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 136906 by leena12345 last updated on 27/Mar/21

$$\underset{9}{\stackrel{\mathrm{\infty }}{\int }}\frac{1}{{(x-8)}^{\frac{3}{2}}}dx$$

Answered by Dwaipayan Shikari last updated on 27/Mar/21

$$=-2{\left[{(x-8)}^{-\frac{1}{2}}\right]}_9^{\mathrm{\infty }}=-2(-1)=2$$

Answered by Mathspace last updated on 27/Mar/21

$$I={\int }_9^{\mathrm{\infty }}\frac{dx}{{(x-8)}^{\frac{3}{2}}}{=}_{x-8=t}{\int }_1^{\mathrm{\infty }}\frac{dt}{t^{\frac{3}{2}}}$$$$={\int }_1^{\mathrm{\infty }}{t}^{-\frac{3}{2}}dt={\left[\frac{1}{-\frac{3}{2}+1}{t}^{-\frac{3}{2}+1}\right]}_1^{\mathrm{\infty }}$$$$={[-2t^{-\frac{1}{2}}]}_1^{\mathrm{\infty }}=-2{\left[\frac{1}{\sqrt{t}}\right]}_1^{\mathrm{\infty }}$$$$=-2(-1)=2$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com