First slide

Introduction to definite integration

Question

$lim_{n \to \infty} (\frac{1}{n + 1} + \frac{1}{n + 2} + \dots + \frac{1}{6 n})$ is equal to

Moderate

A

$\log 2$
B

$\log 3$
C

$\log 5$
D

$\log 6$

Solution

$\begin{array}{l} lim_{n \to \infty} (\frac{1}{n + 1} + \frac{1}{n + 2} + \dots + \frac{1}{6 n}) \\ = lim_{n \to \infty} (\frac{1}{n + 1} + \frac{1}{n + 2} + \dots + \frac{1}{n + 5 n}) \\ = lim_{n \to \infty} \sum_{r = 1}^{5 n} (\frac{1}{n + r}) = lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{5 n} (\frac{1}{1 + \frac{r}{n}}) \end{array}$
$∵ Lower limit of r = 1$
$∴ Lower limit of integration = lim_{n \to \infty} \frac{1}{n} = 0 ∵ Upper limit of r = 5 n ∴ Upper limit of integration = lim_{n \to \infty} \frac{5 n}{n} = 5 from eq (i) \int_{0}^{5} \frac{1}{1 + x} dx = [\log (1 + x)]_{0}^{5} = \log 6 - \log 1 = \log 6$

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