Timeline for Maximizing a Definite Integral Subject to Constraints
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Mar 21, 2011 at 23:22 | vote | accept | Jeff Morrow | ||
| Mar 21, 2011 at 22:46 | comment | added | Will Jagy | $$ 0 \geq \int_0^b (x-a) (f(x) - g(x)) dx = \int_0^b x (f(x) - g(x)) dx = \int_0^b x f(x) dx - \int_0^b x g(x) dx $$ See answer below by Charles Matthews. | |
| Mar 21, 2011 at 22:20 | history | edited | Will Jagy | CC BY-SA 2.5 |
I gave it a shot; added 6 characters in body
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| Mar 21, 2011 at 22:16 | comment | added | Jeff Morrow | I do not know how to get the neat integrals, but the integral of g(x) from 0 to a = [a / (b + a)]. The integral of g(x) from a to b = (1/2)(b^2 - 2ab + a^2) / (b^2 - a^2) = (b - a) / 2(b + a). So the integral from 0 to b = [(2a + b - a) / 2(b + a)] = [(b + a) / 2(b + a)] = (1/2). If I have erred in integrating, I am ashamed because it is essential that the integral = 1/2. | |
| Mar 21, 2011 at 22:00 | comment | added | Jeff Morrow | First, thank you for the type setting. That certainly is easier to read. Second, I believe that you have re-stated the problem correctly except "all possible k" may be confusing because k(x) = xg(x), and g(x) is fully defined so only one possible k(x) exists. | |
| Mar 21, 2011 at 21:10 | answer | added | Charles Matthews | timeline score: 1 | |
| Mar 21, 2011 at 21:04 | comment | added | KConrad | I edited the question so it is easier to read mathematically. If I messed up any of the conditions when I changed the typesetting, please say so. | |
| Mar 21, 2011 at 21:02 | history | edited | KConrad | CC BY-SA 2.5 |
added 59 characters in body; added 1 characters in body
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| Mar 21, 2011 at 16:18 | history | asked | Jeff Morrow | CC BY-SA 2.5 |