November 27th, 2010, 04:00 AM  #1 
Newbie Joined: Nov 2010 Posts: 8 Thanks: 0  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 
November 27th, 2010, 05:46 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,469 Thanks: 2038 
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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