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 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 $\sigma_x$ 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 $\sigma_x$ changed to $\sigma\,$in the second integral, "flow" not italicized, and "p" italicized in the final equation). $\int_{\,D_a}\,^{D_b}\,\frac{dD}{D}\,=\int_{\ \sigma_a}\,^{\sigma_b}\frac{d\sigma}{2B\sigma\,-\,4\tau_{\text{flow}}(1\,+\,B)}$ Given that $\sigma_x\,+\,p\,=\,2\tau_{\text{flow}}\,=\,\sigma_ {\text{flow}}$ prove that $p\,=\,\sigma_{\text{flow}}(1\,+\,1/B)(1\,-\,(D_a/D_b)^2B)$

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