Question

show that integating lnx/x^3 with limits of e and 1, equals (e^2 -3)/4e^2

Answer 1

\[\int\limits_{1}^{e}\frac{ lnx }{ x^3 } = \frac{ e^2 -3 }{ 4e^2 }\]

Answer 2

to begin with, i've integrated by parts, using:
u=lnx
du/dx = 1/x
v=\[-\frac{ 1 }{ 2 } x ^{-2}\]
dv/dx= \[x^-3\]

Answer 3

that's good do far

Answer 4

i've got:
\[\int\limits_{1}^{e}f(x) dx= -\frac{ 1 }{ 2 }x ^{-2} lnx +\frac{ 1 }{ 2 } \int\limits_{1}^{e} x ^{-3} dx\]

Answer 5

yup

Answer 6

but first bit is within interval too -- pedantic

Answer 7

yeah sorry ahaha, so then i get...

Answer 8

\[\left[ \frac{ 1 }{ 4 } e ^{-2} (2lnx-1)\right]\ \lim_{e \rightarrow 1}\]

Answer 9

mmm you mean x^-2 AND check your signs....

Answer 10

yes sorry, and which sign?

Answer 11

.. and how on earth do i get it into the form stated in the question?

Answer 12

get the solution right first.

+1/2 ∫ x^-3 dx = MINUS 1/4 x^-2

both terms in solution have a minus sign

Answer 13

when you have the solution done correctly, knowing that ln e = 1 & ln 1 = 0 gives you the answer

but get the solution all lined up before doing that...

Answer 14

i cant get it into that form, im stuck
i've got
\[(-\frac{ 1 }{ 4e ^{2} }(3)) - (-\frac{ 1 }{ 4})\]

Answer 15

=1/4 - 3/4e^2

Answer 16

the final integral is
\[-\frac{lnx}{2x^{2}} - \frac{1}{4x^{2}}\]
it simplifies to
\[-\frac{1}{4x^2}(2lnx + 1)\]

Answer 17

you are almost done

=1/4 - 3/4e^2
=1/4 * (e^2 / e^2) - 3/4e^2

= e^2/ 4 e^2 - 3 / 4 e^2

= (e^2 - 3) / 4 e^2

Answer 18

ohhhh now i see it! you have to make the denomenators the same to add them, ahaha thank you very much!