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Question Number 205725 by NasaSara last updated on 28/Mar/24
Answered by Berbere last updated on 29/Mar/24
∫02∫03[xy]dydx=xy=t∫02∫03x[t]dtxdx=∫021x∫03x[t]dtdx==∑5k=1{∫k3k+131x{∑k−1j=1∫jj+1[t]+∫k3x[t]}dt=isolatk=1;andk⩾2∫13231x{3x−1}=1−[ln(2))].∑5k=2∫k3k+121x{∑k−1j=1j+∫k3xkdt}dx+1−ln(2)=∑5k=2∫k3k+131x{(k−1)k2+k(3x−k)}dx=∑5k=2k+∑5k=2∫k3k+131x(−(1+k)k2)dx+1−ln(2)=15−∑5k=2(1+k)k2ln(k+1k)−ln(2)=15−3ln(32)−6ln(43)−10ln(54)−15ln(65)−ln(2)=15+3ln(3)+2ln(2)+4ln(4)+5ln(5)−15ln(6)=15−ln(61533.22.44.55)=15−ln(312.25.55)=15−ln(170061123125)
Commented by NasaSara last updated on 04/Apr/24
thank you so much
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