Integration

[Viper915]

I need some help with an integration problem. I have the problem and a conversation i had with a friend uploaded as a txt file, so that should help a bit with the explanation. We came up with 2 sets of answers that seem to be legitimate, so we're stumped. Thanks in advance.

http://members.shaw.ca/w0lf/CalcProblem6.txt

[Damned charming :)]

Isn't the average value of g(x) equal to
frac{1}{2} \int_1^3 g(x) dx =\frac{1}{2} \int_1^3 A -f(x)= A - \frac{1}{2} A= \frac{1}{2}A dx

I really don't understand when you say
g(2~3) = g(3) -g(2)
do you mean \int_2^3 g(x) = G(3) - G(2)
where G(x) is a function that has derivative g(x)?

[cookiemonster]

The answers I reached were:

\int_1^3 f(x)\,dx = A
\overline{g(x)|_1^3} = \frac{A}{2}
k = 4

I can show you how I got those if you want.

And sorry, but I didn't read your little log...

Edit: I should probably note that notation I used in the second line a butchery of a few different notations, and I don't suggest using it.

cookiemonster

[NateTG]

So, a little cleaning of the problem:

g(x)=A-f(x)
\int_1^2f(x)dx=\int_2^3g(x)dx
and
\int_2^3f(x)dx=-3A

a) find \int_1^3f(x) dx
\int_2^3g(x)dx=\int_2^3 A-f(x) dx
right?
b) find the average value of g(x) on [1,3]
How about using \frac{\int_1^3g(x)dx}{3-1}?
c)find the value of k if \int_0^1f(x+1)dx=kA.
If
y=x+1
then that's
\int_1^2f(y)dy

[Damned charming :)]

I really think you need to tidy up that text discussion and
make it very clear what you are claiming.

One line made no sense.
dUKeRuLz 16: wouldnt (2~3)g(x)dx = g(3) - g(2) = A - f(3) - (A - f(2))

[g)] This line seems to be claiming
\int_2^3 g(x) = g(3) - g(2) = A -f(3) -(A -f(2))
How can you say that when you intergrate g(x) you still get g(x) unless you know g(x) = e^x?

[Viper915]

Cookie, that's what i had, also, and damned charming, that second part is what i mean. as for the average value part, i'm unfamiliar with that frac12 notation, but if i can get the correct answer to part a, then i'm set.

[cookiemonster]

Part a is pretty simple if you just make the substitution given by the assertion made in part (i), as NateTG showed. Once you split up the integral, nothing in it is unknown.

cookiemonster

[Viper915]

and charming, about that line, it gets really confusing trying to draw integrals over aim, so we might have lost something in there, hehe. Nate, thanks for cleaning that up. if you guys can show me how you're making those nice looking integrals, i'd really appreciate it

[cookiemonster]

LaTeX is a great thing. Read this thread to see how to use it on physicsforums.

http://www.physicsforums.com/showthread.php?s=&threadid=8997

cookiemonster

[Damned charming :)]

frac12 was a typo I just meant \frac{1}{2}. Sorry I typed it wrong and Now I have fixed it.

[Viper915]

yikes, you guys are replying faster than i can read these! cookie, that's what i thought when i did it at first, but then when my friend started asking me about it, her answers seemed to work too, but i can see how they wouldn't because of the integrating g(x) problem. thanks a bunch, you guys, this problem has been making my head hurt :\