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Jeff12341234
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I'm getting different answers for this integral
Jeff12341234 said:There should only be one correct answer and several ways to write it. In this case, the answers are different, not just written differently. One is right, and the others are wrong. I'm trying to verify which one is truly correct.
Jeff12341234 said:There should only be one correct answer and several ways to write it. In this case, the answers are different, not just written differently. One is right, and the others are wrong. I'm trying to verify which one is truly correct.
Borek said:I have a hard time finding why you expect both TI and WA input to be equivalent.
bringing up the "+c" that gets added to any integration solution doesn't address the original question. It's unrelated. It doesn't in any way address the question of which of the 3 answers is right since the differences in the answers clearly don't have anything to do with the arbitrary constant.I like Serena said:Suppose we calculate:
$$e^{\int dt} = e^{t +C} = e^C e^t \qquad (1)$$
Both ##e^t## and ##2 e^t## are correct answers for (1).
You'll get the first if we pick C=0, and the second if we pick C=ln 2.
You might also say that the correct answer for (1) is:
$$e^{\int dt} = De^t$$
where D is an arbitrary integration constant.
It appears that your Ti-nSpire is not equipped to show integration constants.
Jeff12341234 said:bringing up the "+c" that gets added to any integration solution doesn't address the original question. It's unrelated. It doesn't in any way address the question of which of the 3 answers is right since the differences in the answers clearly don't have anything to do with the arbitrary constant.
To conclude, it appears that the image below is the correct answer + c. I don't know wtf wolfram alpha is doing..
Ray Vickson said:No, thee "+c" issue is not unrelated---it is the crux of most of your problem. The fact is that Wolfram Alpha should have written |x-50| or |50-x| instead of 50-x, but that is the only thing "wrong" about its answer, and even that is not wrong if x < 50. Exactly WHY do you think the +C issue is irrelevant? Just saying it does not make it true; you need to *demonstrate* it.
micromass said:Hmm, I see: http://www.wolframalpha.com/input/?i=e^%28int+5%2F%28100+-+2t%29+dt+%29
Jeff12341234 said:now, +c wasn't added to the first part of this 2 step process. I just took the answer wolfram gave, set it as a power of e, and let wolfram simplify it. Even so, the answer in the first step is wrong/different from what I get from the Ti-nSpire. I get -5/2*ln(x-50)
Jeff12341234 said:yep, that's essentially what I got from WA earlier. It's still different from the Ti's answer. As I've been saying since the beginning, they can't both be right.
Jeff12341234 said:The two different answers have different graphs though.
Am I to assume that either answer will give me the right result when using it during the process of solving a D.E.??
I like Serena said:Regardless what your Ti and W|A are saying (mathematically speaking they are both wrong), the proper calculation is:
micromass said:Well, strictly speaking, WA is not wrong. When wolfram alpha calculates an integral, it just gives one solution and it doesn't give the constant of integration.
micromass said:Well, strictly speaking, WA is not wrong. When wolfram alpha calculates an integral, it just gives one solution and it doesn't give the constant of integration.
Your solutions is of course correct, but it is equal for a correct choice of contants. (I wish to stress that WA allows the constants of integration to be complex!).
The trickery part in this problem is that the original function doesn't exist in the point [itex]t=50[/itex]. This means that the constants of integration will not be unique. Like you did, you found constants [itex]D_1[/itex] and [itex]D_2[/itex].
But if you're just interested in one particular solution (like WA), then the answer of WA is perfectly valid.
I like Serena said:Btw, doesn't the solution to an indefinite integral require the integration constant?
Jeff12341234 said:If you take away D1 and/or D2, the equation matches with what the Ti spit out. Therefore, it seems that the Ti would be the answer to go with.
hogrampage said:-[itex]\frac{5}{2}[/itex]∫[itex]\frac{1}{u}[/itex]du = -[itex]\frac{5}{2}[/itex]ln|u| + C
micromass said:The entire point is that this equality is not true. Although all calculus like to think that it is. There are more functions F such that [itex]F^\prime (u)=\frac{1}{u}[/itex] than just the functions [itex]F(u)=ln|u|[/itex].
It's all good if you eventually take a specific C (like you do). But I stress that these are not all the solutions!
hogrampage said:I understand that, but for the problem he is trying to solve (mixture), it is valid. It is annoying that these books don't explain the things you all are.
micromass said:I know, right? No single calculus book does it the right way. Even Apostol makes the mistake of saying [itex]\int \frac{1}{u}du = ln|u| + C[/itex]. Spivak just doesn't deal with indefinite integration, so I guess that he's clean.
I know that for his problem, the method is perfectly valid. But his confusion comes from the fact that you need integration constants, and that you sometimes need more than one integration constant.
Ray Vickson said:No, you are wrong about that. You have an indefinite integration, so a problem of the formJeff12341234 said:There should only be one correct answer and several ways to write it. In this case, the answers are different, not just written differently. One is right, and the others are wrong. I'm trying to verify which one is truly correct.
[tex] e^{\int(f(x) \, dx}[/tex] will have an answer of the form
[tex] e^{C + F(x)} = k\, e^{F(x)},[/tex] where ##F(x)## is an anti-derivative of ##f(x),## ##C## is an arbitrary constant and ##k = e^C.##
Ray Vickson said:Jeff12341234 said:bringing up the "+c" that gets added to any integration solution doesn't address the original question. It's unrelated. It doesn't in any way address the question of which of the 3 answers is right since the differences in the answers clearly don't have anything to do with the arbitrary constant.
To conclude, it appears that the image below is the correct answer + c. I don't know what wolfram alpha is doing..
No, thee "+c" issue is not unrelated---it is the crux of most of your problem. The fact is that Wolfram Alpha should have written |x-50| or |50-x| instead of 50-x, but that is the only thing "wrong" about its answer, and even that is not wrong if x < 50. Exactly WHY do you think the +C issue is irrelevant? Just saying it does not make it true; you need to *demonstrate* it.
Jeff12341234 said:The + c does need to be there. I'm not arguing that. However, it doesn't play any part in answering the question of which of the 3 answers is correct. It's just a, "by-the-way, you need to have + c in there" technicality. It's irrelevant to my original, specific, specific question.
There could be several reasons for this. One possibility is that you may have entered the wrong function or limits for your integral. Another possibility is that there may be a mistake in your calculator's settings or mode. It is also possible that the function you are trying to integrate may be too complex for the calculator to handle accurately.
To ensure accuracy, double check that you have entered the correct function and limits for your integral. Also, make sure your calculator is in the correct mode (i.e. Radian or Degree) and that your settings are appropriate for the type of integral you are trying to solve. If the function is too complex, try breaking it down into smaller parts or using a different method of integration.
Yes, there is a specific syntax for entering integrals on the Ti-nSpire. The integral symbol is represented by "int(" and the function to be integrated is entered after that, followed by a comma and the variable of integration. The limits of integration are then entered in square brackets after a comma. For example, to integrate the function x^2 from 0 to 5, the input would be "int(x^2,x,0,5)".
While the Ti-nSpire is a powerful calculator, it is not infallible. There may be certain functions or integrals that it cannot accurately solve due to their complexity. In these cases, it is best to check your work with another calculator or by hand to confirm the answer.
One common mistake is forgetting to include the variable of integration in the integrand. Another mistake is entering the limits of integration in the wrong order. It is also important to make sure that your calculator is in the correct mode and that you have entered the function and limits correctly. Additionally, if the function being integrated is discontinuous or has sharp turns, the calculator may struggle to accurately solve the integral.