24
![⌊x+y]≥[x]+[y] [∫_0 ^(2023) ((2dx)/(x+e^x ))]=⌊Σ_(k=0) ^(2022) ∫_k ^(k+1) ((2dx)/(x+e^x ))⌋≤Σ_(k=0) ^(2022) [∫_k ^(k+1) ((2dx)/(x+e^x ))] Σ_(k=0) ^(2022) ⌊∫_k ^(k+1) (2/(x+e^x ))dx⌋=[∫_0 ^1 ((2dx)/(x+e^x ))]+Σ_(k=1) ^(2022) [∫_k ^(k+1) ((2dx)/(x+e^x ))] ∫_k ^(k+1) ((2dx)/(x+e^x ))<∫_k ^(k+1) ((2dx)/(k+e^k ))=(2/(k+e^k ))<1;∀k≥1;“e^1 >2” ⇒Σ_(k=1) ^(2022) [∫_k ^(k+1) ((2dx)/(x+e^x ))]=0 ∫_0 ^1 ((2dx)/(x+e^x )) e^x ≥1+x⇒(1/(e^x +x))≤(1/(2x+1))⇒∫_0 ^1 ((2dx)/(x+e^x ))≤∫_0 ^1 ((2dx)/(2x+1))=[ln(2x+1)]_0 ^1 =ln(3) ⇒[∫_0 ^(2023) ((2dx)/(x+e^x ))]≤[ln(3)]=1 ⇒[∫_0 ^(2023) ((2dx)/(x+e^x ))]≤1 ∫_0 ^1 ((2dx)/(x+e^x ))≥∫_0 ^1 2(dx/(1+e^x ))=2ln((2/(1+e^(−1) ))) ∫_1 ^2 ((2dx)/(x+e^x ))≥∫_1 ^2 ((2dx)/(2+e^x ))=ln(((1+2e^(−1) )/(1+2e^(−2) ))) ∫_0 ^2 ((2dx)/(x+e^x ))≥ln(((4(1+2e^(−1) ))/((1+e^(−1) )^2 (1+2e^(−2) )))) 4(1+2e^(−1) )≥e(1+2e^(−1) +e^(−2) )(1+2e^(−2) ) 4+8e^(−1) ≥2e^(−1) +4e^(−2) +2e^(−3) +e+2+e^(−1) 2+5e^(−1) ≥4e^(−2) +2e^(−3) +e e^4 +2+4e−2e^3 −5e^2 ≤0 e^3 (e−2−3e^(−1) )+2+4e−2e^2 <0...True 3e^(−1) >1⇒e−2−3e^(−1) <e−3<0 2+4e−2e^2 =2(1+2e−e^2 ); Tackx→^p 1+2x−x^2 p′(x)=2−2x x≥1 decreade e>2.5⇒p(e)<p(2.5)=1+5−(2.5)^2 =−0.25<0 ⇒ln(((4(1+2e^(−1) ))/((1+e^(−1) )^2 (1+2e^(−2) ))))≥ln(e)=1 ∫_0 ^(2023) ((2dx)/(x+e^x ))≥∫_0 ^2 ((2dx)/(x+e^x ))≥1 ⇒⌊∫_0 ^(2023) ((2dx)/(x+e^x ))⌋≥1⇒enswer =1](https://www.tinkutara.com/question/Q204415.png)
$$\left.\lfloor\mathrm{x}+\mathrm{y}\right]\geqslant\left[\mathrm{x}\right]+\left[\mathrm{y}\right] \\ $$$$\left[\int_{\mathrm{0}} ^{\mathrm{2023}} \frac{\mathrm{2dx}}{\mathrm{x}+\mathrm{e}^{\mathrm{x}} }\right]=\lfloor\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{2022}} {\sum}}\int_{\mathrm{k}} ^{\mathrm{k}+\mathrm{1}} \frac{\mathrm{2dx}}{\mathrm{x}+\mathrm{e}^{\mathrm{x}} }\rfloor\leqslant\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{2022}} {\sum}}\left[\int_{\mathrm{k}} ^{\mathrm{k}+\mathrm{1}} \frac{\mathrm{2dx}}{\mathrm{x}+\mathrm{e}^{\mathrm{x}} }\right] \\ $$$$\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{2022}} {\sum}}\lfloor\int_{\mathrm{k}} ^{\mathrm{k}+\mathrm{1}} \frac{\mathrm{2}}{\mathrm{x}+\mathrm{e}^{\mathrm{x}} \:\:}\mathrm{dx}\rfloor=\left[\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2dx}}{\mathrm{x}+\mathrm{e}^{\mathrm{x}} }\right]+\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{2022}} {\sum}}\left[\int_{\mathrm{k}} ^{\mathrm{k}+\mathrm{1}} \frac{\mathrm{2dx}}{\mathrm{x}+\mathrm{e}^{\mathrm{x}} }\right] \\ $$$$\int_{\mathrm{k}} ^{\mathrm{k}+\mathrm{1}} \frac{\mathrm{2dx}}{\mathrm{x}+\mathrm{e}^{\mathrm{x}} }<\int_{\mathrm{k}} ^{\mathrm{k}+\mathrm{1}} \frac{\mathrm{2dx}}{\mathrm{k}+\mathrm{e}^{\mathrm{k}} }=\frac{\mathrm{2}}{\mathrm{k}+\mathrm{e}^{\mathrm{k}} }<\mathrm{1};\forall\mathrm{k}\geqslant\mathrm{1};“\mathrm{e}^{\mathrm{1}} >\mathrm{2}'' \\ $$$$\Rightarrow\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{2022}} {\sum}}\left[\int_{\mathrm{k}} ^{\mathrm{k}+\mathrm{1}} \frac{\mathrm{2dx}}{\mathrm{x}+\mathrm{e}^{\mathrm{x}} }\right]=\mathrm{0} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2dx}}{\mathrm{x}+\mathrm{e}^{\mathrm{x}} } \\ $$$$\mathrm{e}^{\mathrm{x}} \geqslant\mathrm{1}+\mathrm{x}\Rightarrow\frac{\mathrm{1}}{\mathrm{e}^{\mathrm{x}} +\mathrm{x}}\leqslant\frac{\mathrm{1}}{\mathrm{2x}+\mathrm{1}}\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2dx}}{\mathrm{x}+\mathrm{e}^{\mathrm{x}} }\leqslant\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2dx}}{\mathrm{2x}+\mathrm{1}}=\left[\mathrm{ln}\left(\mathrm{2x}+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} =\mathrm{ln}\left(\mathrm{3}\right) \\ $$$$\Rightarrow\left[\int_{\mathrm{0}} ^{\mathrm{2023}} \frac{\mathrm{2dx}}{\mathrm{x}+\mathrm{e}^{\mathrm{x}} }\right]\leqslant\left[\mathrm{ln}\left(\mathrm{3}\right)\right]=\mathrm{1} \\ $$$$\Rightarrow\left[\int_{\mathrm{0}} ^{\mathrm{2023}} \frac{\mathrm{2dx}}{\mathrm{x}+\mathrm{e}^{\mathrm{x}} }\right]\leqslant\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2dx}}{\mathrm{x}+\mathrm{e}^{\mathrm{x}} }\geqslant\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{2}\frac{\mathrm{dx}}{\mathrm{1}+\mathrm{e}^{\mathrm{x}} }=\mathrm{2ln}\left(\frac{\mathrm{2}}{\mathrm{1}+\mathrm{e}^{−\mathrm{1}} }\right) \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \frac{\mathrm{2dx}}{\mathrm{x}+\mathrm{e}^{\mathrm{x}} }\geqslant\int_{\mathrm{1}} ^{\mathrm{2}} \frac{\mathrm{2dx}}{\mathrm{2}+\mathrm{e}^{\mathrm{x}} }=\mathrm{ln}\left(\frac{\mathrm{1}+\mathrm{2e}^{−\mathrm{1}} }{\mathrm{1}+\mathrm{2e}^{−\mathrm{2}} }\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}} \frac{\mathrm{2dx}}{\mathrm{x}+\mathrm{e}^{\mathrm{x}} }\geqslant\mathrm{ln}\left(\frac{\mathrm{4}\left(\mathrm{1}+\mathrm{2e}^{−\mathrm{1}} \right)}{\left(\mathrm{1}+\mathrm{e}^{−\mathrm{1}} \right)^{\mathrm{2}} \left(\mathrm{1}+\mathrm{2e}^{−\mathrm{2}} \right)}\right) \\ $$$$\mathrm{4}\left(\mathrm{1}+\mathrm{2e}^{−\mathrm{1}} \right)\geqslant\mathrm{e}\left(\mathrm{1}+\mathrm{2e}^{−\mathrm{1}} +\mathrm{e}^{−\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{2e}^{−\mathrm{2}} \right) \\ $$$$\mathrm{4}+\mathrm{8e}^{−\mathrm{1}} \geqslant\mathrm{2e}^{−\mathrm{1}} +\mathrm{4e}^{−\mathrm{2}} +\mathrm{2e}^{−\mathrm{3}} +\mathrm{e}+\mathrm{2}+\mathrm{e}^{−\mathrm{1}} \\ $$$$\mathrm{2}+\mathrm{5e}^{−\mathrm{1}} \geqslant\mathrm{4e}^{−\mathrm{2}} +\mathrm{2e}^{−\mathrm{3}} +\mathrm{e} \\ $$$$\mathrm{e}^{\mathrm{4}} +\mathrm{2}+\mathrm{4e}−\mathrm{2e}^{\mathrm{3}} −\mathrm{5e}^{\mathrm{2}} \leqslant\mathrm{0} \\ $$$$\mathrm{e}^{\mathrm{3}} \left(\mathrm{e}−\mathrm{2}−\mathrm{3e}^{−\mathrm{1}} \right)+\mathrm{2}+\mathrm{4e}−\mathrm{2e}^{\mathrm{2}} <\mathrm{0}…\mathrm{True} \\ $$$$\mathrm{3e}^{−\mathrm{1}} >\mathrm{1}\Rightarrow\mathrm{e}−\mathrm{2}−\mathrm{3e}^{−\mathrm{1}} <\mathrm{e}−\mathrm{3}<\mathrm{0} \\ $$$$\mathrm{2}+\mathrm{4e}−\mathrm{2e}^{\mathrm{2}} =\mathrm{2}\left(\mathrm{1}+\mathrm{2e}−\mathrm{e}^{\mathrm{2}} \right); \\ $$$$\mathrm{Tackx}\overset{\mathrm{p}} {\rightarrow}\mathrm{1}+\mathrm{2x}−\mathrm{x}^{\mathrm{2}} \\ $$$$\mathrm{p}'\left(\mathrm{x}\right)=\mathrm{2}−\mathrm{2x}\: \\ $$$$\mathrm{x}\geqslant\mathrm{1}\:\mathrm{decreade}\:\:\:\mathrm{e}>\mathrm{2}.\mathrm{5}\Rightarrow\mathrm{p}\left(\mathrm{e}\right)<\mathrm{p}\left(\mathrm{2}.\mathrm{5}\right)=\mathrm{1}+\mathrm{5}−\left(\mathrm{2}.\mathrm{5}\right)^{\mathrm{2}} =−\mathrm{0}.\mathrm{25}<\mathrm{0} \\ $$$$\Rightarrow\mathrm{ln}\left(\frac{\mathrm{4}\left(\mathrm{1}+\mathrm{2e}^{−\mathrm{1}} \right)}{\left(\mathrm{1}+\mathrm{e}^{−\mathrm{1}} \right)^{\mathrm{2}} \left(\mathrm{1}+\mathrm{2e}^{−\mathrm{2}} \right)}\right)\geqslant\mathrm{ln}\left(\mathrm{e}\right)=\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2023}} \frac{\mathrm{2dx}}{\mathrm{x}+\mathrm{e}^{\mathrm{x}} }\geqslant\int_{\mathrm{0}} ^{\mathrm{2}} \frac{\mathrm{2dx}}{\mathrm{x}+\mathrm{e}^{\mathrm{x}} }\geqslant\mathrm{1} \\ $$$$\Rightarrow\lfloor\int_{\mathrm{0}} ^{\mathrm{2023}} \frac{\mathrm{2dx}}{\mathrm{x}+\mathrm{e}^{\mathrm{x}} }\rfloor\geqslant\mathrm{1}\Rightarrow\mathrm{enswer}\:=\mathrm{1} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 17/Feb/24
$${right}\:{answer}!\:{thanks}! \\ $$
Commented by witcher3 last updated on 17/Feb/24
$$\mathrm{withe}\:\mathrm{pleasur} \\ $$