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susus · Mar 14, 2011, 03:54 PM

what is the integral of |lnx| , between (1/e) to e

I know the integral of lnx = xlnx -x

so what I did was [{(elne-e - (1ln1-1)]} + {(1ln1 -1 ) - 1/e(ln1/e)} ) ].. sadly I got a wrong answer..

so how can I calculate it? Thanls

jcaron2 · Mar 14, 2011, 08:42 PM

You're very close, but you missed a minus sign.

You did the right thing by breaking it up into two separate intervals, 1/e to 1 and 1 to e. The reason for choosing 1 as the breaking point is that it's the value where the function ln(x) goes from positive to negative. Everything to the right of that point should be evaluated as usual (as if the absolute value sign wasn't even there). However, everything to the left of 1 (between 1/e and 1) would be negative if not for the absolute value sign. But since the absolute value sign causes it to become positive, it's the same as integrating -ln(x) over that interval. Hence, if you replace your + sign with a - sign where you add the two sub-intervals together, you should get the right answer.

Here it is written with math:

$\int_{\frac 1 e}^{e} {\left | \ln x \right |}=\int_{\frac 1 e}^{1} {\left | \ln x \right |}+\int_{1}^{e} {\left | \ln x \right |}=\int_{\frac 1 e}^{1} {-\ln x}+\int_{1}^{e} {\ln x}=\int_{1}^{e} {\ln x}-\int_{\frac 1 e}^{1} {\ln x}$

Notice the minus sign between the two!

susus · Mar 15, 2011, 07:11 AM

Thanks dear :)