# Evaluating definite integral of ln(x)/x from 1 to e

• May 28th 2009, 05:56 PM
drew.walker
Evaluating definite integral of ln(x)/x from 1 to e
I think I'm getting the hang of evaluating integrals, but I just wanted to check my answer. Is someone able to confirm if this working and result are correct?

$\int^e_1 \frac{ln(x)}{x} dx = \int^e_1 \frac{1}{x}ln(x) dx$

Let

$u = ln(x)$
$du = \frac {1}{x} dx$

Substituting values gives:

$\int^e_1 u du = \left [ \frac{u^2}{2} \right ]^e_1 = \left [ \frac{(ln(x))^2}{2} \right ]^e_1 = \frac {(ln(e))^2}{2} - \frac {(ln(1))^2}{2} = \frac {1}{2} - 0 = \frac{1}{2}$
• May 28th 2009, 06:10 PM
Sampras
Quote:

Originally Posted by drew.walker
I think I'm getting the hang of evaluating integrals, but I just wanted to check my answer. Is someone able to confirm if this working and result are correct?

$\int^e_1 \frac{ln(x)}{x} dx = \int^e_1 \frac{1}{x}ln(x) dx$

Let

$u = ln(x)$
$du = \frac {1}{x} dx$

Substituting values gives:

$\int^e_1 u du = \left [ \frac{u^2}{2} \right ]^e_1 = \left [ \frac{(ln(x))^2}{2} \right ]^e_1 = \frac {(ln(e))^2}{2} - \frac {(ln(1))^2}{2} = \frac {1}{2} - 0 = \frac{1}{2}$

It would be $\int_{0}^{1} u \ du$ since $u = \ln 1 = 0$ and $u = \ln e = 1$.
• May 28th 2009, 06:20 PM
drew.walker
hmmm, so the interval of the integral changes with the substitution? Is there a name for this or somewhere where I can see more examples that you know of?
• May 28th 2009, 06:28 PM
drew.walker
Also, if it's from 0 to 1, ln(1) = 0 and ln(0) is undefined, therefore the answer would now be 0. Is that right?
• May 29th 2009, 04:21 AM
mr fantastic
Quote:

Originally Posted by drew.walker
I think I'm getting the hang of evaluating integrals, but I just wanted to check my answer. Is someone able to confirm if this working and result are correct?

$\int^e_1 \frac{ln(x)}{x} dx = \int^e_1 \frac{1}{x}ln(x) dx$

Let

$u = ln(x)$
$du = \frac {1}{x} dx$

Substituting values gives:

$\int^e_1 u du = \left [ \frac{u^2}{2} \right ]^e_1 = \left [ \frac{(\ln (x))^2}{2} \right ]^e_1 = \frac {(\ln (e))^2}{2} - \frac {(\ln (1))^2}{2} = \frac {1}{2} - 0 = \frac{1}{2}$

You are getting things mixed up and confused. When making a substitution in a definite integral, you can either:

1. Change the integral terminals and then evaluate the new definite integral: $\int_0^1 u \, du$, or

2. Get the anti-derivative in terms of the new variable, re-substitute to get the anti-derivative in terms of the old variable and then evaluate the definite integral in terms of the old variable: $\left [ \frac{(\ln (x))^2}{2} \right ]^e_1$.

You do one or the other. You're mixing both together.

1. $\int_0^1 u \, du = \left[ \frac{u^2}{2} \right]_0^1 = \frac{1}{2}$.
2. $\left [ \frac{(\ln (x))^2}{2} \right ]^e_1 = \frac{1}{2}$.