0-r-1-R-2-r-2-dr-

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

\int_{0}^{r} \frac{1}{R^{2} - r^{2}} d r = ?

Commented by Frix last updated on 14/Dec/22

Error : dependent borders are not allowed .

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

w h y n o t ?

\int_{0}^{r} \frac{1}{R^{2} - r^{2}} d r

= \int_{0}^{r} \frac{1}{R^{2} - x^{2}} d x

= \int_{0}^{r} \frac{1}{R^{2} - y^{2}} d y

= \int_{0}^{r} \frac{1}{R^{2} - t^{2}} d t

= \int_{0}^{r} \frac{1}{R^{2} - λ^{2}} d λ

\dots

Commented by Frix last updated on 14/Dec/22

I^{'} m not sure, it might just be a formal

problem \dots

Commented by Frix last updated on 14/Dec/22

My thoughts :

\int f (x) d x = F (x) + C

gives the antiderivative

\underset{a}{\int^{b}} f (x) d x = F (b) - F (a)

gives the area between f (x) and the x - axis

\underset{a}{\int^{x}} f (x) d x = F (x) - F (a)

gives w h a t ? I never needed this \dots anyway

{it}^{'} s the first case with C = - F (a)

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

\underset{a}{\int^{b}} f (x) d x = F (b) - F (a)

y o u c a n r e p l a c e b w i t h t o r x a s y o u

l i k e a n d g e t

\underset{a}{\int^{t}} f (x) d x = F (t) - F (a)

\underset{a}{\int^{x}} f (x) d x = F (x) - F (a)

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

\underset{a}{\int^{x}} f (x) d x = F (x) - F (a) g i v e s t h e a r e a

u n d e r t h e c u r v e y = f (x) f r o m x = a

t o x = x .

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

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

\int_{0}^{r} \frac{1}{R^{2} - r^{2}} d r

= \frac{1}{2 R} \int_{0}^{r} (\frac{1}{R - r} + \frac{1}{R + r}) d r

= \frac{1}{2 R} {[\ln \frac{R + r}{R - r}]}_{0}^{r}

= \frac{1}{2 R} \ln \frac{R + r}{R - r}

Facebook Tweet Pin