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

**s3a** · February 6th 2010, 10:40 AM

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.

I have no idea how to do this. Someone please help me!

Answer:
5

Any input would be greatly appreciated!
Thanks in advance!

**running-gag** · February 6th 2010, 11:14 AM

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
$\int_{0}^{2} f(2x) dx = \int_{0}^{4} f(u) \frac{du}{2} = \frac12 \int_{0}^{4} f(u) du = \frac12 \times 10 = 5$

**nehme007** · February 6th 2010, 11:14 AM

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

**s3a** · February 8th 2010, 08:20 AM

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).

**HallsofIvy** · February 8th 2010, 10:33 AM

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).

If you are asking why
$\int_a^b f(x)dx= \int_a^b f(u)du$ , 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.

The same thing happens with sums:
$\sum_{i=1}^3 i^2= 1^2+ 2^2+ 3^2= 1+ 4+ 9= 14$
$\sum_{n=1}^3 n^2= 1^2+ 2^2+ 3^2= 1+ 4+ 9= 14$

so $\sum_{i=1}^3 i^2= \sum_{n=1}^3 n^2$ .

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