User Name Remember Me? Password

 Calculus Calculus Math Forum

 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: 20,942 Thanks: 2210 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 Tags integration Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post rose Calculus 8 October 25th, 2017 01:08 AM LousInMaths Calculus 10 September 27th, 2012 02:26 AM gelatine1 Calculus 5 September 26th, 2012 06:38 PM fantom2012 Calculus 3 June 11th, 2012 05:49 PM rose Calculus 7 April 11th, 2010 12:51 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.      