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Question Number 197190 by universe last updated on 10/Sep/23

$${\int }_0^1{}^3\sqrt{1-x^7}dx-{\underset{0}{\int }}^1{}^7\sqrt{1-x^3}dx=?$$

Commented by Frix last updated on 10/Sep/23

$$0$$

Commented by universe last updated on 10/Sep/23

$$solutionsir$$$$$$

Answered by Mathspace last updated on 11/Sep/23

$$I={\int }_0^1{(1-x^7)}^{\frac{1}{3}}dx(x=t^{\frac{1}{7}})$$$$I=\frac{1}{7}{\int }_0^1{(1-t)}^{\frac{1}{3}}{t}^{\frac{1}{7}-1}dt$$$$B(p,q)={\int }_0^1{t}^{p-1}{(1-t)}^{q-1}dt$$$$\Rightarrow I=\frac{1}{7}{\int }_0^1{(1-t)}^{\frac{4}{3}-1}.t^{\frac{1}{7}-1}dt$$$$=\frac{1}{7}B(\frac{4}{3},\frac{1}{7})$$$$J={\int }_0^1{(1-x^3)}^{\frac{1}{7}}dx(x=t^{\frac{1}{3}})$$$$J=\frac{1}{3}{\int }_0^1{(1-t)}^{\frac{1}{7}}{t}^{\frac{1}{3}-1}dt$$$$=\frac{1}{3}{\int }_0^1{(1-t)}^{\frac{8}{7}-1}{t}^{\frac{1}{3}-1}dt$$$$=\frac{1}{3}B(\frac{8}{7}.\frac{1}{3})\Rightarrow$$$$I-J=\frac{1}{7}B(\frac{4}{3},\frac{1}{7})-\frac{1}{3}B(\frac{8}{7},\frac{1}{3})$$

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