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Question Number 182771 by liuxinnan last updated on 14/Dec/22

$$\int\:_{\mathrm{0}} ^{{r}} \:\frac{\mathrm{1}}{{R}^{\mathrm{2}} −{r}^{\mathrm{2}} }{dr}=? \\$$

Commented by Frix last updated on 14/Dec/22

$$\mathrm{Error}:\:\mathrm{dependent}\:\mathrm{borders}\:\mathrm{are}\:\mathrm{not}\:\mathrm{allowed}. \\$$

Commented by mr W last updated on 14/Dec/22

$${why}\:{not}? \\$$$$\int_{\mathrm{0}} ^{{r}} \frac{\mathrm{1}}{{R}^{\mathrm{2}} −{r}^{\mathrm{2}} }{dr} \\$$$$=\int_{\mathrm{0}} ^{{r}} \frac{\mathrm{1}}{{R}^{\mathrm{2}} −{x}^{\mathrm{2}} }{dx} \\$$$$=\int_{\mathrm{0}} ^{{r}} \frac{\mathrm{1}}{{R}^{\mathrm{2}} −{y}^{\mathrm{2}} }{dy} \\$$$$=\int_{\mathrm{0}} ^{{r}} \frac{\mathrm{1}}{{R}^{\mathrm{2}} −{t}^{\mathrm{2}} }{dt} \\$$$$=\int_{\mathrm{0}} ^{{r}} \frac{\mathrm{1}}{{R}^{\mathrm{2}} −\lambda^{\mathrm{2}} }{d}\lambda \\$$$$... \\$$

Commented by Frix last updated on 14/Dec/22

$$\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{sure},\:\mathrm{it}\:\mathrm{might}\:\mathrm{just}\:\mathrm{be}\:\mathrm{a}\:\mathrm{formal} \\$$$$\mathrm{problem}... \\$$

Commented by Frix last updated on 14/Dec/22

$$\mathrm{My}\:\mathrm{thoughts}: \\$$$$\int{f}\left({x}\right){dx}={F}\left({x}\right)+{C} \\$$$$\:\:\:\:\:\mathrm{gives}\:\mathrm{the}\:\mathrm{antiderivative} \\$$$$\underset{{a}} {\overset{{b}} {\int}}{f}\left({x}\right){dx}={F}\left({b}\right)−{F}\left({a}\right) \\$$$$\:\:\:\:\:\mathrm{gives}\:\mathrm{the}\:\mathrm{area}\:\mathrm{between}\:{f}\left({x}\right)\:\mathrm{and}\:\mathrm{the}\:{x}−\mathrm{axis} \\$$$$\underset{{a}} {\overset{{x}} {\int}}{f}\left({x}\right){dx}={F}\left({x}\right)−{F}\left({a}\right) \\$$$$\:\:\:\:\:\mathrm{gives}\:{what}?\:\mathrm{I}\:\mathrm{never}\:\mathrm{needed}\:\mathrm{this}...\:\mathrm{anyway} \\$$$$\:\:\:\:\:\mathrm{it}'\mathrm{s}\:\mathrm{the}\:\mathrm{first}\:\mathrm{case}\:\mathrm{with}\:{C}=−{F}\left({a}\right) \\$$

Commented by mr W last updated on 14/Dec/22

$$\underset{{a}} {\overset{{b}} {\int}}{f}\left({x}\right){dx}={F}\left({b}\right)−{F}\left({a}\right) \\$$$${you}\:{can}\:{replace}\:{b}\:{with}\:{t}\:{or}\:{x}\:{as}\:{you} \\$$$${like}\:{and}\:{get} \\$$$$\underset{{a}} {\overset{{t}} {\int}}{f}\left({x}\right){dx}={F}\left({t}\right)−{F}\left({a}\right) \\$$$$\underset{{a}} {\overset{{x}} {\int}}{f}\left({x}\right){dx}={F}\left({x}\right)−{F}\left({a}\right) \\$$

Commented by mr W last updated on 14/Dec/22

$$\underset{{a}} {\overset{{x}} {\int}}{f}\left({x}\right){dx}={F}\left({x}\right)−{F}\left({a}\right)\:{gives}\:{the}\:{area} \\$$$${under}\:{the}\:{curve}\:{y}={f}\left({x}\right)\:{from}\:{x}={a}\: \\$$$${to}\:{x}={x}. \\$$

Commented by mr W last updated on 14/Dec/22

Answered by mr W last updated on 14/Dec/22

$$\int_{\mathrm{0}} ^{{r}} \frac{\mathrm{1}}{{R}^{\mathrm{2}} −{r}^{\mathrm{2}} }{dr} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}{R}}\int_{\mathrm{0}} ^{{r}} \left(\frac{\mathrm{1}}{{R}−{r}}+\frac{\mathrm{1}}{{R}+{r}}\right){dr} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}{R}}\:\left[\mathrm{ln}\:\frac{{R}+{r}}{{R}−{r}}\right]_{\mathrm{0}} ^{{r}} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}{R}}\:\mathrm{ln}\:\frac{{R}+{r}}{{R}−{r}} \\$$