Find the area between y=ln(x) and the x axis on the interval [1,e^2] I know how to integration, but is this possible? How would I integrate lnx?
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Originally Posted by Sm10389 Find the area between y=ln(x) and the x axis on the interval [1,e^2] I know how to integration, but is this possible? How would I integrate lnx? one way, the most famous, is to integrate by parts. using u = ln(x) as the function you differentiate and dv = dx as the function you integrate ln(x) has actually been integrated several ways on this forum, it might do you good to do a quick search
Ok so use 1 and lnx. My answer is xlnx-1 is that right?
Originally Posted by Sm10389 Ok so use 1 and lnx. My answer is xlnx-1 is that right? no
maybe i just wrote it wrong u=ln(x) v'=1 u'=1/x v= x = lnx*x - x(1/x) = (lnx) - 1 Where is it wrong
Originally Posted by Sm10389 maybe i just wrote it wrong u=ln(x) v'=1 u'=1/x v= x = lnx*x - x(1/x) = (lnx) - 1 Where is it wrong you have to integrate the second term according to the formula... $\displaystyle \int \ln x ~dx = x \ln x - \int x \cdot \frac 1x~dx$
that is integrating... by parts.
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