Supergrunch Posted November 13, 2007 Posted November 13, 2007 The intergral I= 3 S X times square root of X+1 dx. 0 Show that by using the substitution of u=x+1 gives same answer to if you used the integration by parts rule. Now I've attempted this, and gotten 2 different answers, one is a weird answer but the method I follow seems more right whereas the other answer is a nice integer, but the method I used seemed a little risky. So Can someone help? Much appreciated, thanks. I'll write your integral as below, and will generally put limits in square brackets after an expression. (we need some better maths coding stuff) I = int[0,3] x*sqrt(x+1) dx Now, the parts rule is as follows: int f(x)*g'(x) dx = f(x)*g(x) - int f'(x)*g(x) dx f(x) = x, f'(x) = 1 g'(x) = sqrt(x+1), g(x) = (2/3)(x+1)^(3/2) substituting in: I = [(2x/3)(x+1)^(3/2)][0,3] - int[0,3] (2/3)(x+1)^(3/2) dx I = [(2x/3)(x+1)^(3/2)-(4/15)(x+1)^(5/2)][0,3] I = [(2/15)(3x-2)(x+1)^(3/2)][0,3] I = 116/15 And now, the method of substitution: I = int[0,3] x*sqrt(x+1) dx u = x+1, x = u - 1, du/dx = 1, dx = du I = int[0,3] (u-1)*sqrt(u) dx I = int[1,4] u^(3/2) - sqrt(u) du (changing the limits is vital here) I = [(2/15)(3u-5)u^(3/2)][1,4] I = 116/15 Very quickly I scribbled on a bit of paper and it ended up as 116/15 ~=7.73Could be wrong though. The programming made my brain tired. Nope, that's right... see above. :wink:
Ellmeister Posted November 14, 2007 Posted November 14, 2007 Can you help me mr grunch. I don't understand where you go from: I = int[1,4] u^(3/2) - sqrt(u) du (changing the limits is vital here) to I = [(2/15)(3x-5)u^(3/2)][1,4] I got the first line, but then ended up with something different =[ Can you explain how you get it 2/15? Sorry I'm an idiot =[ Thanks for the help though
Supergrunch Posted November 14, 2007 Posted November 14, 2007 Can you help me mr grunch. I don't understand where you go from:I = int[1,4] u^(3/2) - sqrt(u) du (changing the limits is vital here) to I = [(2/15)(3x-5)u^(3/2)][1,4] I got the first line, but then ended up with something different =[ Can you explain how you get it 2/15? Sorry I'm an idiot =[ Thanks for the help though Sorry, I skipped a few lines of working there - after integrating I collected like terms and factorised. So: I = int[1,4] u^(3/2) - sqrt(u) du I = [(2/5)u^(5/2) - (2/3)u^(3/2)][1,4] I = [2u^(3/2)(u/5 - 1/3)][1,4] I = [(2/15)(3u-5)u^(3/2)][1,4] EDIT: Also realised that I put an x where I meant to put a u.
Ellmeister Posted November 14, 2007 Posted November 14, 2007 Sorry, I skipped a few lines of working there - after integrating I collected like terms and factorised. So: I = int[1,4] u^(3/2) - sqrt(u) du I = [(2/5)u^(5/2) - (2/3)u^(3/2)][1,4] I = [2u^(3/2)(u/5 - 1/3)][1,4] I = [(2/15)(3u-5)u^(3/2)][1,4] EDIT: Also realised that I put an x where I meant to put a u. Haha I was just going to ask, where the x comes from :P Thanks a lot
Zell Posted November 20, 2007 Posted November 20, 2007 Can someone please help me with this mathematical fallacy? Working with complex numbers, a student as follows: √(-1)√(-1) = -1 √(-1) = -1/√(-1) = -√1/√(-1) = -√(1/-1) = -√(-1) Hence 1 = -1! Now I know there's obviously a flaw in the logic, but I can't see where it is. I haven't done complex numbers at school yet, so I don't know too much about them. If I was to hazard a guess at where the flaw lies, it would be the application of surd laws to complex numbers.
Ellmeister Posted November 20, 2007 Posted November 20, 2007 A square root times a square root= the number of the square root. That probably made no sense so lets go for: √a times √a= a. Therefore √(-1) times √(-1)=-1 So you have -1=-1 Don't know if thats what you want :S I like complex numbers ^_^ j ftw =p
MoogleViper Posted November 20, 2007 Posted November 20, 2007 A square root times a square root= the number of the square root. That probably made no sense so lets go for: √a times √a= a. Therefore √(-1) times √(-1)=-1 So you have -1=-1 Don't know if thats what you want :S I like complex numbers ^_^ j ftw =p That makes no sense. √(-1)x√(-1) = -1 Where did you get 1 from? EDIT: Sorry I read your post wrong.
Ellmeister Posted November 20, 2007 Posted November 20, 2007 Haha I was like shit I've embarassed myself with maths again ¬_¬ Well at least you agree its right for once =D
MoogleViper Posted November 20, 2007 Posted November 20, 2007 Yeah but I don't think you answered his question.
Ellmeister Posted November 20, 2007 Posted November 20, 2007 What was his question then :S I thought it was just to show where he had gone wrong in his logic in showing -1=-1.
