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

Question

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#).

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Answer 1 · 2018-05-22T00:19:32

The correct equation is:

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

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 · 2018-05-22T00:20:24

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!