Integrating a function that's proportional to...

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• April 12th 2008, 09:43 PM
Boris B
Integrating a function that's proportional to...
I am supposed to integrate a function that is proportional to
$(10+x)^{-2}$
My initial strategy was to treat that as the reciprocal of
$x^2 + 20x + 100$
but that leaves me with the whole thing raised to an exponent of -1. I don't know how to take an antiderivative anything raised to -1!

So I decided to make a new variable, A, and set that equal to x+10. Taking the antiderivative of
$A^{-2}$
was pretty easy, giving me
$-1/A = -(x+10)^{-1}$
I just don't know if what I did was mathematically correct. (Getting a definite integral at that point was easy, and resulted in an answer close to one of my multiple choices.)

The thing is, I'm treating the whole thing as if I were getting the antid. of
$(10+x)^{-2}$.
Really I'm just dealing with an unknown function proportional to that. Is this something I haven't learned yet
(Worried)
or can I just take the antiderivative of what my function is proportional to?
• April 13th 2008, 12:29 AM
Moo
Hello,

I didn't get the whole problem oO

But the antiderivate of $(10+x)^{-2}$ is easy to find.

An antiderivative of $u'(x)u^n(x)$ is $\frac{u^{n+1}(x)}{n+1}$

Here, $u(x)=10+x$
$u'(x)=1$

So $(10+x)^{-2}=1*(10+x)^{-2}=u'(x)u^{-2}(x)$

As you've calculated, an antiderivative will be $-u^{-1}(x)=\frac{-1}{10+x}$

What do you mean by proportional ? You have a function equal to :

$f(x)=a (10+x)^{-2}$ with a a constant ?
If so, it's ok, the antiderivative doesn't take care of the constants...

An antiderivative of f will be a*[antiderivative of 1/(10+x)^2]
• April 13th 2008, 11:23 AM
Boris B
Okay, thanks for helping me with the antiderivative, Moo. Here is the rascally problem in its entirety:

"The lifetime of a part has a continuous distribution on the interval (0, 40) with probability density function f, where f(x) is proportional to
$(10 + x)^{-2}$

Calculate the probability that the lifetime of the part is less than 6."

I've never seen the "proportional to..." language in any other problem, so I'm wondering if I missed something. I am betting that all all density function proportional to the same thing can be translated into a single distribution function. Unfortunately, this requires a deeper understanding of calculus than I have yet developed.
• April 13th 2008, 07:44 PM
Boris B
Okay, I figured it out. You pointed out my antiderivative was correct. In order to get from that to the actual distribution function, we had to correct by a factor: the proportionality constant. I had forgotten to take into account the fact that the integral (0 to 40) f(x) dx = 1, because it's a probability question.
$C \int_0^{40} f(x) dx = 1$
$1 = \int_0^{40} C (10+x)^{-2} dx = -C (10+x)^{-1} = C/10 - C/50$
My answer was off by exactly that factor.
• April 13th 2008, 10:56 PM
Moo
I'm sorry I can't check it, it's out of my abilities :D
• April 13th 2008, 11:03 PM
Jameson
Quote:

Originally Posted by Boris B
Okay, I figured it out. You pointed out my antiderivative was correct. In order to get from that to the actual distribution function, we had to correct by a factor: the proportionality constant. I had forgotten to take into account the fact that the integral (0 to 40) f(x) dx = 1, because it's a probability question.
$C \int_0^40 f(x) dx = 1$
$1 = \int_0^40 C (10+x)^{-2} dx = -C (10+x)^{-1} = C/10 - C/50$
My answer was off by exactly that factor.

I don't quite follow this. The term "0C" is going to make the term you want to integrate 0, thus the integral will be 0. $\int _{a}^{b} 0dx=0$ for all a,b in R I'm pretty sure.
• April 13th 2008, 11:15 PM
Boris B
Oh dear, my bad, I didn't enclose my 40s in braces, so it put 4s in superscript and left the 0 floating around. I fixed it above.

I've pretty much figured this creature out now.
• April 13th 2008, 11:23 PM
Jameson
Yeah that looks good now. Proportional to a function just means some constant times that function like you said. Good work.