# Thread: Something simple about integral theorey but I don't get it.

1. ## Something simple about integral theorey but I don't get it.

Question:
"If f is continuous and if this (integral (f(x)dx) from 0 to 4 - Wolfram|Alpha) integral = 10. Then find, this(integral (f(2x)dx) from 0 to 2 - Wolfram|Alpha) integral.

5

Any input would be greatly appreciated!

2. Hi

We know that $\int_{0}^{4} f(x) dx = 10$ and we want to calculate $\int_{0}^{2} f(2x) dx$

$\int_{0}^{2} f(2x) dx$
Let u = 2x
Then x going from 0 to 2, u is going from 2x0=0 to 2x2=4
and du = 2dx or dx = du/2
Therefore
$\int_{0}^{2} f(2x) dx = \int_{0}^{4} f(u) \frac{du}{2} = \frac12 \int_{0}^{4} f(u) du = \frac12 \times 10 = 5$

3. Let $u = 2x$ and $du = 2dx$, then
$\int_0^2 f(2x) dx = \frac{1}{2}\int_0^4 f(u) du = 5$.

4. These reponses have helped get further along the problem but I am still stuck on the last step. I do not understand how f(u)du = f(x)dx (since the integral is being evulated over the same interval, etc and that is the only thing that differs).

5. Originally Posted by s3a
These reponses have helped get further along the problem but I am still stuck on the last step. I do not understand how f(u)du = f(x)dx (since the integral is being evulated over the same interval, etc and that is the only thing that differs).
$\int_a^b f(x)dx= \int_a^b f(u)du$, that's true because the variable inside the integral is a "dummy variable". $\int_a^b f(x)dx$ is number, not a function of x. Similarly $\int_a^b f(u)du$ is a number not a function of u. Since it doesn't matter what we call the variable, the two numbers must be equal.
$\sum_{i=1}^3 i^2= 1^2+ 2^2+ 3^2= 1+ 4+ 9= 14$
$\sum_{n=1}^3 n^2= 1^2+ 2^2+ 3^2= 1+ 4+ 9= 14$
so $\sum_{i=1}^3 i^2= \sum_{n=1}^3 n^2$.