Question Number 208235 by efronzo1 last updated on 08/Jun/24
Answered by som(math1967) last updated on 08/Jun/24
![here f(x)=f^(−1) (x) ∫_2 ^4 ((1−x)/(1+x))dx =∫_2 ^4 ((2dx)/(1+x)) −∫_2 ^4 ((1+x)/(1+x))dx =[2ln(1+x)−x]_2 ^4 =2ln5−4−2ln3+2 =2ln(5/3) −2](https://www.tinkutara.com/question/Q208236.png)
$$\:{here}\:{f}\left({x}\right)={f}^{−\mathrm{1}} \left({x}\right) \\ $$$$\:\underset{\mathrm{2}} {\overset{\mathrm{4}} {\int}}\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}{dx} \\ $$$$=\underset{\mathrm{2}} {\overset{\mathrm{4}} {\int}}\frac{\mathrm{2}{dx}}{\mathrm{1}+{x}}\:−\underset{\mathrm{2}} {\overset{\mathrm{4}} {\int}}\frac{\mathrm{1}+{x}}{\mathrm{1}+{x}}{dx} \\ $$$$=\left[\mathrm{2}{ln}\left(\mathrm{1}+{x}\right)−{x}\right]_{\mathrm{2}} ^{\mathrm{4}} \\ $$$$=\mathrm{2}{ln}\mathrm{5}−\mathrm{4}−\mathrm{2}{ln}\mathrm{3}+\mathrm{2} \\ $$$$=\mathrm{2}{ln}\frac{\mathrm{5}}{\mathrm{3}}\:−\mathrm{2} \\ $$
Answered by shunmisaki007 last updated on 08/Jun/24
![Let y=f^(−1) (x) f(y)=x=((1−y)/(1+y)) x(1+y)=1−y x+xy=1−y xy+y=1−x y(1+x)=1−x ⇒ y=f^(−1) (x)=((1−x)/(1+x)) ∫_2 ^4 f^(−1) (x)dx=∫_2 ^4 ((1−x)/(1+x))dx=∫_2 ^4 ((1−x+1−1)/(1+x))dx =∫_2 ^4 ((2−x−1)/(1+x))dx=∫_2 ^4 ((2−(x+1))/(1+x))dx =∫_2 ^4 (2/(1+x))dx−∫_2 ^4 ((1+x)/(1+x))dx =[2ln(1+x)]_2 ^4 −[x]_2 ^4 =(2ln(1+4)−2ln(1+2))−(4−2) =2(ln(5)−ln(3))−2 ∴ ∫_2 ^4 f^(−1) (x)dx=2ln((5/3))−2≈−0.987 ★](https://www.tinkutara.com/question/Q208237.png)
$$\mathrm{Let}\:{y}={f}^{−\mathrm{1}} \left({x}\right) \\ $$$$\:\:\:{f}\left({y}\right)={x}=\frac{\mathrm{1}−{y}}{\mathrm{1}+{y}} \\ $$$$\:\:\:{x}\left(\mathrm{1}+{y}\right)=\mathrm{1}−{y} \\ $$$$\:\:\:{x}+{xy}=\mathrm{1}−{y} \\ $$$$\:\:\:{xy}+{y}=\mathrm{1}−{x} \\ $$$$\:\:\:{y}\left(\mathrm{1}+{x}\right)=\mathrm{1}−{x} \\ $$$$\Rightarrow\:{y}={f}^{−\mathrm{1}} \left({x}\right)=\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}} \\ $$$$\:\:\:\underset{\mathrm{2}} {\overset{\mathrm{4}} {\int}}{f}^{−\mathrm{1}} \left({x}\right){dx}=\underset{\mathrm{2}} {\overset{\mathrm{4}} {\int}}\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}{dx}=\underset{\mathrm{2}} {\overset{\mathrm{4}} {\int}}\frac{\mathrm{1}−{x}+\mathrm{1}−\mathrm{1}}{\mathrm{1}+{x}}{dx} \\ $$$$\:\:\:\:\:=\underset{\mathrm{2}} {\overset{\mathrm{4}} {\int}}\frac{\mathrm{2}−{x}−\mathrm{1}}{\mathrm{1}+{x}}{dx}=\underset{\mathrm{2}} {\overset{\mathrm{4}} {\int}}\frac{\mathrm{2}−\left({x}+\mathrm{1}\right)}{\mathrm{1}+{x}}{dx} \\ $$$$\:\:\:\:\:=\underset{\mathrm{2}} {\overset{\mathrm{4}} {\int}}\frac{\mathrm{2}}{\mathrm{1}+{x}}{dx}−\underset{\mathrm{2}} {\overset{\mathrm{4}} {\int}}\frac{\mathrm{1}+{x}}{\mathrm{1}+{x}}{dx} \\ $$$$\:\:\:\:\:=\left[\mathrm{2ln}\left(\mathrm{1}+{x}\right)\right]_{\mathrm{2}} ^{\mathrm{4}} −\left[{x}\right]_{\mathrm{2}} ^{\mathrm{4}} \\ $$$$\:\:\:\:\:=\left(\mathrm{2ln}\left(\mathrm{1}+\mathrm{4}\right)−\mathrm{2ln}\left(\mathrm{1}+\mathrm{2}\right)\right)−\left(\mathrm{4}−\mathrm{2}\right) \\ $$$$\:\:\:\:\:=\mathrm{2}\left(\mathrm{ln}\left(\mathrm{5}\right)−\mathrm{ln}\left(\mathrm{3}\right)\right)−\mathrm{2} \\ $$$$\therefore\:\underset{\mathrm{2}} {\overset{\mathrm{4}} {\int}}{f}^{−\mathrm{1}} \left({x}\right){dx}=\mathrm{2ln}\left(\frac{\mathrm{5}}{\mathrm{3}}\right)−\mathrm{2}\approx−\mathrm{0}.\mathrm{987}\:\bigstar \\ $$$$\:\:\:\: \\ $$