Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 169559 by TOTTI last updated on 03/May/22

Commented by cortano1 last updated on 03/May/22

2^(nd) way determinant ((x,((1−x)^(1/2) )),(1,(−(2/3)(1−x)^(3/2) )),(0,(−(4/(15))(1−x)^(5/2) ))) I=∫ x (√(1−x)) dx = −(2/3)x (√((1−x)^3 )) +(4/(15)) (√((1−x)^5 )) + c

$$\:\mathrm{2}^{{nd}} \:{way}\: \\ $$$$\:\:\begin{array}{|c|c|c|}{{x}}&\hline{\left(\mathrm{1}−{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }\\{\mathrm{1}}&\hline{−\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}−{x}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\\{\mathrm{0}}&\hline{−\frac{\mathrm{4}}{\mathrm{15}}\left(\mathrm{1}−{x}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} }\\\hline\end{array} \\ $$$$\:{I}=\int\:{x}\:\sqrt{\mathrm{1}−{x}}\:{dx}\:=\:−\frac{\mathrm{2}}{\mathrm{3}}{x}\:\sqrt{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }\:+\frac{\mathrm{4}}{\mathrm{15}}\:\sqrt{\left(\mathrm{1}−{x}\right)^{\mathrm{5}} }\:+\:{c} \\ $$

Answered by greougoury555 last updated on 03/May/22

∫ x(√(1−x)) dx let (√(1−x)) = u → { ((x=1−u^2 )),((dx=−2u du)) :} I=∫ (1−u^2 )(u)(−2u)du =−2∫(1−u^2 )u^2 du =−2∫(u^2 −u^4 )du =−2[ (1/3)u^3 −(1/5)u^5 ]+c =−(2/(15))u^3 (5−3u^2 )+c =−(2/(15)) (√((1−x)^3 )) (5−3(1−x))+c =−(2/(15)) (3x+2)(√((1−x)^3 )) + c

$$\:\int\:{x}\sqrt{\mathrm{1}−{x}}\:{dx}\: \\ $$$$\:{let}\:\sqrt{\mathrm{1}−{x}}\:=\:{u}\:\rightarrow\begin{cases}{{x}=\mathrm{1}−{u}^{\mathrm{2}} }\\{{dx}=−\mathrm{2}{u}\:{du}}\end{cases} \\ $$$$\:{I}=\int\:\left(\mathrm{1}−{u}^{\mathrm{2}} \right)\left({u}\right)\left(−\mathrm{2}{u}\right){du} \\ $$$$\:\:=−\mathrm{2}\int\left(\mathrm{1}−{u}^{\mathrm{2}} \right){u}^{\mathrm{2}} \:{du} \\ $$$$\:\:=−\mathrm{2}\int\left({u}^{\mathrm{2}} −{u}^{\mathrm{4}} \right){du} \\ $$$$\:\:=−\mathrm{2}\left[\:\frac{\mathrm{1}}{\mathrm{3}}{u}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{5}}{u}^{\mathrm{5}} \:\right]+{c} \\ $$$$\:\:=−\frac{\mathrm{2}}{\mathrm{15}}{u}^{\mathrm{3}} \left(\mathrm{5}−\mathrm{3}{u}^{\mathrm{2}} \right)+{c} \\ $$$$\:=−\frac{\mathrm{2}}{\mathrm{15}}\:\sqrt{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }\:\left(\mathrm{5}−\mathrm{3}\left(\mathrm{1}−{x}\right)\right)+{c} \\ $$$$\:=−\frac{\mathrm{2}}{\mathrm{15}}\:\left(\mathrm{3}{x}+\mathrm{2}\right)\sqrt{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }\:+\:{c} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com