Integrals

Nas

Apr 2010
11
0
I'm having a problem trying to figure out how to integrate the following integral: ln(y)/y with respect to dy with limits of e and 1.

Please can someone assist me

Thanks
 
May 2009
959
362
let \(\displaystyle u = \ln y \)

then \(\displaystyle du = \frac{dy}{y} \)

so the integral becomes \(\displaystyle \int u \ du = \frac{u^{2}}{2} = \frac{(\ln y)^{2}}{2} \Big|^{e}_{1} \)

\(\displaystyle = \frac{1}{2} \Big((\ln e)^{2} - (\ln 1)^{2}\Big) = \frac{1}{2} \Big(1-0\Big) = \frac{1}{2}\)
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
I'm having a problem trying to figure out how to integrate the following integral: ln(y)/y with respect to dy with limits of e and 1.

Please can someone assist me

Thanks
\(\displaystyle \int{\frac{\ln{y}}{y}\,dy} = \int{\ln{y}\left(\frac{1}{y}\right)\,dy}\).

Now make the substitution \(\displaystyle u = \ln{y}\) so that \(\displaystyle \frac{du}{dy} = \frac{1}{y}\), the integral becomes

\(\displaystyle \int{u\,\frac{du}{dy}\,dy}\)

\(\displaystyle = \int{u\,du}\)

\(\displaystyle = \frac{1}{2}u^2 + C\)

\(\displaystyle = \frac{1}{2}(\ln{y})^2 + C\).



Therefore:

\(\displaystyle \int_1^e{\frac{\ln{y}}{y}\,dy}= \left[\frac{1}{2}(\ln{y})^2\right]_1^e\)

\(\displaystyle = \left[\frac{1}{2}(\ln{e})^2\right] - \left[\frac{1}{2}(\ln{1})^2\right]\)

\(\displaystyle = \frac{1}{2}\).