Expert: Paul Klarreich - 3/31/2006

Question
Hi Paul,
Let f be a continous function such that the integral of [f(2x)] dx from 2 to 3 =8. What is the value of the integral of [f(x)] dx from 4 to 6. How are you supposed to think through this problem?
I was able to get the answer 16 through a lot of substituting. I can only guess that the reason the answer is 16 is because it's double the first integral[the first integral goes from 2 to3(1 unit) and the 2nd goes double,4 to 6(2 units)].But if this is true, why doesn't the change that one function is f(2x) and the other is f(x) afect the answer?

Answer
Hi, Jeff,
Subject:  integration given 1 definite integral
Question:  Hi Paul,

Let f be a continous function such that the integral of [f(2x)] dx from 2 to 3 = 8.

I was able to get the answer 16 through a lot of substituting. I can only guess that the reason the answer is 16 is because it's double the first integral[the first integral goes from 2 to3(1 unit) and the 2nd goes double,4 to 6(2 units)].  But if this is true, why doesn't the change that one function is f(2x) and the other is f(x) afect the answer?

What is the value of the integral of [f(x)] dx from 4 to 6. How are you supposed to think through this problem?

>> Substitution of variables is a powerful technique.  In this case you have  f(2x) when you would like it to say f(x).  So make it say that by the following substitution:

You have:
{3
| f(2x) dx = 8,  as given.
}2

Let  t = 2x, (Don't worry that 't' isn't 'x'.) then:

x = 2  ->  t = 4
x = 3  ->  t = 6
dt = 2dx,  ->  dx = dt/2

So the integral becomes:

{6
| f(t) dt /2  = 8, still the same.
}4

1  {6
--- | f(t) dt  = 8
2  }4

{6
| f(t) dt  = 16
}4

And of course,

{6
| f(x) dx  = 16
}4

BECAUSE the variable of integration is a dummy variable.  All of these are exactly the same:

{6
| f(t) dt
}4

{6
| f(x) dx
}4

{6
| f(w) dw
}4

and so on, to the 23 other letters of the alphabet, along with a few others from the Greek, Russian, etc, alphabets.