Menu Close

calculate-1-3-x-e-x-1-dx-

Tinku Tara 0 Comments
Pin




Question Number 35675 by abdo imad last updated on 21/May/18
calculate ∫_1 ^3 (x/(e^x −1))dx ..
calculate∫13xex−1dx..
Commented by prof Abdo imad last updated on 25/May/18
I = ∫_1 ^3 ((x e^(−x) )/(1−e^(−x) ))dx =∫_1 ^3 ( Σ_(n=0) ^∞ e^(−nx) )x e^(−x) dx = Σ_(n=0) ^∞ ∫_1 ^3 x e^(−(n+1)x) dx = Σ_(n=0) ^∞ A_n with A_n = ∫_1 ^3 x.e^(−(n+1)x) dx .changement (n+1)x=t give A_n =∫_(n+1) ^(3(n+1)) (t/(n+1)) e^(−t) (dt/(n+1)) = (1/((n+1)^2 )) ∫_(n+1) ^(3(n+1)) t^ e^(−t) dt by parts ∫_(n+1) ^(3(n+1)) t e^(−t) dt = [ −t e^(−t) ]_(n+1) ^(3(n+1)) +∫_(n+1) ^(3(n+1)) e^(−t) dt = (n+1)e^(−(n+1)) −3(n+1) e^(−3(n+1)) +[ −e^(−t) ]_(n+1) ^(3(n+1)) =(n+1) e^(−(n+1)) −3(n+1) e^(−3(n+1)) +e^(−(n+1)) −e^(−3(n+1)) A_n = (e^(−(n+1)) /(n+1)) −3(e^(−3(n+1)) /(n+1)) + (e^(−(n+1)) /((n+1)^2 )) −(e^(−3(n+1)) /((n+1)^2 )) Σ_(n=0) ^∞ A_n = Σ_(n=0) ^∞ (e^(−(n+1)) /(n+1)) −3 Σ_(n=0) ^∞ (e^(−3(n+1)) /(n+1)) + Σ_(n=0) ^∞ (e^(−(n+1)) /((n+1)^2 )) −Σ_(n=0) ^∞ (e^(−3(n+1)) /((n+1)^2 )) = Σ_(n=1) ^∞ (e^(−n) /n) −3 Σ_(n=1) ^∞ (e^(−3n) /n) +Σ_(n=1) ^∞ (e^(−n) /n^2 ) −Σ_(n=1) ^∞ (e^(−3n) /n^2 ) ....be continued....
I=∫13xe−x1−e−xdx=∫13(∑n=0∞e−nx)xe−xdx=∑n=0∞∫13xe−(n+1)xdx=∑n=0∞AnwithAn=∫13x.e−(n+1)xdx.changement(n+1)x=tgiveAn=∫n+13(n+1)tn+1e−tdtn+1=1(n+1)2∫n+13(n+1)te−tdtbyparts∫n+13(n+1)te−tdt=[−te−t]n+13(n+1)+∫n+13(n+1)e−tdt=(n+1)e−(n+1)−3(n+1)e−3(n+1)+[−e−t]n+13(n+1)=(n+1)e−(n+1)−3(n+1)e−3(n+1)+e−(n+1)−e−3(n+1)An=e−(n+1)n+1−3e−3(n+1)n+1+e−(n+1)(n+1)2−e−3(n+1)(n+1)2∑n=0∞An=∑n=0∞e−(n+1)n+1−3∑n=0∞e−3(n+1)n+1+∑n=0∞e−(n+1)(n+1)2−∑n=0∞e−3(n+1)(n+1)2=∑n=1∞e−nn−3∑n=1∞e−3nn+∑n=1∞e−nn2−∑n=1∞e−3nn2….becontinued….

Pin

Leave a Reply

Your email address will not be published.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