Question

I'm trying to solve a definite integral but I'm making some mistake and end up with ln(0). Here's the problem with my so-called solution. Can't find my mistake. I'll appreciate it if someone manages to point it out. Thanks!

Answer 1

\[\int\limits_{0}^{e}(x-e)*\ln(x)dx\]

Answer 2

So, I'm trying to solve it by parts. I'll use the matrix notation to mark the interval.
\[I1=\int\limits_{0}^{e}(x-e)*\ln(x)dx=\int\limits_{0}^{e}(x-e)*\ln(x)d(x-e)=\frac{1}{2} \int\limits_{0}^{e}\ln(x)d((x-e)^2)\]
\[I1 = \frac{1}{2}\ln(x)*(x-e)^2 |\left(\begin{matrix}e\\ 0\end{matrix}\right) - \frac{1}{2}\int\limits_{0}^{e}(x-e)^2d(\ln(x))\]
I won't go on with the second term in details, my result is
\[I2 =( \frac{ x^2 }{ 2 } - 2ex +e^2\ln(x) )\left(\begin{matrix}e \\ 0\end{matrix}\right)\]

So, when I try solving I1, I get:
\[I1 = \ln(e)*0-\ln(0)*e^2\]

Answer 3

Ugh website crashed on me earlier >:c
Lemme try to retype that stuff..

Answer 4

\[\Large\bf\sf \int\limits_0^e(x-e)\ln x\;dx\]
Hmm this is how I would approach the problem:
\[\Large\bf\sf u=\ln x, \qquad\qquad\qquad dv=(x-e)\;dx\]\[\Large\bf\sf du=\frac{1}{x}\;dx, \qquad\qquad v=\left(\frac{1}{2}x^2-ex\right)\]

So doing by parts gives us:\[\Large\bf\sf \left(\frac{1}{2}x^2-ex\right)\ln x-\int\limits \left(\frac{1}{2}x^2-ex\right)\frac{1}{x}\;dx\]Simplifying then integrating that second part gives us:\[\Large\bf\sf \left(\frac{1}{2}x^2-ex\right)\ln x-\left(\frac{1}{4}x^2-ex\right)\]

Answer 5

Before evaluating, it might be good to group things a little differently:\[\Large\bf\sf \frac{1}{4}x^2(2\ln x-1)-ex(\ln x - 1)\]

Answer 6

Evaluated at the upper limit:\[\Large\bf\sf \frac{1}{4}e^2(2\ln e-1)-\cancel{e^2(\ln e-1)}\quad=\quad \frac{1}{4}e^2\]

Answer 7

Evaluated at the lower limit:\[\Large\bf\sf \frac{1}{4}0^2(2\ln 0-1)-e\cdot 0(\ln 0 - 1)\]And I guess I'm realizing now that ln0 is undefined, so we should really rewrite this as a limit to be proper.

Answer 8

\[\Large\bf\sf \lim_{b\to0}\frac{1}{4}b^2(2\ln b-1)-e\cdot b(\ln b - 1)\]So we have a couple of troublesome spots.

We need to show that: \[\Large\bf\sf \lim_{b\to0}b\ln b\quad=\quad 0\]

Answer 9

Any power of x will grow and shrink faster than the log function.
That is a fun little fact to remember.
But if you need to be a little more formal about it, you can move things around and apply L'Hopital's Rule.

Answer 10

I think the thing we were disagreeing on is our v.\[\Large\bf\sf dv=(x-e)dx, \qquad\qquad v=\frac{1}{2}(x-e)^2\]I think that's what you came up with right? :o

Answer 11

well, I had \[v = (x - e)^2 \] so that \[d(x-e)^2 = 2(x-e)dx\]

Answer 12

okay, why is your \[v=\frac{ 1 }{ 2 }x^2-ex \]

Answer 13

\[\Large\bf\sf v\quad=\quad \int\limits (x-e)\;dx\]And I integrated each term separately.