∫_0 ^2 lim_((1/n)→0) (((2−x)(x+x^n ))/(1+x^n ))dx= ?


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Question Number 58937 by Tony Lin last updated on 02/May/19

$\int_{0}^{2} \lim_{\frac{1}{n} \to 0} \frac{(2 - x) (x + x^{n})}{1 + x^{n}} d x = ?$
Commented by maxmathsup by imad last updated on 02/May/19
$let A_n =∫_0 ^2 lim_(n→+∞) (((2−x)(x+x^n ))/(1+x^n )) dx and f_n (x) =(((2−x)(x+x^n ))/(1+x^n )) (f_n )are continues on[0,2] and ∣f_n (x)∣ ≤(2−x)(x+x^n ) on [0,2] ⇒ A_n =lim_(n→+∞) ∫_0 ^2 f_n (x)dx but ∫_0 ^2 f_n (x)dx =∫_0 ^1 (((2−x)(x+x^n ))/(1+x^n ))dx +∫_1 ^2 (((2−x)(x+x^n ))/(1+x^n )) dx we have lim_(n→+∞) ∫_0 ^1 (((2−x)(x+x^n ))/(1+x^n )) dx =∫_0 ^1 x(2−x)dx =∫_0 ^1 (2x−x^2 )dx =[x^2 −(x^3 /3)]_0 ^1 =1−(1/3) =(2/3) ∫_1 ^2 (((2−x)(x+x^n ))/(1+x^n )) dx =_(x =1+t) ∫_0 ^1 (((2−1−t)(1+t +(1+t)^n ))/(1+(1+t)^n )) dt =∫_0 ^1 (((1−t)(1+t +(1+t)^n ))/((1+t)^n +1)) dt ⇒lim_(n→+∞) ∫_0 ^1 (...)dt =∫_0 ^1 lim_(n→+∞) (((1−t)(1+t +(1+t)^n ))/((1+t)^n +1)) dt (((1−t)(1+t +(t+1)^n ))/((t+1)^n +1)) =((1−t^2 +(1−t)(t+1)^n )/((t+1)^n +1)) =(((t+1)^n {((1−t^2 )/((t+1)^n )) +1−t})/((t+1)^n {1+(1/((t+1)^n ))})) =((((1−t^2 )/((t+1)^n )) +1−t)/(1 +(1/((t+1)^n )))) →1−t ⇒lim_(n→+∞) ∫_1 ^2 (((2−x)(x+x^n ))/(1+x^n )) dx =∫_0 ^1 (1−t)dt =[t−(t^2 /2)]_0 ^1 =1−(1/2) =(1/2) ⇒ A_n = (2/3) +(1/2) =(7/6)$
$l e t A_{n} = \int_{0}^{2} {l i m}_{n \to + \infty} \frac{(2 - x) (x + x^{n})}{1 + x^{n}} d x a n d f_{n} (x) = \frac{(2 - x) (x + x^{n})}{1 + x^{n}}$ $(f_{n}) a r e c o n t i n u e s o n [0, 2] a n d ∣ f_{n} (x) ∣ ⩽ (2 - x) (x + x^{n}) o n [0, 2] \Rightarrow$ $A_{n} = {l i m}_{n \to + \infty} \int_{0}^{2} f_{n} (x) d x b u t \int_{0}^{2} f_{n} (x) d x = \int_{0}^{1} \frac{(2 - x) (x + x^{n})}{1 + x^{n}} d x$ $+ \int_{1}^{2} \frac{(2 - x) (x + x^{n})}{1 + x^{n}} d x w e h a v e {l i m}_{n \to + \infty} \int_{0}^{1} \frac{(2 - x) (x + x^{n})}{1 + x^{n}} d x$ $= \int_{0}^{1} x (2 - x) d x = \int_{0}^{1} (2 x - x^{2}) d x = {[x^{2} - \frac{x^{3}}{3}]}_{0}^{1} = 1 - \frac{1}{3} = \frac{2}{3}$ $\int_{1}^{2} \frac{(2 - x) (x + x^{n})}{1 + x^{n}} d x =_{x = 1 + t} \int_{0}^{1} \frac{(2 - 1 - t) (1 + t + {(1 + t)}^{n})}{1 + {(1 + t)}^{n}} d t$ $= \int_{0}^{1} \frac{(1 - t) (1 + t + {(1 + t)}^{n})}{{(1 + t)}^{n} + 1} d t \Rightarrow {l i m}_{n \to + \infty} \int_{0}^{1} (. . .) d t$ $= \int_{0}^{1} {l i m}_{n \to + \infty} \frac{(1 - t) (1 + t + {(1 + t)}^{n})}{{(1 + t)}^{n} + 1} d t$ $\frac{(1 - t) (1 + t + {(t + 1)}^{n})}{{(t + 1)}^{n} + 1} = \frac{1 - t^{2} + (1 - t) {(t + 1)}^{n}}{{(t + 1)}^{n} + 1} = \frac{{(t + 1)}^{n} {\frac{1 - t^{2}}{{(t + 1)}^{n}} + 1 - t}}{{(t + 1)}^{n} {1 + \frac{1}{{(t + 1)}^{n}}}}$ $= \frac{\frac{1 - t^{2}}{{(t + 1)}^{n}} + 1 - t}{1 + \frac{1}{{(t + 1)}^{n}}} \to 1 - t \Rightarrow {l i m}_{n \to + \infty} \int_{1}^{2} \frac{(2 - x) (x + x^{n})}{1 + x^{n}} d x = \int_{0}^{1} (1 - t) d t$ $= {[t - \frac{t^{2}}{2}]}_{0}^{1} = 1 - \frac{1}{2} = \frac{1}{2} \Rightarrow A_{n} = \frac{2}{3} + \frac{1}{2} = \frac{7}{6}$
Commented by Tony Lin last updated on 03/May/19

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