# Homework Help: Ti-nSpire giving the wrong answer for integral? Which one is right?

1. Feb 23, 2013

### Jeff12341234

I'm getting different answers for this integral

2. Feb 23, 2013

### SteamKing

Staff Emeritus
Re: Ti-nSpire giving the wrong answer for integral? Which one is right

Are you able to check which one is right by hand?

3. Feb 23, 2013

### Jeff12341234

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

I don't trust it by hand.

It looks like this one is right:

4. Feb 23, 2013

### Staff: Mentor

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

I have a hard time finding why you expect both TI and WA input to be equivalent.

5. Feb 23, 2013

### Jeff12341234

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

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.

6. Feb 23, 2013

### iRaid

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

You can get more than one correct answer for an indefinite integral, if you didn't know that.

7. Feb 23, 2013

### Jeff12341234

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

Can you show me a simple example? because what you're calling a different answer, I'm thinking is just an *equivalent* answer. In other words, the same answer, but written differently which is not what's going on here.

8. Feb 23, 2013

### Ray Vickson

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

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 $k = e^C.$

9. Feb 23, 2013

### I like Serena

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

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.

Last edited: Feb 23, 2013
10. Feb 23, 2013

### micromass

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

I don't see it either. The correct wolfram alpha entry should be: https://www.physicsforums.com/showthread.php?t=673855 and this coincides with the other answers. Well, except for the absolute value signs which really should be there.

11. Feb 23, 2013

### Jeff12341234

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

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..

Last edited by a moderator: Feb 23, 2013
12. Feb 23, 2013

### Ray Vickson

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

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.

13. Feb 23, 2013

### Jeff12341234

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

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.

14. Feb 23, 2013

### micromass

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

Wolfram alpha never even made a mistake. The OP never asked wolfram alpha to evaluate an integral, he asked wolfram alpha to simplify an expression. Wolfram alpha did that correctly since the OP wrote the expression without absolute value.

15. Feb 23, 2013

### Jeff12341234

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

I don't know about "mistake", but WA seems to be wrong. I'll do it again and post the images.

Last edited: Feb 23, 2013
16. Feb 23, 2013

### Jeff12341234

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

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)

17. Feb 23, 2013

### micromass

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

Hmm, I see: http://www.wolframalpha.com/input/?i=e^%28int+5%2F%28100+-+2t%29+dt+%29

18. Feb 23, 2013

### Jeff12341234

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

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.

19. Feb 23, 2013

### micromass

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

OK, but the two answers are equal up to a constant:

$$\frac{-5}{2}\log(-2(x-50)) = \frac{-5}{2}\log(2) + \frac{-5}{2}\log(50-x)$$

So both answers are right. Wolfram alpha just implemented his integral calculations differently from you pocket calculator.

20. Feb 23, 2013

### micromass

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

The thing is that they are both right. An integral is not a unique function, it is determined up to a constant. So if

$$\int f(x)dx = F(x) + C$$

then both F(x), F(x)+1 and F(x) + 200 are valid primitives.

21. Feb 23, 2013

### Jeff12341234

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

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.??

22. Feb 23, 2013

### micromass

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

Sure, I don't see a problem. In fact, if you take all the solutions of a DE (with some conditions), then all solutions should be disjoint and they should fill the plane. This is called the phase portrait of the DE: http://en.wikipedia.org/wiki/Phase_portrait

Yes, it will get you the right result.

In fact, a DE usually has infinitely many solutions. The easy DE $y^\prime = y$ has a solutions $y(x) = Ce^x$. For each number C, we get a solution.

23. Feb 23, 2013

### I like Serena

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

Regardless what your Ti and W|A are saying (mathematically speaking they are both wrong), the proper calculation is:

$$e^{\int \frac{5}{100-2t}dt} = \left[ \begin{matrix} e^{-\frac 5 2 \ln(100-2t) + C_1} & \qquad\text{ if } 100-2t > 0 \\ e^{-\frac 5 2 \ln(2t - 100) + C_2} & \qquad\text{ if } 100-2t < 0 \end{matrix}\right.$$

$$e^{\int \frac{5}{100-2t}dt} = \left[ \begin{matrix} e^{C_1} \cdot (100-2t)^{-5/2} & \qquad\text{ if } t < 50 \\ e^{C_2} \cdot (2t - 100)^{-5/2} & \qquad\text{ if } t > 50 \end{matrix}\right.$$

$$e^{\int \frac{5}{100-2t}dt} = \left[ \begin{matrix} \frac{e^{C_1}}{(100-2t)^{5/2}} & \qquad\text{ if } t < 50 \\ \frac{e^{C_2}}{(2t - 100)^{5/2}} & \qquad\text{ if } t > 50 \end{matrix}\right.$$

We can simplify this with new integration constants to a final answer:

$$e^{\int \frac{5}{100-2t}dt} = \left[ \begin{matrix} \frac{1}{D_1 \cdot |t - 50|^{5/2}} & \qquad\text{ if } t < 50 \\ \frac{1}{D_2 \cdot |t - 50|^{5/2}} & \qquad\text{ if } t > 50 \end{matrix}\right.$$

This answer matches both your Ti and W|M with specific choices for $D_1$ and $D_2$ (which shouldn't have been made).

24. Feb 23, 2013

### micromass

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

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 $t=50$. This means that the constants of integration will not be unique. Like you did, you found constants $D_1$ and $D_2$.

But if you're just interested in one particular solution (like WA), then the answer of WA is perfectly valid.

25. Feb 23, 2013

### I like Serena

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

Usually W|A gives the integration constant in light gray.
See for instance here.

It appears that W|A gave up on that for the more complicated expression.

Btw, doesn't the solution to an indefinite integral require the integration constant?