Question

How do you evaluate the integral `int lnxdx` from `[0,1]`?

Answer 1

`int_0^1ln(x)\ dx=-1`

Explanation:

First we need to work out the antiderivative. To do this, we will use integration by parts. I will let `f=ln(x)` and `g'=1`.

We know:
`f'=1/x`
`g=x`

So we can rewrite the integral like so:
`int\ ln(x)\ dx=xln(x)-int\ x*1/x\ dx=xln(x)-int\ 1\ dx`
`=xln(x)-x`

Now we know the antiderivative, we can plug in the limits of integration to compute the definite integral:
`int_0^1 ln(x)\ dx=[xln(x)-x]_0^1=1ln(1)-1-(0ln(0)-0)`

This gives us a problem, since `ln(0)` isn't defined, but we can get around this by taking the limit as the value approaches `0`:
`lim_(x->0)xln(x)=0`

This is quite clearly equal to `0` (because `0` multiplied by anything is `0`), so we can now carry on evaluating our definite integral:
`ln(1)-1-(0)=0-1-0=-1`

So, the answer is:
`int_0^1ln(x)\ dx=-1`