let-I-0-1-x-ln-1-x-dx-determine-a-approximate-value-of-I-

Question Number 68879 by mathmax by abdo last updated on 16/Sep/19

l e t I = \int_{0}^{1} \frac{x}{l n (1 + x)} d x d e t e r m i n e a a p p r o x i m a t e v a l u e o f I

Commented by Abdo msup. last updated on 21/Sep/19

t h e r e a e r r o r o f c a l c u l u s i n t h e a n s w e r i w i l l h i d

t h i s p o s t \dots

Commented by mathmax by abdo last updated on 22/Sep/19

w e h a v e {l n}^{^{'}} (1 + x) = \frac{1}{1 + x} = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{n} \Rightarrow

l n (1 + x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} x^{n + 1}}{n + 1} + c (c = 0) \Rightarrow

l n (1 + x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \dots \Rightarrow x - \frac{x^{2}}{2} ⩽ l n (1 + x) ⩽ x \Rightarrow

\frac{1}{x} ⩽ \frac{1}{l n (1 + x)} ⩽ \frac{1}{x - \frac{x^{2}}{2}} \Rightarrow 1 ⩽ \frac{x}{l n (1 + x)} ⩽ \frac{1}{1 - \frac{x}{2}} \Rightarrow

1 ⩽ \int_{0}^{1} \frac{x}{l n (1 + x)} d x ⩽ \int_{0}^{1} \frac{d x}{1 - \frac{x}{2}} w e h a v e

\int_{0}^{1} \frac{d x}{1 - \frac{x}{2}} = \int_{0}^{1} \frac{2 d x}{2 - x} = - 2 \int_{0}^{1} \frac{d x}{x - 2} = - 2 {[l n ∣ x - 2 ∣]}_{0}^{1} = - 2 (- l n 2)

= 2 l n (2) \Rightarrow 1 ⩽ I ⩽ 2 l n (2) s o v_{0} = \frac{1 + 2 l n (2)}{2} i s a a p p r o x i m a t e

v a l u e o f I .