Menu Close

calvulste-A-n-0-n-t-2-1-t-1-3-dt-and-lim-n-A-n-

Tinku Tara 0 Comments
Pin




Question Number 43938 by abdo.msup.com last updated on 18/Sep/18
calvulste A_n =∫_0 ^n t^2 [(1/((t+1)^3 ))]dt and lim_(n→+∞) A_n
calvulsteAn=∫0nt2[1(t+1)3]dtandlimn→+∞An
Commented by maxmathsup by imad last updated on 20/Sep/18
changement (1/((t+1)^3 )) =x give (t+1)^3 =(1/x) ⇒t+1 = x^(−(1/3)) ⇒ A_n = ∫_0 ^n (x^(−(1/3)) −1)^2 [x] (−(1/3)) x^(−(4/3)) dx =−(1/3) ∫_0 ^n (x^(−(2/3)) −2x^(−(1/3)) +1)x^(−(4/3)) [x] dx =−(1/3) ∫_0 ^n {x^(−2) −2x^(−(5/3)) +x^(−(4/3)) }[x]dx ⇒ −3A_n =Σ_(k=0) ^(n−1) ∫_k ^(k+1) (k x^(−2) dx −2k x^(−(5/3)) +k x^(−(4/3)) )dx =Σ_(k=0) ^(n−1) k ∫_k ^(k+1) x^(−2) dx −2 Σ_(k=0) ^(n−1) k ∫_k ^(k+1) x^(−(5/3)) dx +Σ_(k=0) ^(n−1) k ∫_k ^(k+1) x^(−(4/3)) dx =Σ_(k=0) ^(n−1) k [−(1/x)]_k ^(k+1) −2 Σ_(k=0) ^(n−1) k [(1/(−(5/3)+1))x^(−(5/3)+1) ]_k ^(k+1) +Σ_(k=0) ^(n−1) k [ (1/(−(4/3)+1)) x^(−(4/3)+1) ]_k ^(k+1) =Σ_(k=0) ^(n−1) k{ (1/k)−(1/(k+1))} +3 Σ_(k=0) ^(n−1) k{ (k+1)^(−(2/3)) −k^(−(2/3)) } −3Σ_(k=0) ^(n−1) k { (k+1)^(−(1/3)) −k^(−(1/3)) } = Σ_(k=0) ^(n−1) (1/(k+1)) +3 { Σ_(k=0) ^(n−1) (k/((k+1)^(2/3) )) −Σ_(k=0) ^(n−1) (k/k^(2/3) )} −3 { Σ_(k=0) ^(n−1) (k/((k+1)^(1/3) )) −Σ_(k=0) ^(n−1) (k/k^(1/3) )} ⇒ A_n =Σ_(k=1) ^n (1/k) +3 { Σ_(k=0) ^(n−1) (k/((k+1)^(2/3) )) −Σ_(k=0) ^(n−1) k^(1/3) } −3 { Σ_(k=0) ^(n−1) (k/((k+1)^(1/3) )) −Σ_(k=0) ^(n−1) k^(2/3) } .
changement1(t+1)3=xgive(t+1)3=1x⇒t+1=x−13⇒An=∫0n(x−13−1)2[x](−13)x−43dx=−13∫0n(x−23−2x−13+1)x−43[x]dx=−13∫0n{x−2−2x−53+x−43}[x]dx⇒−3An=∑k=0n−1∫kk+1(kx−2dx−2kx−53+kx−43)dx=∑k=0n−1k∫kk+1x−2dx−2∑k=0n−1k∫kk+1x−53dx+∑k=0n−1k∫kk+1x−43dx=∑k=0n−1k[−1x]kk+1−2∑k=0n−1k[1−53+1x−53+1]kk+1+∑k=0n−1k[1−43+1x−43+1]kk+1=∑k=0n−1k{1k−1k+1}+3∑k=0n−1k{(k+1)−23−k−23}−3∑k=0n−1k{(k+1)−13−k−13}=∑k=0n−11k+1+3{∑k=0n−1k(k+1)23−∑k=0n−1kk23}−3{∑k=0n−1k(k+1)13−∑k=0n−1kk13}⇒An=∑k=1n1k+3{∑k=0n−1k(k+1)23−∑k=0n−1k13}−3{∑k=0n−1k(k+1)13−∑k=0n−1k23}.

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *