Ti-nSpire giving the wrong answer for integral? Which one is right?

[micromass]

-\frac{5}{2}∫\frac{1}{u}du = -\frac{5}{2}ln|u| + C

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 F^\prime (u)=\frac{1}{u} than just the functions F(u)=ln|u|.

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]

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 F^\prime (u)=\frac{1}{u} than just the functions F(u)=ln|u|.

It's all good if you eventually take a specific C (like you do). But I stress that these are not all the solutions!

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]

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.

I know, right? No single calculus book does it the right way. Even Apostol makes the mistake of saying \int \frac{1}{u}du = ln|u| + C. 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.

 

[I like Serena]

I know, right? No single calculus book does it the right way. Even Apostol makes the mistake of saying \int \frac{1}{u}du = ln|u| + C. 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.

I guess we need a special notation for this.
Something like:

\int \frac{1}{u}du = ln|u| + \mathfrak C
where $\mathfrak C$ denotes an arbitrary constant that can be different in disconnected parts of the domain.

Perhaps we can still write
\int \frac{1}{u}du = ln|u| \color{gray}{+ C}
with the implicit understanding that C represents a class of possibly different constants depending on the part of the domain.

 

[SammyS]

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.

No, you are wrong about that. You have an indefinite integration, so a problem of the form
e^{\int(f(x) \, dx} will have an answer of the form
e^{C + F(x)} = k\, e^{F(x)}, where $F(x)$ is an anti-derivative of $f(x),$ $C$ is an arbitrary constant and $k = e^C.$

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.

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.

I realize that this thread is already quite long and the subject has been quite thoroughly hashed over, ... but back in some fairly early posts, Ray Vickson (RVG), did a quite good job of pointing out that the only difference between the results from the TI n-spire CAS and the Wolfram Mathematica Online Integrator was a different choice for the constant of integration.

Looking at your (Jeff12341234) screen shots, it's apparent that the TI's result for the integration was

\displaystyle \int \frac{5}{1-2x}\,dx=\frac{-5\cdot\log(x-50)}{2}+C_1

\displaystyle \quad\quad\quad\ =\frac{-5\cdot\log(x-50)}{2}\ \with C₁ chosen to be zero.​

By ignoring the absolute value, this intermediate result is valid only for x > 50, but that's not the point I'm addressing in this post. (I really wanted to put the absolute values in there --- still working on my OCD --- it's really more of an affliction that a disorder!)​

Your screen shot from the W-M Integrator shows a somewhat different result.

\displaystyle \int \frac{5}{1-2x}\,dx=\frac{-5\cdot\log((-2)(x-50))}{2}

However, \displaystyle \ \ \frac{-5\cdot\log((-2)(x-50))}{2}=\frac{-5\cdot\log(50-x))}{2}+\frac{-5\cdot\log(2)}{2}\ .

This is equivalent to having

\displaystyle \quad \quad \quad \int \frac{5}{1-2x}\,dx=\frac{-5\cdot\log(50-x)}{2}+C_2\,, \ where C₂ is chosen to be \displaystyle \ \ \frac{-5\cdot\log(2)}{2}\ .​

Of course, \ \ \ e^{\left(\displaystyle \frac{-5\cdot\log(2)}{2}\right)}=\displaystyle \frac{1}{4\sqrt{2\,}}\ .

RVG had some very good advice indeed !