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Question-190624




Question Number 190624 by Rupesh123 last updated on 07/Apr/23
Answered by mr W last updated on 07/Apr/23
Commented by mr W last updated on 07/Apr/23
geometrically:  rectangle bf(b)=A_1 +A_2 +A_3   bf(b)=af(a)+∫_a ^b f(x)dx+∫_(f(a)) ^(f(b)) f^(−1) (y)dy  ⇒∫_a ^b f(x)dx+∫_(f(a)) ^(f(b)) f^(−1) (x)dx=bf(b)−af(a)
$${geometrically}: \\ $$$${rectangle}\:{bf}\left({b}\right)={A}_{\mathrm{1}} +{A}_{\mathrm{2}} +{A}_{\mathrm{3}} \\ $$$${bf}\left({b}\right)={af}\left({a}\right)+\int_{{a}} ^{{b}} {f}\left({x}\right){dx}+\int_{{f}\left({a}\right)} ^{{f}\left({b}\right)} {f}^{−\mathrm{1}} \left({y}\right){dy} \\ $$$$\Rightarrow\int_{{a}} ^{{b}} {f}\left({x}\right){dx}+\int_{{f}\left({a}\right)} ^{{f}\left({b}\right)} {f}^{−\mathrm{1}} \left({x}\right){dx}={bf}\left({b}\right)−{af}\left({a}\right) \\ $$
Commented by Rupesh123 last updated on 09/Apr/23
Excellent!
Answered by mehdee42 last updated on 08/Apr/23
f^(−1) (x)=t⇒x=f(t)⇒dx=f ′(t)dt  ⇒∫_(f(a)) ^(f(b))  f^(−1) (x)dx=∫_a ^b  tf ′(t)dt  t=u⇒dt=du    &   f ′(t)dt=dv ⇒f(t)=v  ⇒∫_(f(a)) ^(f(b))  f^(−1) (x)dx=∫_a ^b tf ′(t)=tf(t)∣_a ^b −∫_a ^b  f(t)dt  ⇒∫_a ^b  f(x)dx +∫_(f(a)) ^(f(b))  f^(−1) (x)dx=bf(b)−af(a)
$${f}^{−\mathrm{1}} \left({x}\right)={t}\Rightarrow{x}={f}\left({t}\right)\Rightarrow{dx}={f}\:'\left({t}\right){dt} \\ $$$$\Rightarrow\int_{{f}\left({a}\right)} ^{{f}\left({b}\right)} \:{f}^{−\mathrm{1}} \left({x}\right){dx}=\int_{{a}} ^{{b}} \:{tf}\:'\left({t}\right){dt} \\ $$$${t}={u}\Rightarrow{dt}={du}\:\:\:\:\&\:\:\:{f}\:'\left({t}\right){dt}={dv}\:\Rightarrow{f}\left({t}\right)={v} \\ $$$$\Rightarrow\int_{{f}\left({a}\right)} ^{{f}\left({b}\right)} \:{f}^{−\mathrm{1}} \left({x}\right){dx}=\int_{{a}} ^{{b}} {tf}\:'\left({t}\right)={tf}\left({t}\right)\mid_{{a}} ^{{b}} −\int_{{a}} ^{{b}} \:{f}\left({t}\right){dt} \\ $$$$\Rightarrow\int_{{a}} ^{{b}} \:{f}\left({x}\right){dx}\:+\int_{{f}\left({a}\right)} ^{{f}\left({b}\right)} \:{f}^{−\mathrm{1}} \left({x}\right){dx}={bf}\left({b}\right)−{af}\left({a}\right) \\ $$
Commented by Rupesh123 last updated on 09/Apr/23
Excellent!

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