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

• Feb 6th 2010, 10:40 AM
s3a
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!
• Feb 6th 2010, 11:14 AM
running-gag
Hi

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

$\displaystyle \int_{0}^{2} f(2x) dx$
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
$\displaystyle \int_{0}^{2} f(2x) dx = \int_{0}^{4} f(u) \frac{du}{2} = \frac12 \int_{0}^{4} f(u) du = \frac12 \times 10 = 5$
• Feb 6th 2010, 11:14 AM
nehme007
Let $\displaystyle u = 2x$ and $\displaystyle du = 2dx$, then
$\displaystyle \int_0^2 f(2x) dx = \frac{1}{2}\int_0^4 f(u) du = 5$.
• Feb 8th 2010, 08:20 AM
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).
• Feb 8th 2010, 10:33 AM
HallsofIvy
Quote:

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).

$\displaystyle \int_a^b f(x)dx= \int_a^b f(u)du$, that's true because the variable inside the integral is a "dummy variable". $\displaystyle \int_a^b f(x)dx$ is number, not a function of x. Similarly $\displaystyle \int_a^b f(u)du$ is a number not a function of u. Since it doesn't matter what we call the variable, the two numbers must be equal.
$\displaystyle \sum_{i=1}^3 i^2= 1^2+ 2^2+ 3^2= 1+ 4+ 9= 14$
$\displaystyle \sum_{n=1}^3 n^2= 1^2+ 2^2+ 3^2= 1+ 4+ 9= 14$
so $\displaystyle \sum_{i=1}^3 i^2= \sum_{n=1}^3 n^2$.