# 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 $\left(2 x\right)$ then I should only have to swipe half the interval (from $\setminus {\int}_{0}^{2}$ to $\setminus {\int}_{0}^{1}$).

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2
May 22, 2018

The correct equation is:

${\int}_{0}^{2} f \left(x\right) \mathrm{dx} = 2 {\int}_{0}^{1} f \left(2 x\right) \mathrm{dx}$

#### Explanation:

Using the substitution of variables, let $t = \frac{x}{2}$, $x = 2 t$, $\mathrm{dx} = 2 \mathrm{dt}$.

Note that if $x \in \left[0 , 2\right]$ then the range of $t = \frac{x}{2}$ is $\left[0 , 1\right]$, so:

${\int}_{0}^{2} f \left(x\right) \mathrm{dx} = {\int}_{0}^{1} f \left(2 t\right) \left(2 \mathrm{dt}\right) = 2 {\int}_{0}^{1} f \left(2 t\right) \mathrm{dt}$

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

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Noah G Share
May 22, 2018

Let $\int f \left(x\right) \mathrm{dx} = F \left(x\right)$. Then

${\int}_{0}^{2} f \left(x\right) \mathrm{dx} = F \left(2\right) - F \left(0\right)$

As for the second integral, if we let $u = 2 x$, then $\mathrm{du} = 2 \mathrm{dx}$ and $\frac{1}{2} \mathrm{du} = \mathrm{dx}$, then we must also adjust the bounds of integration.

${\int}_{0}^{1} f \left(2 x\right) \mathrm{dx} = \frac{1}{2} {\int}_{0}^{2} f \left(u\right) \mathrm{du} = \frac{1}{2} \left(F \left(2\right) - F \left(0\right)\right)$

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

Hopefully this helps!

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