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Question Number 213656 by efronzo1 last updated on 12/Nov/24

Answered by golsendro last updated on 13/Nov/24

$$\left(\mathrm{i}\right)\mathrm{g}(4-\mathrm{x})=-\mathrm{g}\left(\mathrm{x}\right)$$$$\underset{0}{\stackrel{4}{\int }}\mathrm{g}\left(\mathrm{x}\right)\mathrm{dx}=\underset{0}{\stackrel{4}{\int }}\mathrm{g}(4-\mathrm{x})\mathrm{dx}$$$$\underset{0}{\stackrel{4}{\int }}-\mathrm{g}(4-\mathrm{x})\mathrm{dx}=\underset{0}{\stackrel{4}{\int }}\mathrm{g}(4-\mathrm{x})\mathrm{dx}$$$$\mathrm{it}\mathrm{folow}\mathrm{that}\underset{0}{\stackrel{4}{\int }}\mathrm{g}(4-\mathrm{x})\mathrm{dx}=0$$$$\mathrm{and}\underset{0}{\stackrel{4}{\int }}\mathrm{g}\left(\mathrm{x}\right)\mathrm{dx}=0$$$$\left(\mathrm{ii}\right)\mathrm{f}\left({\mathrm{x}}^2\right)=\mathrm{f}\left({(-\mathrm{x})}^2\right),\mathrm{f}\left(\mathrm{x}\right)\mathrm{even}\mathrm{function}$$$$\underset{-4}{\stackrel{4}{\int }}\mathrm{f}\left({\mathrm{x}}^2\right)\mathrm{dx}=2\underset{0}{\stackrel{4}{\int }}\mathrm{f}\left({\mathrm{x}}^2\right)\mathrm{dx}$$$$=2\underset{0}{\stackrel{4}{\int }}({4\mathrm{x}}^3-\mathrm{g}\left(\mathrm{x}\right))\mathrm{dx}$$$$=2\underset{0}{\stackrel{4}{\int }}{4\mathrm{x}}^3\mathrm{dx}-0$$$$=2{\left[{\mathrm{x}}^4\right]}_0^4=2\times 256=512$$

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