Question Number 118679 by Algoritm last updated on 19/Oct/20
Answered by 1549442205PVT last updated on 19/Oct/20
![∫_0 ^( 1) (([2x].x)/([2x]+x^2 ))dx=∫_0 ^(1/2) (([2x].x)/([2x]+x^2 ))dx+∫_(1/2) ^1 (([2x].x)/([2x]+x^2 ))dx =∫_(1/2) ^( 1) (x/(1+x^2 ))dx=(1/2)∫((d(x^2 +1))/(x^2 +1))=(1/2)ln(x^2 +1)∣_(1/2) ^1 =(1/2)ln2−ln(√(5/4))=(3/2)ln2−ln(√5) ≈0.2350 Since [2x]=0 for 0≤x<0.5 [2x]=1 for 0.5≤x<1,[2x]=2 for x=1](https://www.tinkutara.com/question/Q118696.png)
$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\left[\mathrm{2x}\right].\mathrm{x}}{\left[\mathrm{2x}\right]+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}=\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\left[\mathrm{2x}\right].\mathrm{x}}{\left[\mathrm{2x}\right]+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}+\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{\left[\mathrm{2x}\right].\mathrm{x}}{\left[\mathrm{2x}\right]+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$=\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\:\mathrm{1}} \frac{\mathrm{x}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{d}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\mid_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln2}−\mathrm{ln}\sqrt{\frac{\mathrm{5}}{\mathrm{4}}}=\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ln2}−\mathrm{ln}\sqrt{\mathrm{5}}\:\approx\mathrm{0}.\mathrm{2350} \\ $$$$\mathrm{Since}\:\left[\mathrm{2x}\right]=\mathrm{0}\:\mathrm{for}\:\mathrm{0}\leqslant\mathrm{x}<\mathrm{0}.\mathrm{5} \\ $$$$\left[\mathrm{2x}\right]=\mathrm{1}\:\mathrm{for}\:\mathrm{0}.\mathrm{5}\leqslant\mathrm{x}<\mathrm{1},\left[\mathrm{2x}\right]=\mathrm{2}\:\mathrm{for}\:\mathrm{x}=\mathrm{1} \\ $$
Commented by Dwaipayan Shikari last updated on 19/Oct/20
$$\frac{\mathrm{3}}{\mathrm{2}}{log}\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}{log}\mathrm{5}=\frac{\mathrm{1}}{\mathrm{2}}{log}\mathrm{8}−\frac{\mathrm{1}}{\mathrm{2}}{log}\mathrm{5}=\frac{\mathrm{1}}{\mathrm{2}}{log}\left(\frac{\mathrm{8}}{\mathrm{5}}\right) \\ $$
Answered by mathmax by abdo last updated on 19/Oct/20
![I =∫_0 ^1 ((x[2x])/(x^2 +[2x]))dx ⇒I =_(2x=t) ∫_0 ^2 (t/2)×(([t])/((t^2 /4)+[t]))×(dt/2) =∫_0 ^2 ((t[t])/(t^2 +4[t]))dt ={ ∫_0 ^1 o dt +∫_1 ^2 (t/(t^2 +4))dt} =(1/2)∫_1 ^2 ((2t)/(t^2 +4))dt =(1/2)[ln(t^2 +4)]_1 ^2 =(1/2){ln(8)−ln5}=(1/2)ln((8/5)) answer B is correct](https://www.tinkutara.com/question/Q118729.png)
$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{x}\left[\mathrm{2x}\right]}{\mathrm{x}^{\mathrm{2}} +\left[\mathrm{2x}\right]}\mathrm{dx}\:\:\Rightarrow\mathrm{I}\:=_{\mathrm{2x}=\mathrm{t}} \:\:\int_{\mathrm{0}} ^{\mathrm{2}} \:\:\frac{\mathrm{t}}{\mathrm{2}}×\frac{\left[\mathrm{t}\right]}{\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{4}}+\left[\mathrm{t}\right]}×\frac{\mathrm{dt}}{\mathrm{2}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}} \:\:\frac{\mathrm{t}\left[\mathrm{t}\right]}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{4}\left[\mathrm{t}\right]}\mathrm{dt}\:=\left\{\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\mathrm{o}\:\mathrm{dt}\:+\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{\mathrm{t}}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{4}}\mathrm{dt}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{\mathrm{2t}}{\mathrm{t}^{\mathrm{2}} +\mathrm{4}}\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{ln}\left(\mathrm{t}^{\mathrm{2}} +\mathrm{4}\right)\right]_{\mathrm{1}} ^{\mathrm{2}} \:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{ln}\left(\mathrm{8}\right)−\mathrm{ln5}\right\}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\frac{\mathrm{8}}{\mathrm{5}}\right) \\ $$$$\mathrm{answer}\:\mathrm{B}\:\mathrm{is}\:\mathrm{correct} \\ $$$$ \\ $$