The value of sumlimits_{r = 2}^{16} {intlimit | vedclass

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Question

(D) Let $I = \sum\limits_{r = 2}^{16} {\int_r^{r + 1} {\frac{{dx}}{{(2r - x)(2r + 2 - x)}}} }$. For each integral,let $u = x - (2r + 1)$. Then $du = d

Vedclass · Accepted Answer

$\ln \left( \frac{2\sqrt{2}}{\sqrt{3}} \right)$

Vedclass · Answer

(D) Let $I = \sum\limits_{r = 2}^{16} {\int_r^{r + 1} {\frac{{dx}}{{(2r - x)(2r + 2 - x)}}} }$. For each integral,let $u = x - (2r + 1)$. Then $du = dx$. When $x = r$,$u = r - 2r - 1 = -r - 1$. When $x = r + 1$,$u = r + 1 - 2r - 1 = -r$. The denominator becomes $(2r - (u + 2r + 1))(2r + 2 - (u + 2r + 1)) = (-u - 1)(1 - u) = (u + 1)(u - 1) = u^2 - 1$. Thus,$\int_r^{r+1} \frac{dx}{(2r-x)(2r+2-x)} = \int_{-r-1}^{-r} \frac{du}{u^2 - 1} = \left[ \frac{1}{2} \ln \left| \frac{u-1}{u+1} ight| ight]_{-r-1}^{-r}$. $= \frac{1}{2} \left( \ln \left| \frac{-r-1}{-r+1} ight| - \ln \left| \frac{-r-1-1}{-r-1+1} ight| ight) = \frac{1}{2} \left( \ln \left( \frac{r+1}{r-1} ight) - \ln \left( \frac{r+2}{r} ight) ight)$. Summing from $r=2$ to $16$: $\frac{1}{2} \sum_{r=2}^{16} \left( \ln \frac{r+1}{r-1} - \ln \frac{r+2}{r} ight)$. This is a telescoping sum: $\frac{1}{2} [(\ln \frac{3}{1} - \ln \frac{4}{2}) + (\ln \frac{4}{2} - \ln \frac{5}{3}) + \dots + (\ln \frac{17}{15} - \ln \frac{18}{16})]$. $= \frac{1}{2} (\ln 3 - \ln \frac{18}{16}) = \frac{1}{2} \ln \left( \frac{3 	imes 16}{18} ight) = \frac{1}{2} \ln \left( \frac{48}{18} ight) = \frac{1}{2} \ln \left( \frac{8}{3} ight) = \ln \sqrt{\frac{8}{3}} = \ln \left( \frac{2\sqrt{2}}{\sqrt{3}} ight)$.

The value of $\sum\limits_{r = 2}^{16} {\int\limits_r^{r + 1} {\frac{{dx}}{{\left( {2r - x} \right)\left( {2r + 2 - x} \right)}}} }$ is equal to

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