Q72 / 160 · TS EAMCET 2014 · Mathematics

By the definition of the definite integral, the value of $\lim _{n \rightarrow \infty}\left(\frac{1^4}{1^5+n^5}+\frac{2^4}{2^5+n^5}+\frac{3^4}{3^5+n^5}+\ldots+\frac{n^4}{n^5+n^5}\right)$ is

  • A
    (a) $\log 2$
  • Correct answer
    B
    $\frac{1}{5} \log 2$
  • C
    $\frac{1}{4} \log 2$
  • D
    $\frac{1}{3} \log 2$

Solution

$\begin{aligned} & \begin{array}{l}\lim _{n \rightarrow \infty}\left(\frac{1^4}{1^5+n^5}+\frac{2^4}{2^5+n^5}+\frac{3^4}{3^5+n^5}+\ldots+\frac{n^4}{n^5+n^5}\right) \\ \qquad \sum_{r=0}^{r=n} \frac{r^4}{r^5+n^5}=\int_0^1 \frac{x^4}{1+x^5} d x \\ \text { Put } 1+x^5=t \Rightarrow 5 x^4 d x=d t \\ =\int_1^2 \frac{1}{5 t} d t=\frac{1}{5}[\log t]_1^2 \\ =\frac{1}{5}[\log 2-\log 1]=\frac{\log 2}{5}\end{array}\end{aligned}$

Hence, the correct answer is option: 'B'.

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