Question

Let $P_k$ be a point on the curve $y=\ln x$ in $xy-$plane whose $x$ coordinate is $1+\frac{k}{n},$ $k=1,2,3,\dots ,n.$ If $A$ is $(1,0),$ then $\underset{n\to \infty }{lim}\frac{1}{n}\sum _{k=1}^n{\left(A{P}_k\right)}^2$ equals
(Here, $A{P}_k$ is the distance between points $A$ and $P_k$)

• A. $\frac{1}{3}+2(\ln 2{)}^2$
• B. $\frac{1}{3}+2{\left(\ln \left(\frac{2}{e}\right)\right)}^2$
• C. $\frac{1}{3}+{\left(\ln \left(\frac{2}{e}\right)\right)}^2$
• D. $\frac{1}{3}+2\ln \left(\frac{2}{e}\right)$

Answer 1

The correct option is B $\frac{1}{3}+2{\left(\ln \left(\frac{2}{e}\right)\right)}^2$

Distance between points $A(1,0)$ and $P_k(1+\frac{k}{n},\ln (1+\frac{k}{n}))$ is
$A{P}_k=\sqrt{{(1+\frac{k}{n}-1)}^2+{(\ln (1+\frac{k}{n})-0)}^2}$
$\Rightarrow A{P}_k=\sqrt{{\left(\frac{k}{n}\right)}^2+{\left(\ln (1+\frac{k}{n})\right)}^2}$
$\Rightarrow {\left(A{P}_k\right)}^2={\left(\frac{k}{n}\right)}^2+{\left(\ln (1+\frac{k}{n})\right)}^2$

Now, $\underset{n\to \infty }{lim}\frac{1}{n}\sum _{k=1}^n{\left(A{P}_k\right)}^2$
$=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{k=1}^n\left[{\left(\frac{k}{n}\right)}^2+{\left(\ln (1+\frac{k}{n})\right)}^2\right]$
$=\underset{0}{\overset{1}{\int }}(x^2+(\ln (1+x){)}^2)dx$
$=\frac{1}{3}+\underset{1}{\overset{2}{\int }}(\ln t{)}^{2}dt$
$=\frac{1}{3}+{\left[t\right(\ln t{)}^2]}_1^2-\underset{1}{\overset{2}{\int }}2\ln \left(t\right)dt$
$=\frac{1}{3}+2(\ln 2{)}^2-2[t(\ln t-1){]}_1^2$
$=\frac{1}{3}+2(\ln 2{)}^2-2[2\left(\ln 2\right)-1]$
$=\frac{1}{3}+2(\ln 2-1{)}^2$
$=\frac{1}{3}+2{\left(\ln \left(\frac{2}{e}\right)\right)}^2$