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Question Number 31145 by Joel578 last updated on 03/Mar/18

$$\mathrm{Given}$$$${\int }_0^{1}f\left(x\right)dx=\left(\begin{array}{c}2018\\ 0\end{array}\right)+\frac{1}{2}\left(\begin{array}{c}2018\\ 1\end{array}\right)+\frac{1}{3}\left(\begin{array}{c}2018\\ 2\end{array}\right)+...+\frac{1}{2019}\left(\begin{array}{c}2018\\ 2018\end{array}\right)$$$${\int }_0^{1}g\left(x\right)dx=\left(\begin{array}{c}2018\\ 0\end{array}\right)-\frac{1}{2}\left(\begin{array}{c}2018\\ 1\end{array}\right)+\frac{1}{3}\left(\begin{array}{c}2018\\ 2\end{array}\right)-...+\frac{1}{2019}\left(\begin{array}{c}2018\\ 2018\end{array}\right)$$$$h\left(x\right)\mathrm{is}\mathrm{an}\mathrm{odd}\mathrm{function}$$$$\mathrm{Then}\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{value}\mathrm{of}{\int }_{-3}^{3}f\left(x\right).g\left(x\right).h\left(x\right)dx?$$

Answered by ajfour last updated on 03/Mar/18

$$f\left(x\right)={(1+x)}^{2018},g\left(x\right)={(1-x)}^{2018}$$$$\Rightarrow f\left(x\right)g\left(x\right)={(1-x^2)}^{2018}\left(evenfunction\right)$$$$sof\left(x\right)g\left(x\right)h\left(x\right)isodd$$$$\Rightarrow {\int }_{-3}^{3}f\left(x\right)g\left(x\right)h\left(x\right)dx=0.$$

Commented by Joel578 last updated on 04/Mar/18

$$Thankyouverymuch$$

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