`h(u) = (3 - 2u)^-1 , [-1, 1]` Find the average value of the function on the given interval.

The average value is the integral of a function over an interval divided by the length of an interval. The length here is 2, let's found the integral:

`int_(-1)^1 1/(3-2u) du = -1/2 ln(3-2u)|_(u=-1)^1 = -1/2(ln1-ln5)=ln5/2.`

So the average is `ln5/4 approx 0.40.`

Note that the given interval [-1, 1] doesn't contain the point 1.5 where `3-2u=0.`

The average value is the integral of a function over an interval divided by the length of an interval. The length here is 2, let's found the integral:

`int_(-1)^1 1/(3-2u) du = -1/2 ln(3-2u)|_(u=-1)^1 = -1/2(ln1-ln5)=ln5/2.`

So the average is `ln5/4 approx 0.40.`

Note that the given interval [-1, 1] doesn't contain the point 1.5 where `3-2u=0.`