Question Number 180388 by universe last updated on 11/Nov/22
Answered by aleks041103 last updated on 12/Nov/22
![∫_0 ^3 f^(−1) (x)dx=∫_(f^(−1) (0)) ^(f^(−1) (3)) f^(−1) (f(y))d(f(y))= =∫_(f^(−1) (0)) ^(f^(−1) (3)) y d(f(y))=[yf(y)]_(f^(−1) (0)) ^(f^(−1) (3)) −∫_(f^(−1) (0)) ^(f^(−1) (3)) f(y)dy= =3f^(−1) (3)−∫_(f^(−1) (0)) ^(f^(−1) (3)) f(x)dx f:[0,∞)→[0,∞) f:surjective ⇒∃b:f(b)=0 but if b>0, then since f is strictly increasing f(b)>f(0)⇒0>f(0)≥0→contradiction ⇒b=0⇒f(0)=0 ⇒s[f]=∫_0 ^5 f(x)dx+∫_0 ^3 f^(−1) (x)dx=3f^(−1) (3)+∫_(f^(−1) (3)) ^( 5) f(x)dx let f^(−1) (3)<5⇒s[f]>3f^(−1) (3)+(5−f^(−1) (3))f(f^(−1) (3))=15 let f^(−1) (3)>5⇒s[f]=3f^(−1) (3)−∫_5 ^(f^(−1) (3)) f(x)dx ⇒s[f]>3f^(−1) (3)−(f^(−1) (3)−5)f(f^(−1) (3))=15 if f^(−1) (3)=5: s[f]=3.5−0=15 ⇒a=15](https://www.tinkutara.com/question/Q180404.png)
$$\int_{\mathrm{0}} ^{\mathrm{3}} {f}^{−\mathrm{1}} \left({x}\right){dx}=\int_{{f}^{−\mathrm{1}} \left(\mathrm{0}\right)} ^{{f}^{−\mathrm{1}} \left(\mathrm{3}\right)} {f}^{−\mathrm{1}} \left({f}\left({y}\right)\right){d}\left({f}\left({y}\right)\right)= \\ $$$$=\int_{{f}^{−\mathrm{1}} \left(\mathrm{0}\right)} ^{{f}^{−\mathrm{1}} \left(\mathrm{3}\right)} {y}\:{d}\left({f}\left({y}\right)\right)=\left[{yf}\left({y}\right)\right]_{{f}^{−\mathrm{1}} \left(\mathrm{0}\right)} ^{{f}^{−\mathrm{1}} \left(\mathrm{3}\right)} −\int_{{f}^{−\mathrm{1}} \left(\mathrm{0}\right)} ^{{f}^{−\mathrm{1}} \left(\mathrm{3}\right)} {f}\left({y}\right){dy}= \\ $$$$=\mathrm{3}{f}^{−\mathrm{1}} \left(\mathrm{3}\right)−\int_{{f}^{−\mathrm{1}} \left(\mathrm{0}\right)} ^{{f}^{−\mathrm{1}} \left(\mathrm{3}\right)} {f}\left({x}\right){dx} \\ $$$${f}:\left[\mathrm{0},\infty\right)\rightarrow\left[\mathrm{0},\infty\right) \\ $$$${f}:{surjective} \\ $$$$\Rightarrow\exists{b}:{f}\left({b}\right)=\mathrm{0} \\ $$$${but}\:{if}\:{b}>\mathrm{0},\:{then}\:{since}\:{f}\:{is}\:{strictly}\:{increasing} \\ $$$${f}\left({b}\right)>{f}\left(\mathrm{0}\right)\Rightarrow\mathrm{0}>{f}\left(\mathrm{0}\right)\geqslant\mathrm{0}\rightarrow{contradiction} \\ $$$$\Rightarrow{b}=\mathrm{0}\Rightarrow{f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\Rightarrow{s}\left[{f}\right]=\int_{\mathrm{0}} ^{\mathrm{5}} {f}\left({x}\right){dx}+\int_{\mathrm{0}} ^{\mathrm{3}} {f}^{−\mathrm{1}} \left({x}\right){dx}=\mathrm{3}{f}^{−\mathrm{1}} \left(\mathrm{3}\right)+\int_{{f}^{−\mathrm{1}} \left(\mathrm{3}\right)} ^{\:\mathrm{5}} {f}\left({x}\right){dx} \\ $$$${let}\:{f}^{−\mathrm{1}} \left(\mathrm{3}\right)<\mathrm{5}\Rightarrow{s}\left[{f}\right]>\mathrm{3}{f}^{−\mathrm{1}} \left(\mathrm{3}\right)+\left(\mathrm{5}−{f}^{−\mathrm{1}} \left(\mathrm{3}\right)\right){f}\left({f}^{−\mathrm{1}} \left(\mathrm{3}\right)\right)=\mathrm{15} \\ $$$${let}\:{f}^{−\mathrm{1}} \left(\mathrm{3}\right)>\mathrm{5}\Rightarrow{s}\left[{f}\right]=\mathrm{3}{f}^{−\mathrm{1}} \left(\mathrm{3}\right)−\int_{\mathrm{5}} ^{{f}^{−\mathrm{1}} \left(\mathrm{3}\right)} {f}\left({x}\right){dx} \\ $$$$\Rightarrow{s}\left[{f}\right]>\mathrm{3}{f}^{−\mathrm{1}} \left(\mathrm{3}\right)−\left({f}^{−\mathrm{1}} \left(\mathrm{3}\right)−\mathrm{5}\right){f}\left({f}^{−\mathrm{1}} \left(\mathrm{3}\right)\right)=\mathrm{15} \\ $$$${if}\:{f}^{−\mathrm{1}} \left(\mathrm{3}\right)=\mathrm{5}:\:{s}\left[{f}\right]=\mathrm{3}.\mathrm{5}−\mathrm{0}=\mathrm{15} \\ $$$$\Rightarrow{a}=\mathrm{15} \\ $$
Commented by universe last updated on 12/Nov/22
$${thank}\:{you}\:{sir} \\ $$