Question Number 183112 by CrispyXYZ last updated on 20/Dec/22
$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\frac{\partial}{\partial{x}}\:\underset{\mathrm{0}} {\overset{{x}} {\int}}{f}\left({s}\right)\mathrm{d}{s}={f}\left({x}\right) \\ $$
Answered by Emrice last updated on 20/Dec/22
$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\frac{\partial}{\partial{x}}\:\underset{\mathrm{0}} {\overset{{x}} {\int}}{f}\left({s}\right)\mathrm{d}{s}={f}\left({x}\right) \\ $$$${on}\:{sait}\:{que} \\ $$$$\:\int_{{a}} ^{{b}} \:{f}\left({x}\right){dx}=\int_{{a}} ^{{b}} {f}\left({a}+{b}−{x}\right){dx} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{{x}\:} {f}\left({s}\right){ds}=\int_{\mathrm{0}} ^{{x}} {f}\left(\mathrm{0}+{x}−{s}\right){ds} \\ $$$$\Rightarrow\frac{\partial}{\partial{x}}\int_{\mathrm{0}} ^{{x}} {f}\left({s}\right){ds}=\int_{\mathrm{0}} ^{{x}} \frac{\partial{f}\left({x}−{s}\right)}{\partial{x}}{ds} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\int_{\mathrm{0}} ^{{x}} {f}'\left({x}−{s}\right){ds} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\int_{\mathrm{0}} ^{{x}} {f}'\left({s}\right){ds} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{f}\left({x}\right) \\ $$$$ \\ $$
Answered by mr W last updated on 21/Dec/22
$${say}\:\frac{{dF}\left({x}\right)}{{dx}}={f}\left({x}\right) \\ $$$${then}\:\int_{\mathrm{0}} ^{{x}} {f}\left({s}\right){ds}={F}\left({x}\right)−{F}\left(\mathrm{0}\right) \\ $$$$\frac{{d}}{{dx}}\int_{\mathrm{0}} ^{{x}} {f}\left({s}\right){ds}=\frac{{dF}\left({x}\right)}{{dx}}−\frac{{dF}\left(\mathrm{0}\right)}{{dx}}={f}\left({x}\right)\:\checkmark \\ $$