Question

For a continuous function `f`, is it generally true that `\int_0^2 f(x)dx = \int_0^1 f(2x)dx`?

This seems to be true because if I "speed up" the function by going at twice the speed `(2x)` then I should only have to swipe half the interval (from `\int_0^2` to `\int_0^1`).

Answer 1

The correct equation is:

`int_0^2 f(x) dx = 2 int_0^1 f(2x)dx`

Explanation:

Using the substitution of variables, let `t=x/2`, `x=2t`, `dx= 2dt`.

Note that if `x in [0,2]` then the range of `t=x/2` is `[0,1]`, so:

`int_0^2 f(x) dx = int_0^1 f(2t) (2dt) = 2 int_0^1 f(2t)dt`

So, in fact you need to swipe only the interval `[0,1]` but you have to consider that in the Riemann sums the lengths of the intervals is half.

Answer 2

Let `int f(x) dx = F(x)`. Then

`int_0^2 f(x) dx = F(2) - F(0)`

As for the second integral, if we let `u = 2x`, then `du = 2dx` and `1/2du = dx`, then we must also adjust the bounds of integration.

`int_0^1 f(2x)dx = 1/2int_0^2 f(u) du = 1/2(F(2) - F(0))`

Thus the given integral property is false (the right hand side will be half the value of the left).

Hopefully this helps!