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 124675 by sdfg last updated on 05/Dec/20

$${\int }_1^3{(x-1)}^3{(3-x)}^{2}dx$$

Commented by mohammad17 last updated on 05/Dec/20

$$put:m=x-1\Rightarrow x=m+1\Rightarrow dx=dm$$$$$$$$x=3\Rightarrow m=2,x=1\Rightarrow m=0$$$$$$$$\therefore {\int }_1^3{(x-1)}^3{(3-x)}^{2}dx={\int }_0^2{m}^3{(2-m)}^{2}dm$$$$$$$$={\int }_0^2(4m^3-4m^4+m^5)dm$$$$$$$$={(m^4-\frac{4m^5}{5}+\frac{m^6}{6})}_0^2=16-\frac{128}{5}+\frac{64}{6}=\frac{16}{15}$$$$$$$$\left(mohammadAldolaimy\right)$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com