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Question Number 196522 by Erico last updated on 26/Aug/23

Prove that \forall n \in N

{\int_{n}}^{n + 1} \ln t dt ⩽ \ln ({\int_{n}}^{n + 1} t dt)

Answered by aleks041103 last updated on 26/Aug/23

\int l n (t) d t = t l n (t) - t

\Rightarrow \int_{n}^{n + 1} l n (t) d t =

= (n + 1) l n (n + 1) - n - 1 - (n l n (n) - n) =

= n l n (1 + \frac{1}{n}) + l n (n + 1) - 1

l n (\int_{n}^{n + 1} t d t) = l n (n + \frac{1}{2})

l n (1 + \frac{1}{n}) < \frac{1}{n} - \frac{1}{2 n^{2}}

\Rightarrow \int_{n}^{n + 1} l n (t) d t - l n (\int_{n}^{n + 1} t d t) < - \frac{1}{2 n} + l n (\frac{n + 1}{n + \frac{1}{2}}) =

= l n (1 + \frac{1}{2 n + 1}) - \frac{1}{2 n} < \frac{1}{2 n + 1} - \frac{1}{2 {(2 n + 1)}^{2}} - \frac{1}{2 n} < 0

\Rightarrow \int_{n}^{n + 1} l n (t) d t ⩽ l n (\int_{n}^{n + 1} t d t)