# Basic Integration

• Dec 11th 2008, 04:46 PM
Sm10389
Basic Integration
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?
• Dec 11th 2008, 05:02 PM
Jhevon
Quote:

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
• Dec 11th 2008, 05:07 PM
Sm10389
Ok so use 1 and lnx.

is that right?
• Dec 11th 2008, 05:08 PM
Jhevon
Quote:

Originally Posted by Sm10389
Ok so use 1 and lnx.

is that right?

no
• Dec 11th 2008, 05:11 PM
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
• Dec 11th 2008, 05:12 PM
Jhevon
Quote:

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...

$\displaystyle \int \ln x ~dx = x \ln x - \int x \cdot \frac 1x~dx$
• Dec 11th 2008, 05:13 PM
Sm10389
that is integrating... by parts.