Re: Indefinate integrals, erroneus Natural log?
- To: mathgroup at smc.vnet.net
- Subject: [mg77921] Re: Indefinate integrals, erroneus Natural log?
- From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
- Date: Tue, 19 Jun 2007 07:01:47 -0400 (EDT)
- Organization: The Open University, Milton Keynes, UK
- References: <email@example.com>
David Rees wrote: > Hi, > > In preparation for a major exam tomorrow, I was just checking some of my > answers to past-papers with mathematica, I fed it this input: > > In:= > \!\(\[Integral]\((xE\^\(2 x\))\) \[DifferentialD]x\) > Out= > \!\(xE\^\(2\ x\)\/\(2\ Log[xE]\)\) > > I marked myself wrong and moved on to the next question, but I happened > accross the actual mark-scheme which said I was correct. It gave this > answer: > > ((1/2)xE^(2x))-(1/4)e^(2x) > > My Integral calculus isn't so strong, so I don't know if the additional > natural log should be there, or if the two expressions are identical. > > Thanks They are not. However, I strongly suspect that you made some typos while entering the expressions. I am almost certain that, for the integrand, what you wanted was "x times the base of the natural logarithm raised to the power of 2 times x, that is "x*E^(2*x)" and not "the variable called xE raised to the power of 2 times x," that is what you wrote. Moreover, the regular lowercase letter "e" denotes nothing in Mathematica: the base of the natural logarithm is written E (capital e). (Note that you can also use some special character to enter or display it. See the Basic Input palette.) Also, multiplication is denoted by a space between the variable names or by a star. Having said that, In:= expr1 = Integrate[x*E^(2*x), x] expr2 = (1/2)*x*E^(2*x) - (1/4)*E^(2*x) Simplify[expr1 == expr2] Out= 2 x 1 x E (-(-) + -) 4 2 Out= 2 x E 1 2 x -(----) + - E x 4 2 Out= True Regards, Jean-Marc