Question Number 213114 by efronzo1 last updated on 30/Oct/24
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Answered by issac last updated on 30/Oct/24
$$\:\:\:{f}\left({x}\right)=−{C}\left({x}−\mathrm{1}\right)+\frac{\boldsymbol{{i}}}{\pi}\left({x}−\mathrm{1}\right)^{\mathrm{5000}} \mathrm{ln}\left({x}−\mathrm{1}\right) \\ $$$$\left(\mathrm{thx}\:\mathrm{wolfram}\:\mathrm{alpha}!!\right) \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{0}\:,\:\mathrm{cus}\:\:{f}\left(\mathrm{1}+\mathrm{0}\right)+{f}\left(\mathrm{1}−\mathrm{0}\right)=\mathrm{0}^{\mathrm{5000}} =\mathrm{0} \\ $$$${f}^{\left(\mathrm{1}\right)} \left({x}+\mathrm{1}\right)−{f}^{\left(\mathrm{1}\right)} \left(\mathrm{1}−{x}\right)=\mathrm{5000}{x}^{\mathrm{4999}} \\ $$$$…… \\ $$$$??????????\: \\ $$$$\mathrm{Damn}\:\mathrm{i}\:\mathrm{was}\:\mathrm{fooled}\:\mathrm{XD} \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \:{f}\left({x}+\mathrm{1}\right)\mathrm{d}{x}+\int_{\mathrm{0}} ^{\:\mathrm{1}} \:{f}\left(\mathrm{1}−{x}\right)\mathrm{d}{x}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \:{x}^{\mathrm{5000}} \mathrm{d}{x} \\ $$$$\int_{\mathrm{1}} ^{\:\mathrm{2}} \:{f}\left({u}\right)−\int_{\mathrm{1}} ^{\:\mathrm{0}} \:{f}\left({u}\right)\mathrm{d}{u}=\int_{\mathrm{0}} ^{\:\mathrm{2}} \:{f}\left({u}\right)\mathrm{d}{u}=\frac{\mathrm{1}}{\mathrm{5001}} \\ $$$$\mathrm{Explain}. \\ $$$$\mathrm{integral}\:\mathrm{both}\:\mathrm{side} \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} {f}\left({x}+\mathrm{1}\right)\mathrm{d}{x}+\int_{\mathrm{0}} ^{\:\mathrm{1}} {f}\left(\mathrm{1}−{x}\right)\mathrm{d}{x}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \:{x}^{\mathrm{5000}} \mathrm{d}{x} \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \:{f}\left({x}+\mathrm{1}\right)\mathrm{d}{x}=\int_{\mathrm{1}} ^{\:\mathrm{2}} \:{f}\left({u}\right)\mathrm{d}{u} \\ $$$$\mathrm{cus}\:\mathrm{substitude}\:{x}+\mathrm{1}={u} \\ $$$$\mathrm{d}{u}=\mathrm{d}{x} \\ $$$$\mathrm{similarly}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{f}\left(\mathrm{1}−{x}\right)\mathrm{d}{x}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \:{f}\left({u}\right)\mathrm{d}{u} \\ $$$$\mathrm{1}−{x}={u}\:\:\:\mathrm{d}{u}=−\mathrm{d}{x} \\ $$$$\int_{{a}} ^{\:{b}} +\int_{\:{b}} ^{\:{c}} \:=\int_{\:{a}} ^{\:{c}} \:\: \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{2}} \:{f}\left({u}\right)\mathrm{d}{u}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \:{x}^{\mathrm{5000}} \mathrm{d}{x}=\frac{\mathrm{1}}{\mathrm{5001}} \\ $$$$\mathcal{Y}{eah}!! \\ $$
Answered by mr W last updated on 30/Oct/24
![∫_0 ^2 f(x)dx =∫_0 ^1 f(x)dx+∫_1 ^2 f(x)dx =∫_(−1) ^0 f(1+t)d(1+t)+∫_0 ^(−1) f(1−t)d(1−t) =∫_(−1) ^0 f(1+t)dt+∫_(−1) ^0 f(1−t)dt =∫_(−1) ^0 [f(1+t)+f(1−t)]dt =∫_(−1) ^0 t^(5000) dt =[(t^(5001) /(5001))]_(−1) ^0 =(1/(5001))](https://www.tinkutara.com/question/Q213117.png)
$$\int_{\mathrm{0}} ^{\mathrm{2}} {f}\left({x}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}+\int_{\mathrm{1}} ^{\mathrm{2}} {f}\left({x}\right){dx} \\ $$$$=\int_{−\mathrm{1}} ^{\mathrm{0}} {f}\left(\mathrm{1}+{t}\right){d}\left(\mathrm{1}+{t}\right)+\int_{\mathrm{0}} ^{−\mathrm{1}} {f}\left(\mathrm{1}−{t}\right){d}\left(\mathrm{1}−{t}\right) \\ $$$$=\int_{−\mathrm{1}} ^{\mathrm{0}} {f}\left(\mathrm{1}+{t}\right){dt}+\int_{−\mathrm{1}} ^{\mathrm{0}} {f}\left(\mathrm{1}−{t}\right){dt} \\ $$$$=\int_{−\mathrm{1}} ^{\mathrm{0}} \left[{f}\left(\mathrm{1}+{t}\right)+{f}\left(\mathrm{1}−{t}\right)\right]{dt} \\ $$$$=\int_{−\mathrm{1}} ^{\mathrm{0}} {t}^{\mathrm{5000}} {dt} \\ $$$$=\left[\frac{{t}^{\mathrm{5001}} }{\mathrm{5001}}\right]_{−\mathrm{1}} ^{\mathrm{0}} =\frac{\mathrm{1}}{\mathrm{5001}} \\ $$