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Question Number 213114 by efronzo1 last updated on 30/Oct/24

$$\underset{―}{\mathbf{÷}}$$

Answered by issac last updated on 30/Oct/24

$$f\left(x\right)=-C(x-1)+\frac{\mathit{i}}{\pi }{(x-1)}^{5000}\ln (x-1)$$$$(\mathrm{thx}\mathrm{wolfram}\mathrm{alpha}!!)$$$$f\left(1\right)=0,\mathrm{cus}f(1+0)+f(1-0)=0^{5000}=0$$$$f^{\left(1\right)}(x+1)-f^{\left(1\right)}(1-x)=5000x^{4999}$$$$......$$$$??????????$$$$\mathrm{Damn}\mathrm{i}\mathrm{was}\mathrm{fooled}\mathrm{XD}$$$${\int }_0^{1}f(x+1)\mathrm{d}x+{\int }_0^{1}f(1-x)\mathrm{d}x={\int }_0^1{x}^{5000}\mathrm{d}x$$$${\int }_1^{2}f\left(u\right)-{\int }_1^{0}f\left(u\right)\mathrm{d}u={\int }_0^{2}f\left(u\right)\mathrm{d}u=\frac{1}{5001}$$$$\mathrm{Explain}.$$$$\mathrm{integral}\mathrm{both}\mathrm{side}$$$${\int }_0^{1}f(x+1)\mathrm{d}x+{\int }_0^{1}f(1-x)\mathrm{d}x={\int }_0^1{x}^{5000}\mathrm{d}x$$$${\int }_0^{1}f(x+1)\mathrm{d}x={\int }_1^{2}f\left(u\right)\mathrm{d}u$$$$\mathrm{cus}\mathrm{substitude}x+1=u$$$$\mathrm{d}u=\mathrm{d}x$$$$\mathrm{similarly}{\int }_0^{1}f(1-x)\mathrm{d}x={\int }_0^{1}f\left(u\right)\mathrm{d}u$$$$1-x=u\mathrm{d}u=-\mathrm{d}x$$$${\int }_a^b+{\int }_b^c={\int }_a^c$$$${\int }_0^{2}f\left(u\right)\mathrm{d}u={\int }_0^1{x}^{5000}\mathrm{d}x=\frac{1}{5001}$$$$\mathcal{Y}eah!!$$

Answered by mr W last updated on 30/Oct/24

$${\int }_0^{2}f\left(x\right)dx$$$$={\int }_0^{1}f\left(x\right)dx+{\int }_1^{2}f\left(x\right)dx$$$$={\int }_{-1}^{0}f(1+t)d(1+t)+{\int }_0^{-1}f(1-t)d(1-t)$$$$={\int }_{-1}^{0}f(1+t)dt+{\int }_{-1}^{0}f(1-t)dt$$$$={\int }_{-1}^0[f(1+t)+f(1-t)]dt$$$$={\int }_{-1}^0{t}^{5000}dt$$$$={\left[\frac{t^{5001}}{5001}\right]}_{-1}^0=\frac{1}{5001}$$

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