For a continuous function $f$ $f$ , is it generally true that $\int_{0}^{2} f (x) d x = \int_{0}^{1} f (2 x) d x$ $\int_0^2 f(x)dx = \int_0^1 f(2x)dx$ ?

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For a continuous function $f$ $f$ , is it generally true that $\int_{0}^{2} f (x) d x = \int_{0}^{1} f (2 x) d x$ $\int_0^2 f(x)dx = \int_0^1 f(2x)dx$ ?

This seems to be true because if I "speed up" the function by going at twice the speed $(2 x)$ $(2x)$ then I should only have to swipe half the interval (from $\int_{0}^{2}$ $\int_0^2$ to $\int_{0}^{1}$ $\int_0^1$ ).

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Answer 1 · 2018-05-22T00:19:32

The correct equation is:

$\int_{0}^{2} f (x) d x = 2 \int_{0}^{1} f (2 x) d x$ $int_0^2 f(x) dx = 2 int_0^1 f(2x)dx$

Explanation:

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

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

$\int_{0}^{2} f (x) d x = \int_{0}^{1} f (2 t) (2 d t) = 2 \int_{0}^{1} f (2 t) d t$ $int_0^2 f(x) dx = int_0^1 f(2t) (2dt) = 2 int_0^1 f(2t)dt$

So, in fact you need to swipe only the interval $[0, 1]$ $[0,1]$ 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 $\int f (x) d x = F (x)$ $int f(x) dx = F(x)$ . Then

$\int_{0}^{2} f (x) d x = F (2) - F (0)$ $int_0^2 f(x) dx = F(2) - F(0)$

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

$\int_{0}^{1} f (2 x) d x = \frac{1}{2} \int_{0}^{2} f (u) d u = \frac{1}{2} (F (2) - F (0))$ $int_0^1 f(2x)dx = 1/2int_0^2 f(u) du = 1/2(F(2) - F(0))$

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

Hopefully this helps!

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For a continuous function $f$ $f$ , is it generally true that $\int_{0}^{2} f (x) d x = \int_{0}^{1} f (2 x) d x$ $\int_0^2 f(x)dx = \int_0^1 f(2x)dx$ ?

This seems to be true because if I "speed up" the function by going at twice the speed $(2 x)$ $(2x)$ then I should only have to swipe half the interval (from $\int_{0}^{2}$ $\int_0^2$ to $\int_{0}^{1}$ $\int_0^1$ ).

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Explanation:

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For a continuous function ff, is it generally true that ∫20f(x)dx=∫10f(2x)dx\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)(2x) then I should only have to swipe half the interval (from ∫20\int_0^2 to ∫10\int_0^1).

2 Answers

Explanation:

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For a continuous function $f$ $f$ , is it generally true that $\int_{0}^{2} f (x) d x = \int_{0}^{1} f (2 x) d x$ $\int_0^2 f(x)dx = \int_0^1 f(2x)dx$ ?

This seems to be true because if I "speed up" the function by going at twice the speed $(2 x)$ $(2x)$ then I should only have to swipe half the interval (from $\int_{0}^{2}$ $\int_0^2$ to $\int_{0}^{1}$ $\int_0^1$ ).