multiple integrals

[CPL.Luke]

is it possible to have multiple integrals of the same function for instance

 

 

when you differentiate a function like f(x)g(x)

 

then you get f(x)g'(x)+g(x)f'(x)

 

but then integrating that function becomes integral of (f(x)g'(x)+ integral of f'(x)g(x)

 

which is most certainly not f(x)g(x)

[Dave]

There's no general rule for integrating [math]f(x)g(x)[/math]. You have to work it out by hand, I'm afraid. The closest you're going to get is integration by parts.

[Johnny5]

[math] \text{Integration By Parts Lecture} [/math]

 

Suppose you are given two functions at random:

 

1. f(x)

2. g(x)

 

Now, take their product:

 

[math] \text{f(x)} \cdot \text{g(x)} [/math]

 

Which can also be written without the dot, multiplication implicit instead of explicit, as follows:

 

[math] \text{f(x) g(x)} [/math]

 

Now, take the derivative with respect to x, of their product:

 

[math] \text{f(x) g(x)} [/math]

 

[math] \frac{d}{dx} [\text{f(x) g(x)} ] = \frac{df}{dx} g(x) + f(x) \frac{dg}{dx} [/math]

 

Where I have made use of the product rule of the differential calculus, to be proven at the end of the lecture.

 

Now multiply both sides of the equation above by dx, to obtain:

 

[math] d[\text{f(x) g(x)} ] = df g(x) + f(x) dg [/math]

 

Now, integrate both sides of the formula above:

 

[math] \int d[\text{f(x) g(x)} ] = \int[df g(x) + f(x) dg] [/math]

 

Now, the integral of a sum is the sum of the integrals, that is:

 

[math] \int[df g(x) + f(x) dg] = \int df g(x) + \int f(x) dg [/math]

 

Therefore, using the transitive property of equality:

 

[math] \int d[\text{f(x) g(x)} ] = \int df g(x) + \int f(x) dg [/math]

 

Integral d(anything) = anything + C

 

Ignoring the arbitrary constant of integration we have: