Constant of IntegrationDate: 03/04/2003 at 10:08:28 From: Denz Subject: What's wrong with this proof? Using integration by parts integration of (1/x)dx = [x * (1/x)]+ integration of (1/x)dx After simplifying by using the addition property of equality and multiplication, the answer would lead to 0 = 1, which should be wrong. The proof seems correct. Date: 03/04/2003 at 10:23:50 From: Doctor Jacques Subject: Re: What's wrong with this proof? Hi Denz, This looks like a paradox indeed, but try to see what you get if you evaluate the integral over an interval [a,b]... Please feel free to write back if you are still stuck. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/ Date: 03/04/2003 at 10:31:11 From: Denz Subject: What's wrong with this proof? The problem is possible in the interval [a,b] but my teacher insists that there is a problem with the proof, while every one of us thinks the proof is correct. Date: 03/05/2003 at 02:10:15 From: Doctor Jacques Subject: Re: What's wrong with this proof? Hi Denz, The problem is that an indefinite integral (antiderivative) is only defined up to an additive constant. More technically, it is not a single function, but an equivalence class of functions. For example, INT(0 dx) = C, where C is any constant, since the derivative of a constant is 0. In this case, we should have written: (INT{dx/x} + C_1) = 1 + (INT{dx/x} + C_2) and this merely shows that the constants must satisfy C_1 = 1 + C_2. When you compute a "real" integral, i.e. between limits, these constants disappear. Does this make sense? Please write back if you want to discuss this further. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/ Date: 03/05/2003 at 04:39:58 From: Denz Subject: Re: What's wrong with this proof? As I read your explanation I got confused with the constants. Isn't it that the constant of int(dx/x) on the left-hand side of the equation is equal to the constant at the right-hand side of the equation? Date: 03/05/2003 at 06:46:19 From: Doctor Jacques Subject: Re: What's wrong with this proof? Hi Denz, These constants have no actual meaning - they are artificial. When we write INT{f(x)dx} = g(x) we simply mean that the derivative of g(x) is f(x). Of course, the derivative of g(x) + C, where C is _any_ constant, is also f(x). The particular constant that comes out depends on the method of integration. The whole point of the exercise is to show that different calculations can yield functions that differ by a constant. We can even illustrate this with the function 1/x in another simpler way. We know that the "true" integral is ln(x). Now, in INT{dx/x}, if we make the substitution ax = u, with a > 0, you will easily see that the result is ln(ax) = ln(x) + ln(a) = ln(x) + constant and, as we can take any positive number for a, we can make the constant ln(a) anything we wish. There is no contradiction in writing: INT{dx/x} = ln(x) INT{dx/x} = ln(x) + C because these are not true equalities between functions. This is exactly the same as modular arithmetic. When we write 2 = 7 (mod 5) the numbers 2 and 7 are not simple numbers. 2 represents all the numbers that are a multiple of 5 + 2, and 7 represents all the numbers that are a multiple of 5 + 7, and these sets of numbers are the same - that is what the equality means (in this case, we often use a special symbol instead of =, to mean congruence). In a similar way, an expression like INT{f(x)dx} represents, not a single function, but the set of all functions whose derivative is f(x). An equality between integrals is an equality between sets of functions. Does this clarify things ? - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/ Date: 03/05/2003 at 07:37:13 From: Denz Subject: Thank you (What's wrong with this proof?) Thanks for the clarification. |
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