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

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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# ).

This seems to be true because if I "speed up" the function by going at twice the speed

##### 2 Answers

The correct equation is:

#### Explanation:

Using the substitution of variables, let

Note that if

So, in fact you need to swipe only the interval

Let

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

As for the second integral, if we let

#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!