prove-that-0-1-1-x-7-1-3-dx-0-1-1-x-3-1-7-dx-

Question Number 76680 by Tony Lin last updated on 29/Dec/19

$${prove}\:{that}: \\ $$$$\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}^{\mathrm{7}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} {dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}^{\mathrm{3}} \right)^{\frac{\mathrm{1}}{\mathrm{7}}} {dx} \\ $$

Commented by mr W last updated on 29/Dec/19

see Q#76232, 76146 ∫_a ^b y(x) dx=∫_c ^d x(y) dy y=(1−x^7 )^(1/3) ⇒y^3 =1−x^7 ⇒x=(1−y^3 )^(1/7) y(a=0)=1=d y(b=1)=0=c ∫_0 ^1 y(x)dx=∫_0 ^1 (1−x^7 )^(1/3) dx =∫_0 ^1 x(y)dy=∫_0 ^1 (1−y^3 )^(1/7) dy =∫_0 ^1 (1−x^3 )^(1/7) dx

$${see}\:{Q}#\mathrm{76232},\:\mathrm{76146} \\ $$$$\int_{{a}} ^{{b}} {y}\left({x}\right)\:{dx}=\int_{{c}} ^{{d}} {x}\left({y}\right)\:{dy} \\ $$$${y}=\left(\mathrm{1}−{x}^{\mathrm{7}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\Rightarrow{y}^{\mathrm{3}} =\mathrm{1}−{x}^{\mathrm{7}} \\ $$$$\Rightarrow{x}=\left(\mathrm{1}−{y}^{\mathrm{3}} \right)^{\frac{\mathrm{1}}{\mathrm{7}}} \\ $$$${y}\left({a}=\mathrm{0}\right)=\mathrm{1}={d} \\ $$$${y}\left({b}=\mathrm{1}\right)=\mathrm{0}={c} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {y}\left({x}\right){dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}^{\mathrm{7}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} {dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {x}\left({y}\right){dy}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{y}^{\mathrm{3}} \right)^{\frac{\mathrm{1}}{\mathrm{7}}} {dy} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}^{\mathrm{3}} \right)^{\frac{\mathrm{1}}{\mathrm{7}}} {dx} \\ $$

Commented by mr W last updated on 29/Dec/19