Question Number 159938 by ZiYangLee last updated on 22/Nov/21
$$\mathrm{Evaluate}\:\int_{\mathrm{1}} ^{\:\mathrm{4}} \sqrt{\frac{{x}−\mathrm{1}}{{x}^{\mathrm{5}} }}\:{dx}. \\ $$
Answered by Ar Brandon last updated on 22/Nov/21
![I=∫_1 ^4 (√((x−1)/x^5 ))dx=∫_1 ^4 (1/x^2 )(√(1−(1/x)))dx u=1−(1/x)⇒du=(1/x^2 )dx I=∫_0 ^(3/4) (√u)du=(2/3)[u^(3/2) ]_0 ^(3/4) =(2/3)(((√3)/2))^3 =((√3)/4)](https://www.tinkutara.com/question/Q159948.png)
$${I}=\int_{\mathrm{1}} ^{\mathrm{4}} \sqrt{\frac{{x}−\mathrm{1}}{{x}^{\mathrm{5}} }}{dx}=\int_{\mathrm{1}} ^{\mathrm{4}} \frac{\mathrm{1}}{{x}^{\mathrm{2}} }\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}}{dx} \\ $$$${u}=\mathrm{1}−\frac{\mathrm{1}}{{x}}\Rightarrow{du}=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{dx} \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\mathrm{3}}{\mathrm{4}}} \sqrt{{u}}{du}=\frac{\mathrm{2}}{\mathrm{3}}\left[{u}^{\frac{\mathrm{3}}{\mathrm{2}}} \right]_{\mathrm{0}} ^{\frac{\mathrm{3}}{\mathrm{4}}} =\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{3}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$