Don't know how to approach this one

[Caldus]

Let f be a function satisfying f'(x) = -3xf(x), f(1) = 4 and lim x->infinity f(x) = 0.

Evaluate integral of -3x*f(x)*dx between 1 and infinity.

Does this have to do with using comparison properties or something? I really don't know where to start. Thanks for any help.

 

[Muzza]

$$\int -3xf(x) dx = \int f'(x) dx.$$

Surely you know how to integrate f'(x).

 

[M98Ranger]

To approach this problem, we can use the Fundamental Theorem of Calculus which states that the integral of a function f(x) from a to b can be evaluated by finding the antiderivative of f(x) and evaluating it at b and a, then taking the difference of the two values. In this case, we are given the derivative of f(x) and its limits at 1 and infinity, so we can use this information to find the antiderivative and evaluate the integral.

First, we can rewrite the given derivative as f'(x) = -3x*f(x) as this will make it easier to find the antiderivative. We can then use the method of separation of variables to solve for f(x):

f'(x)/f(x) = -3x
ln(f(x)) = -3x^2/2 + C
f(x) = e^(-3x^2/2 + C)
f(x) = Ce^(-3x^2/2)

Next, we can use the given limit at infinity to solve for the constant C:
0 = lim x->infinity f(x) = lim x->infinity Ce^(-3x^2/2)
0 = C*lim x->infinity e^(-3x^2/2)
Since the limit of e^(-3x^2/2) as x approaches infinity is 0, we can solve for C and get C = 0.

Now, we can plug in the values of f(x) and C into the antiderivative and evaluate it at 1 and infinity:
Integral of -3x*f(x)*dx = Integral of -3x*Ce^(-3x^2/2)dx
= C*Integral of -3x*e^(-3x^2/2)dx
= 0*Integral of -3x*e^(-3x^2/2)dx
= 0

Therefore, the integral of -3x*f(x)*dx between 1 and infinity is equal to 0. This does not require the use of comparison properties, but rather the use of the Fundamental Theorem of Calculus and solving for the constant C. I hope this helps guide you in approaching similar problems in the future.