Question

Find `int_4^5f(5x+2) dx` if `int_22^27f(x)dx` ?

Answer 1

No, there's a factor `1/5` missing

Explanation:

`int_4^5 f(5x+2) dx = 1/5 int_22^27 f(x) dx`

To see this, consider `f(x) = 1`.

Then:

`int_4^5 f(5x+2) = [ x ]_4^5 = 5-4 = 1`

and:

`int_22^27 f(x) dx = [ x ]_22^27 = 27-22 = 5`

When we perform a substitution, we need to consider the derivative.

So letting `u=5x+2` we have:

`int_4^5 f(5x+2) dx = int_22^27 f(u) (dx)/(du) du = 1/5 int_22^27 f(u) du = 1/5 int_22^27 f(x) dx`