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Question Number 197190 by universe last updated on 10/Sep/23
∫_0 ^1 ^3 (√(1−x^7 )) dx − ∫^1 _0 ^7 (√(1−x^3 )) dx = ?
$$\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:^{\mathrm{3}} \sqrt{\mathrm{1}−{x}^{\mathrm{7}} }\:{dx}\:−\:\underset{\mathrm{0}} {\int}^{\mathrm{1}} \:^{\mathrm{7}} \sqrt{\mathrm{1}−{x}^{\mathrm{3}} }\:{dx}\:\:=\:\:? \\ $$
Commented by Frix last updated on 10/Sep/23
0
$$\mathrm{0} \\ $$
Commented by universe last updated on 10/Sep/23
solution sir
$${solution}\:{sir} \\ $$$$ \\ $$
Answered by Mathspace last updated on 11/Sep/23
I=∫_0 ^1 (1−x^7 )^(1/3) dx (x=t^(1/7) ) I=(1/7)∫_0 ^1 (1−t)^(1/3) t^((1/7)−1) dt B(p,q)=∫_0 ^1 t^(p−1) (1−t)^(q−1) dt ⇒I=(1/7)∫_0 ^1 (1−t)^((4/3)−1) .t^((1/7)−1) dt =(1/7)B((4/3),(1/7)) J=∫_0 ^1 (1−x^3 )^(1/7) dx (x=t^(1/3) ) J=(1/3)∫_0 ^1 (1−t)^(1/7) t^((1/3)−1) dt =(1/3)∫_0 ^1 (1−t)^((8/7)−1) t^((1/3)−1) dt =(1/3)B((8/7).(1/3)) ⇒ I−J=(1/7)B((4/3),(1/7))−(1/3)B((8/7),(1/3))
$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}^{\mathrm{7}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} {dx}\:\:\:\:\left({x}={t}^{\frac{\mathrm{1}}{\mathrm{7}}} \right) \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{7}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:{t}^{\frac{\mathrm{1}}{\mathrm{7}}−\mathrm{1}} {dt} \\ $$$${B}\left({p},{q}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{p}−\mathrm{1}} \left(\mathrm{1}−{t}\right)^{{q}−\mathrm{1}} {dt} \\ $$$$\Rightarrow{I}=\frac{\mathrm{1}}{\mathrm{7}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}\right)^{\frac{\mathrm{4}}{\mathrm{3}}−\mathrm{1}} .{t}^{\frac{\mathrm{1}}{\mathrm{7}}−\mathrm{1}} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{7}}{B}\left(\frac{\mathrm{4}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{7}}\right) \\ $$$${J}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}^{\mathrm{3}} \right)^{\frac{\mathrm{1}}{\mathrm{7}}} {dx}\:\:\left({x}={t}^{\frac{\mathrm{1}}{\mathrm{3}}} \right) \\ $$$${J}=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}\right)^{\frac{\mathrm{1}}{\mathrm{7}}} \:{t}^{\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}} \:{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}\right)^{\frac{\mathrm{8}}{\mathrm{7}}−\mathrm{1}} {t}^{\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}{B}\left(\frac{\mathrm{8}}{\mathrm{7}}.\frac{\mathrm{1}}{\mathrm{3}}\right)\:\Rightarrow \\ $$$${I}−{J}=\frac{\mathrm{1}}{\mathrm{7}}{B}\left(\frac{\mathrm{4}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{7}}\right)−\frac{\mathrm{1}}{\mathrm{3}}{B}\left(\frac{\mathrm{8}}{\mathrm{7}},\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$

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