Question

What is the antiderivative of `v^-1` from 2 to 4?

Answer 1

`int_2^4 v^(-1)dv=ln2`

Explanation:

This problem becomes very simple once you realize that `v^(-1)` is equivalent to `1/v`. Therefore the problem is:
`int_2^4 1/vdv`

Now our task is to find a function whose derivative is `1/v`. That's the natural log function, of course! All that's left is to evaluate it from `2` to `4`:
`int_2^4 1/vdv=[lnv]_2^4`
`=ln4-ln2`

Using the property of logs that states `lna-lnb=ln(a/b)`, this becomes:
`ln(4/2)=ln2~~0.693...`

Note
Normally `int1/xdx=lnabs(x)`. The reason for this is that `1/x` is defined for all `x` except `x=0`, but `lnx` is defined for only `x>0`. This means that while `1/x` is defined for negative numbers, its antiderivative is not. The simple solution is to use the absolute value of `x` rather than just plain `x`, making the natural log function defined for all `x` except`x=0`, which is what we want. However, in my answer I didn't use the absolute value bars because we were only dealing with positive values (`2` to `4`) and it wasn't necessary.