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Question Number 57488 by Abdo msup. last updated on 05/Apr/19

l e t A_{n} = \int_{0}^{n} \frac{t [t]}{3 + t^{2}} d t

1) c a l c u l a t e {l i m}_{n \to + \infty} A_{n}

2) f i n d n a t u r e i f Σ A_{n}

Commented by maxmathsup by imad last updated on 06/Apr/19

1) w e h a v e A_{n} = \sum_{k = 0}^{n - 1} \int_{k}^{k + 1} \frac{t k}{3 + t^{2}} d t = \sum_{k = 0}^{n - 1} k \int_{k}^{k + 1} \frac{t d t}{t^{2} + 3}

= \frac{1}{2} \sum_{k = 0}^{n - 1} k {[l n (t^{2} + 3)]}_{k}^{k + 1} = \frac{1}{2} \sum_{k = 0}^{n - 1} k l n {\frac{{(k + 1)}^{2} + 3}{k^{2} + 3}}

= \frac{1}{2} \sum_{k = 0}^{n - 1} k l n {\frac{k^{2} + 3 + 2 k}{k^{2} + 3}} = \frac{1}{2} \sum_{k = 1}^{n - 1} k l n {1 + \frac{2 k}{k^{2} + 3}} .

= \frac{1}{2} {l n (1 + \frac{2}{4}) + 2 l n (1 + \frac{4}{7}) + 3 l n (1 + \frac{6}{12}) + \dots . + (n - 1) l n (1 + \frac{2 n - 2}{{(n - 1)}^{2} + 3})}

Answered by tanmay.chaudhury50@gmail.com last updated on 06/Apr/19

\int_{0}^{n} \frac{t [t]}{3 + t^{2}} d t

\frac{1}{2} \int_{0}^{1} \frac{d (3 + t^{2})}{3 + t^{2}} \times 0 + \frac{1}{2} \int_{1}^{2} \frac{d (3 + t^{2})}{3 + t^{2}} \times 1 + \frac{1}{2} \int_{2}^{3} \frac{d (3 + t^{2})}{3 + t^{2}} \times 2 d t + \dots + \frac{1}{2} \int_{n - 1}^{n} \frac{d (3 + t^{2})}{3 + t^{2}} \times (n - 1) d t

= \frac{1}{2} [0 \times ∣ l n (3 + t^{2}) ∣_{0}^{1} + 1 \times ∣ l n (3 + t^{2}) ∣_{1}^{2} + 2 \times ∣ l n (3 + t^{2}) ∣_{2}^{3} + . . + (n - 1) ∣ l n (3 + t^{2}) ∣_{n - 1}^{n}]

= \frac{1}{2} \underset{n = 1}{\sum^{n}} (n - 1) l n {\frac{3 + n^{2}}{3 + {(n - 1)}^{2}}}