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Question Number 70818 by oyemi kemewari last updated on 08/Oct/19

$$\mathrm{what}\mathrm{the}\mathrm{prove}\mathrm{that}$$$${\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}(\mathrm{a}+\mathrm{b}-\mathrm{x})\mathrm{dx}$$

Commented by kaivan.ahmadi last updated on 08/Oct/19

$$u=a+b-x\Rightarrow du=-dx$$$$\{\begin{array}{l}x=a\Rightarrow u=b\\ x=b\Rightarrow u=a\end{array}$$$$\Rightarrow {\int }_a^{b}f(a+b-x)dx={\int }_b^a-f\left(u\right)du={\int }_a^{b}f\left(u\right)du={\int }_a^{b}f\left(x\right)dx$$

Answered by MJS last updated on 08/Oct/19

$$\underset{a}{\stackrel{b}{\int }}f\left(x\right)dx={\left[F\left(x\right)\right]}_a^b=F\left(b\right)-F\left(a\right)$$$$\underset{a}{\stackrel{b}{\int }}f(a+b-x)dx=$$$$[t=a+b-x\Rightarrow dx=-dt]$$$$=-\underset{b}{\stackrel{a}{\int }}f\left(t\right)dt=-{\left[F\left(t\right)\right]}_b^a={\left[F\left(t\right)\right]}_a^b=F\left(b\right)-F\left(a\right)$$

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