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Evaluate the integral of ln(x) dx from 1 to 2
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\[\int\limits_{1}^{2}\ln x dx=(xlnx-x)|_{1}^{2}=2\ln2-3\]
I got 2ln2 -1
The latter is correct \[2\ln 2 - 2 - (1\ln 1 - 1) = 2\ln 2 - 2 + 1 = 2\ln 2 -1\]
Ok thanks!
intg(0 to2)lnxdx=xlnx -x]=x(lnx-x)]=2\(ln2-2)\-(1ln1-1) =2ln2 -2+1=2ln2-1
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