MoogleViper Posted November 20, 2007 Posted November 20, 2007 √(-1)√(-1) = -1 √(-1) = -1/√(-1) = -√1/√(-1) I think this is your flaw. = -√(1/-1) = -√(-1) Hence 1 = -1! I may be wrong but I don't see how he got from -1/√(-1) to -√1/√(-1)
The fish Posted November 21, 2007 Posted November 21, 2007 Working with complex numbers, a student as follows: √(-1)√(-1) = -1 √(-1) = -1/√(-1) = -√1/√(-1) = -√(1/-1) = -√(-1) Hence 1 = -1! The part highlighted in red is the flaw. -1/√(-1) / √(-1) = √(-1)/√(-1), which are the steps highlighted, equals -1/[√(-1) x √(-1)] = 1, or -1/-1 =1, which equals 1=1. I think I've explained that right, but badly. However, I'm prone to totally screw up relatively simple fractions, though, so someone else should check I've done it right.
DCK Posted November 21, 2007 Posted November 21, 2007 Can somebody show me how to evaluate the limits of sqrt(9x^6 - x) ) / (x^3 + 1 ) To both positive and negative infinity? EDIT: Hmmm, seems like squaring the nominator and then taking the root again seems like a good idea for positive infinity, negative infinity should be similar but more tricky. EDIT2: Limits are 3 and -3, right?
thirtynine. Posted December 6, 2007 Posted December 6, 2007 VB.net problem any help? Here is my problem: I have 2 forms: The first form contains an Image box. The second Form contains a text box. I need the input of the text box to be what is loaded into the image box i.e The first form loads the user clicks load the second form appears they write C://n-europe.jpg and close the form then the image box in the first form displays that image. Anyone?
Mr_Odwin Posted December 6, 2007 Posted December 6, 2007 Use stuff like: Form1.Close() Form1.PictureBox1.Image = Nothing Form1.PictureBox1.ImageLocation = Form2.TextBox1.Text Form1.Show() Form1.BringToFront() I just lifted this from some of FAVC's code so some of the stuff won't make sense, but you can fill in the gaps.
ReZourceman Posted December 7, 2007 Posted December 7, 2007 Oh my fucking god. My eyes bleed. My eyes literally bleed. I feel for you folks!
Sooj Posted January 3, 2008 Posted January 3, 2008 Hey well cut to the chase, I need notes, past essays anything on the socio-imagination. I have a 2000 word essay to be done in 12 hours (roughly) and if I don't get it done then I'm basically kicked out the course which would be sooo shit. Loads of reasons why I didn't do it earlier, one of them being that I have a pretty large lump from my arm pit to my elbow, been in a lot of pain but nothing too serious and secondly basically its christmas/new year season and, well, you know haha But yeah, I would really be indebted to anyone who is able to help me with my Sociological imagination essay. my email address is [email protected] Thanks.
Ginger_Chris Posted January 3, 2008 Posted January 3, 2008 I would but I have 5000 words to write for tomorrow. Work over Christmas should be banned. PS its about Gifted and talented education if anyone is interested in offering :P
Iun Posted January 3, 2008 Posted January 3, 2008 I would but I have 5000 words to write for tomorrow. Work over Christmas should be banned. PS its about Gifted and talented education if anyone is interested in offering :P I was in an ALP, and I teach gifted children, if you want quotes.
Supergrunch Posted January 3, 2008 Posted January 3, 2008 I would but I have 5000 words to write for tomorrow. Work over Christmas should be banned. PS its about Gifted and talented education if anyone is interested in offering :P Just be pleased you don't have collections any more. ...just found out I have mine on the 15th and 16th.
Ginger_Chris Posted January 3, 2008 Posted January 3, 2008 Just be pleased you don't have collections any more. ...just found out I have mine on the 15th and 16th. No it's worse, I have to write essays and do social science. The lack of fact and peer review is killing me. I was in an ALP, and I teach gifted children, if you want quotes. Cheers for the offer. It's a literacy review unfortunately, so lots of reading and social science "research". Its just to go towards my dissertation (another pointless exercise - why bother doing a study when it's only relevant to one school?). Hopefully it might help me get a job in a 'better' school. Hopefully. Sigh.
Supergrunch Posted January 6, 2008 Posted January 6, 2008 So, what happens to acetyl fluoride when it reacts with BF3? The resulting salt has carbon-13 NMR peaks at 7.5 and 150.3 ppm, and a strong IR absorption at 2297/cm. So obviously acetyl fluoride loses it's fluorine to BF3, but what does it turn into? The oxygen is surely still there, to shift one of the carbons so much, and clearly the peak at 7.5 means there is a methyl group or something similarly unshifted. But I have no idea what to make of the IR data - 2297 suggests some kind of triple bond, but surely that can't be right? If anyone can explain this, thanks.
MoogleViper Posted January 6, 2008 Posted January 6, 2008 I would but I have 5000 words to write for tomorrow. Work over Christmas should be banned. PS its about Gifted and talented education if anyone is interested in offering :P I was in Gifted and Talented. We didn't do much. Basically the top 30 people in the year (I was #1 ) got together and pissed about at Kingswood. Then we did this shitty forensic thing which was a load of bollocks. Then in year 9 the top 4 went to London for this leacture thing. It was quite good but the theatre was old and didn't have air con and it was the middle of summer so I fell asleep during the last lecture. Apparently I snored loudly.
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