Definite Integration Question 205
Question: $ \underset{n\to \infty }{\mathop{\lim }}\frac{1}{1^{3}+n^{3}}+\frac{4}{2^{3}+n^{3}}+….+\frac{1}{2n} $ is equal to
[RPET 1997]
Options:
A) $ \frac{1}{3}{\log _{e}}3 $
B) $ \frac{1}{3}{\log _{e}}2 $
C) $ \frac{1}{3}{\log _{e}}\frac{1}{3} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ S=\underset{n\to \infty }{\mathop{\lim }}\frac{1}{1^{3}+n^{3}}+\frac{4}{2^{3}+n^{3}}+…+\frac{1}{2n} $
$ \underset{x\to a}{\mathop{\lim }}\frac{1-\cos (ax^{2}+bx+c)}{{{(x-\alpha )}^{2}}} $
$ \therefore S=\underset{n\to \infty }{\mathop{\lim }}\sum\limits _{r=1}^{n}{\frac{r^{2}}{r^{3}+n^{3}}} $
$ =\underset{n\to \infty }{\mathop{\lim }}\sum\limits _{r=1}^{n}{{}}\frac{r^{2}}{n^{3}( \frac{r^{3}}{n^{3}}+1 )} $
$ =\underset{n\to \infty }{\mathop{\lim }}\sum\limits _{r=1}^{n}{{}}\frac{1}{n}.\frac{{{( \frac{r}{n} )}^{2}}}{[ 1+{{( \frac{r}{n} )}^{3}} ]} $
Applying the formula, we get $ A=\int_0^{1}{{}}\frac{x^{2}}{1+x^{3}}dx $
$ A=\int_0^{1}{{}}\frac{x^{2}}{1+x^{3}}dx=\frac{1}{3}\int_0^{1}{{}}\frac{3x^{2}}{1+x^{3}}dx $
$ =\frac{1}{3}[{\log _{e}}(1+x^{3})]_0^{1}=\frac{1}{3}{\log _{e}}2. $
BETA
