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Question Number 67197 by necxxx last updated on 23/Aug/19

$${\int }_0^2{x}^5{(1-\frac{x}{2})}^{4}dx$$

Answered by turbo msup by abdo last updated on 24/Aug/19

$$I=\frac{1}{16}{\int }_0^2{x}^5{(x-2)}^{4}dx$$$$=\frac{1}{16}{\int }_0^2{x}^5\sum _{k=0}^4{C}_4^k{x}^k{(-2)}^{4-k}$$$$16I=\sum _{k=0}^4{C}_4^k{(-2)}^{4-k}{\int }_0^2{x}^{k+5}dx$$$$=16\sum _{k=0}^4{C}_4^k{(-2)}^{-k}{\left[\frac{1}{k+6}{x}^{k+6}\right]}_0^2$$$$\Rightarrow I={\sum ^4}_{k=0}{(-2)}^{-k}{C}_4^k\times \frac{2^{k+6}}{k+6}$$

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