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

∫ _0^r (1/(R^2 −r^2 ))dr=?

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

Commented by Frix last updated on 14/Dec/22

Error: dependent borders are not allowed.

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

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

why not? ∫_0 ^r (1/(R^2 −r^2 ))dr =∫_0 ^r (1/(R^2 −x^2 ))dx =∫_0 ^r (1/(R^2 −y^2 ))dy =∫_0 ^r (1/(R^2 −t^2 ))dt =∫_0 ^r (1/(R^2 −λ^2 ))dλ ...

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

Commented by Frix last updated on 14/Dec/22

I′m not sure, it might just be a formal problem...

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

Commented by Frix last updated on 14/Dec/22

My thoughts: ∫f(x)dx=F(x)+C gives the antiderivative ∫_a ^b f(x)dx=F(b)−F(a) gives the area between f(x) and the x−axis ∫_a ^x f(x)dx=F(x)−F(a) gives what? I never needed this... anyway it′s the first case with C=−F(a)

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

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

∫_a ^b f(x)dx=F(b)−F(a) you can replace b with t or x as you like and get ∫_a ^t f(x)dx=F(t)−F(a) ∫_a ^x f(x)dx=F(x)−F(a)

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

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

∫_a ^x f(x)dx=F(x)−F(a) gives the area under the curve y=f(x) from x=a to x=x.

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

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

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

∫_0 ^r (1/(R^2 −r^2 ))dr =(1/(2R))∫_0 ^r ((1/(R−r))+(1/(R+r)))dr =(1/(2R)) [ln ((R+r)/(R−r))]_0 ^r =(1/(2R)) ln ((R+r)/(R−r))

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

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