Question Number 208129 by efronzo1 last updated on 06/Jun/24
$$\:\:\:\:\:\underbrace{\pm \cancel{} } \\ $$
Answered by mr W last updated on 06/Jun/24
![f(0)=1 f(1)=379 ∫_1 ^(379) f^(−1) (x)dx=379×1−1×0−∫_0 ^1 f(x)dx =379−[x^(304) +x^(74) +x]_0 ^1 =379−3=376](Q208131.png)
$${f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{379} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{379}} {f}^{−\mathrm{1}} \left({x}\right){dx}=\mathrm{379}×\mathrm{1}−\mathrm{1}×\mathrm{0}−\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{379}−\left[{x}^{\mathrm{304}} +{x}^{\mathrm{74}} +{x}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{379}−\mathrm{3}=\mathrm{376} \\ $$
Commented by efronzo1 last updated on 06/Jun/24
$$\: \\ $$$$\:\mathrm{not}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\:\mathrm{376} \\ $$
Commented by MM42 last updated on 06/Jun/24
![∫_a ^b f(x)dx≠∫_(f^(−1) (a)) ^(f^(−1) (b)) f^(−1) (x)dx for example f(x)=x^3 f(1)=1 & f(2)=8 ∫_1 ^2 x^3 dx=(1/4)x^4 ]_1 ^2 =((15)/4) ∫_1 ^8 f^(−1) (x)dx=∫_1 ^8 x^((1/3) ) dx=(3/4)x^(4/3) ]_1 ^8 =(3/4)(16−1)=((45)/4) ⇒∫_1 ^8 x^(1/3) dx≠∫_1 ^2 x^3 dx](Q208141.png)
$$\int_{{a}} ^{{b}} {f}\left({x}\right){dx}\neq\int_{{f}^{−\mathrm{1}} \left({a}\right)} ^{{f}^{−\mathrm{1}} \left({b}\right)} {f}^{−\mathrm{1}} \left({x}\right){dx} \\ $$$${for}\:{example} \\ $$$${f}\left({x}\right)={x}^{\mathrm{3}} \:\:{f}\left(\mathrm{1}\right)=\mathrm{1}\:\:\:\&\:\:{f}\left(\mathrm{2}\right)=\mathrm{8} \\ $$$$\left.\int_{\mathrm{1}} ^{\mathrm{2}} {x}^{\mathrm{3}} {dx}=\frac{\mathrm{1}}{\mathrm{4}}{x}^{\mathrm{4}} \right]_{\mathrm{1}} ^{\mathrm{2}} =\frac{\mathrm{15}}{\mathrm{4}} \\ $$$$\left.\int_{\mathrm{1}} ^{\mathrm{8}} {f}^{−\mathrm{1}} \left({x}\right){dx}=\int_{\mathrm{1}} ^{\mathrm{8}} \:{x}^{\frac{\mathrm{1}}{\mathrm{3}}\:} {dx}=\frac{\mathrm{3}}{\mathrm{4}}{x}^{\frac{\mathrm{4}}{\mathrm{3}}} \right]_{\mathrm{1}} ^{\mathrm{8}} =\frac{\mathrm{3}}{\mathrm{4}}\left(\mathrm{16}−\mathrm{1}\right)=\frac{\mathrm{45}}{\mathrm{4}} \\ $$$$\Rightarrow\int_{\mathrm{1}} ^{\mathrm{8}} {x}^{\frac{\mathrm{1}}{\mathrm{3}}} {dx}\neq\int_{\mathrm{1}} ^{\mathrm{2}} {x}^{\mathrm{3}} {dx} \\ $$$$ \\ $$
Commented by mr W last updated on 06/Jun/24
$${i}\:{have}\:{corrected}. \\ $$
Commented by mr W last updated on 06/Jun/24
Commented by mr W last updated on 06/Jun/24
$$\int_{{c}} ^{{d}} {f}^{−\mathrm{1}} \left({x}\right){dx}+\int_{{a}} ^{{b}} {f}\left({x}\right){dx}={bd}−{ac} \\ $$
Commented by MM42 last updated on 06/Jun/24
$$\:\cancel{\lesseqgtr} \\ $$
Answered by efronzo1 last updated on 06/Jun/24
![let f^(−1) (x) = u ⇒x = f(u) dx = f^′ (u) du = (304)(303)u^(302) +(74)(73)u^(72) du x=1⇒f(u)=1 ; u = 0 x=379⇒f(u)=379 ; u=1 I = ∫_1 ^(379) f^(−1) (x)dx = ∫_0 ^1 u ((304)(303)u^(302) +(74)(73)u^(72) )du = ∫_0 ^1 (304)(303)u^(303) +(74)(73)u^(73) du = [ 303 u^(304) + 73 u^(74) ]_0 ^1 = 303 + 73 = 376](Q208138.png)
$$\:\mathrm{let}\:\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right)\:=\:\mathrm{u}\:\Rightarrow\mathrm{x}\:=\:\mathrm{f}\left(\mathrm{u}\right)\: \\ $$$$\:\mathrm{dx}\:=\:\mathrm{f}\:^{'} \left(\mathrm{u}\right)\:\mathrm{du}\:=\:\left(\mathrm{304}\right)\left(\mathrm{303}\right)\mathrm{u}^{\mathrm{302}} +\left(\mathrm{74}\right)\left(\mathrm{73}\right)\mathrm{u}^{\mathrm{72}} \:\mathrm{du} \\ $$$$\:\:\mathrm{x}=\mathrm{1}\Rightarrow\mathrm{f}\left(\mathrm{u}\right)=\mathrm{1}\:;\:\mathrm{u}\:=\:\mathrm{0} \\ $$$$\:\mathrm{x}=\mathrm{379}\Rightarrow\mathrm{f}\left(\mathrm{u}\right)=\mathrm{379}\:;\:\mathrm{u}=\mathrm{1} \\ $$$$\:\mathrm{I}\:=\:\underset{\mathrm{1}} {\overset{\mathrm{379}} {\int}}\:\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right)\mathrm{dx}\:=\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\mathrm{u}\:\left(\left(\mathrm{304}\right)\left(\mathrm{303}\right)\mathrm{u}^{\mathrm{302}} +\left(\mathrm{74}\right)\left(\mathrm{73}\right)\mathrm{u}^{\mathrm{72}} \right)\mathrm{du} \\ $$$$\:\:\:=\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left(\mathrm{304}\right)\left(\mathrm{303}\right)\mathrm{u}^{\mathrm{303}} +\left(\mathrm{74}\right)\left(\mathrm{73}\right)\mathrm{u}^{\mathrm{73}} \:\mathrm{du} \\ $$$$\:\:\:=\:\left[\:\mathrm{303}\:\mathrm{u}^{\mathrm{304}} \:+\:\mathrm{73}\:\mathrm{u}^{\mathrm{74}} \:\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\:\:\:=\:\mathrm{303}\:+\:\mathrm{73}\:=\:\mathrm{376}\: \\ $$$$ \\ $$