Multiplication in a Definite Integral Equation

  • #1
If

[itex]f(x)=\int_0^\infty g(x) dx[/itex]

and I wanted to multiply the integral by, say, a, what would I multiply the left side by? In other words,

[itex]? \times f(x) = a \int_0^\infty g(x) dx[/itex]

Thanks in advance!
 

Answers and Replies

  • #2
778
0
I'm a little confused, why/what do you want to multiply? If you want to keep the equation the same, you multiply both sides by the same thing, if you want to change the integral, you wouldn't do anything to the other side.
 
  • #3
296
0
Call f(x)=u and the integral v. Then, do the multiplication as you would in algebra, and then substitute the integral and the function back in.
 
  • #4
chiro
Science Advisor
4,797
133
Hey drewfstr314 and welcome to the forums.

If a is independent of the integral (which it should be), then yes that it what it does. It doesn't have to be a constant, it can be a function of any variable, but it can't be something that changes the integral: it has to be orthogonal or independent of the integral unless you specify otherwise in a special constraint of some sort.
 
  • #5
Hey drewfstr314 and welcome to the forums.

If a is independent of the integral (which it should be), then yes that it what it does. It doesn't have to be a constant, it can be a function of any variable, but it can't be something that changes the integral: it has to be orthogonal or independent of the integral unless you specify otherwise in a special constraint of some sort.

What if a did change the integral. I guess I mean something like

[itex]? \times f(x) = h(x) \int_0^\infty \left( \frac{g(x)}{h(x)}\right) dx[/itex]

since multiplying the integral by h(x) would make the integral of only g(x).
 
Last edited:
  • #6
543
1
Just so you know, if your definite integral integrates with respect to x, then your final answer won't be a function of x (other than being a constant function of x), so in your first equation, f(x) = C. However, if you multiply both sides by h(x), you can't put h(x) in the denominator inside the integral.
 

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