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Question Number 155488 by SANOGO last updated on 01/Oct/21

Answered by puissant last updated on 01/Oct/21

$$Q={\int }_a^b\frac{f\left(x\right)}{f\left(x\right)+f(a+b-x)}dx\left(1\right)$$$$u=a+b-x\to x=a+b-u\to dx=-du$$$$\Rightarrow Q={\int }_b^a\frac{f(a+b-u)}{f(a+b-u)+f\left(u\right)}(-du)$$$$={\int }_a^b\frac{f(a+b-u)}{f\left(u\right)+f(a+b-u)}du\left(2\right)$$$$\frac{\left(1\right)+\left(2\right)}{2}\Rightarrow Q=\frac{1}{2}{\int }_a^b\frac{f\left(x\right)+f(a+b-x)}{f\left(x\right)+f(a+b-x)}dx$$$$=\frac{1}{2}{\int }_a^{b}dx=\frac{b-a}{2}$$$$Divers:{\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(u\right)du=.....$$$$\because \therefore Q={\int }_a^b\frac{f\left(x\right)}{f\left(x\right)+f(a+b-x)}dx=\frac{b-a}{2}$$

Commented by peter frank last updated on 01/Oct/21

$$\mathrm{good}$$

Commented by SANOGO last updated on 01/Oct/21

$$mercilegeneral$$

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