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November 27th, 2010, 04:00 AM   #1
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Integration help

Can you please prove this integration in the photo?
http://s401.photobucket.com/albums/p...gggggggggg.jpg



this may helb
The left hand side is pretty easy. You should recognize that as the natural logarithm of D. So we get

ln(D) | Da, Db

where I'm using the | sign to indicate the limits of integration. Then, we get

ln(Db) - ln(Da)

When you subtract logs, it's the same as taking the log of the quotient, so

ln(Db / Da)

Now let's work with the right hand side. Let's simplify the integral a little bit, by defining a = 2B, and b = -4?(1+B). Then, we have

int(d? / (a?+b), ?a, ?b)

Now, to do the integral, let's make the substitution x = a?+b. Then,

dx = a * d?
d? = dx / a

So we can write

int((dx/a) / x, ?a, ?b)
1/a * int(dx / x, ?a, ?b)

Now, the integral is just the natural log of x, so we have

1/a * ln(x) | ?a, ?b

Now, if we substitute back in for x, we have

1/a * ln(a? + b) | ?a, ?b

Plugging in the limits of integration, we get

1/a * ln(a * ?a + b) - 1/a * ln(a * ?b + b)
1/a * [ln(a * ?a + b) - ln(a * ?b + b)]

We'll reuse the rule of logarithms, that when you subtract them, it's the same as taking the log of the quotient, so

1/a * ln[(a * ?a + b) / (a * ?b + b)]

Now, another rule of logarithms is that log(a^b) = b * log(a). We can use that in reverse, to write

ln[(a * ?a + b)^(1/a) / (a * ?b + b)^(1/a)]

Now, put the left and right sides together, and you get

ln(Db / Da) = ln{[(a * ?a + b) / (a * ?b + b)]^(1/a)}

Since the logs are equal, the arguments must be equal. That means

Db / Da = [(a * ?a + b) / (a * ?b + b)]^(1/a)

If we raise both sides to the power of a, we get

(Db / Da)^a = {[(a * ?a + b) / (a * ?b + b)]^(1/a)}^a
(Db / Da)^a = [(a * ?a + b) / (a * ?b + b)]^(1/a * a)
(Db / Da)^a = [(a * ?a + b) / (a * ?b + b)]^1
(Db / Da)^a = (a * ?a + b) / (a * ?b + b)

It's all pretty basic algebra from here. Let's plug a and b back in.

(Db / Da)^(2B) = [2B * ?a + -4?(1+B)] / [2B * ?b + -4?(1+B)]
(Db / Da)^(2B) = [2B * ?a - 4?(1+B)] / [2B * ?b - 4?(1+B)]

Isolate ?a.

[2B * ?b - 4?(1+B)] * (Db / Da)^(2B) = 2B * ?a - 4?(1+B)
[2B * ?b - 4?(1+B)] * (Db / Da)^(2B) + 4?(1+B) = 2B * ?a
{[2B * ?b - 4?(1+B)] * (Db / Da)^(2B) + 4?(1+B)} / (2B) = ?a
[?b - 2?(1+B)/B] * (Db / Da)^(2B) + 2?(1+B)/B = ?a

Divide by 2?.

{[?b - 2?(1+B)/B] * (Db / Da)^(2B) + 2?(1+B)/B} / (2?) = ?a / (2?)
[?b/(2?) - (1+B)/B] * (Db / Da)^(2B) + (1+B)/B = ?a / (2?)
?b/(2?) * (Db / Da)^(2B) - (1+B)/B * (Db / Da)^(2B) + (1+B)/B = ?a / (2?)
?b/(2?) * (Db / Da)^(2B) + (1+B)/B * [-(Db / Da)^(2B) + 1] = ?a / (2?)
?b/(2?) * (Db / Da)^(2B) + (1+B)/B * [1 - (Db / Da)^(2B)] = ?a / (2?)

Rearranging a little,

?a / (2?) = (1+B)/B * [1 - (Db / Da)^(2B)] + ?b/(2?) * (Db / Da)^(2B)

Thanks in advance
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November 27th, 2010, 05:46 AM   #2
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I didn't understand in the second equation or why the limits in the second integral disappear.

To avoid the sheer torture of photobucket, below is a LaTeX version of the problem (with changed to in the second integral, "flow" not italicized, and "p" italicized in the final equation).



Given that



prove that

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