# How do you evaluate the definite integral of |ln x|, if the lower limit of integration is x=1/e and the upper limit is x=e?

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### 1 Answer

We notice that if x is located in the interval [1/e , 1], the values of ln x are negative.

If x is located in the interval [1 , e], the values of ln x are positive.

According to these, we'll solve the definite integral:

Int (-ln x)dx(1/e->1) + Int ln x dx (1->e)

We'll solve the integrals above by parts:

Int (-ln x)dx(1/e->1)

u = ln x => u' = dx/x

v' = (x)' => v = x

Int (-ln x)dx = x*ln x - Int dx = x(ln x - 1),(1/e->1)

We'll apply Leibniz Newton:

x(ln x - 1),(1/e->1) = (1/e)(ln1 - lne - 1) - 1(ln 1 - 1) (1)

Int ln x dx (1->e) = x(ln x - 1)

x(ln x - 1)(1->e) = e(lne - 1) - 1(ln 1 - 1)(2)

Int (-ln x)dx(1/e->1) + Int ln x dx (1->e) = (1) + (2)

Int (-ln x)dx(1/e->1) + Int ln x dx (1->e) = 2 - 2/e

**The definite integral Int |ln x|dx, x = 1/e to x = e, is Int |ln x|dx = 2 - 2/e.**