Express the following limit as definite integral $\underset{n \to \propto}{L t} (\frac{1}{n + 1} + \frac{1}{n + 2} + .... + \frac{1}{6 n})$

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$\int_{0}^{1} \frac{1}{x} d x$

$\int_{0}^{5} \frac{1}{x} d x$

$\int_{0}^{5} \frac{1}{1 + x} d x$

$\int_{0}^{1} \frac{1}{1 + x} d x$

answer is A.

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Detailed Solution

$\begin{array}{l} = \underset{n \to \infty}{L t} (\frac{1}{n + 1} + \frac{1}{n + 2} + \dots . + \frac{1}{6 n}) \\ = \underset{n \to \infty}{L t} (\frac{1}{n + 1} + \frac{1}{n + 2} + \dots . + \frac{1}{n + 5 n}) \\ = \underset{n \to \infty}{L t} \sum_{n = 1}^{5 n} (\frac{1}{n + r}) = \underset{n \to \infty}{L t} \frac{5 n}{n} \sum_{r - 1}^{5 n} (\frac{1}{1 + \frac{r}{n}}) = \int_{0}^{5} \frac{1}{1 + x} d x \end{array}$