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 November 4th, 2009, 09:32 PM #1 Newbie   Joined: Nov 2009 Posts: 1 Thanks: 0 Help on an exam question. I was doing some past exams and I came across this: Code: integral of : e^(2x) dx = 5/2 between the terminals 0 and C (C integrates first) find C The answer is c = 1/2 loge(6) or loge(root(6)) The working out shown on the exam is really obscure, is it really the answer and if so could you tell me how you did it near the end, i keep getting similar but wrong answers.
 November 5th, 2009, 08:29 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond Re: Help on an exam question. $\int^{C}_{0}\,e^{2x}\,dx\,=\,\frac{1}{2}e^{2C}\,-\,\frac{1}{2}e^{2(0)}$ $\frac{1}{2}e^{2C}\,-\,\frac{1}{2}e^{2(0)}\,=\,\frac{5}{2}$ $\frac{1}{2}e^{2C}\,-\,\frac{1}{2}\,=\,\frac{5}{2}$ $\frac{1}{2}e^{2C}\,=\,3$ $e^{2C}\,=\,6$ Take the natural log of both sides: $2C\,=\,\ln{6}$ $C\,=\,\frac{1}{2}\ln{6}\,=\,\ln{\sqrt{6}}$
November 5th, 2009, 03:11 PM   #3
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Re: Help on an exam question.

Quote:
 Originally Posted by Stilianos The working out shown on the exam is really obscure, is it really the answer and if so could you tell me how you did it near the end, i keep getting similar but wrong answers.
Again I must add that it would be much more honest if people were asked to show their work rather than vaguely imply that they have done some before giving hand-out solutions. If the exam "working out" is really so obscure, why not offer it here for us to decide that for ourselves. If you keep getting similar but wrong answers, is it any more effort for you than for us to post your "working out" here for us to determine just where you are failing in the solution?

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