Why Does My Calculated Integral Differ from My Calculator's Result?

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SUMMARY

The forum discussion centers on the integral of the function (4-x)x^{-3} and the discrepancies between the user's manual calculations and the results from a TI-89 calculator. The user initially attempted U-Substitution but ultimately found that their method complicated the problem unnecessarily. The correct integral can be derived simply by distributing x^{-1} through the expression, leading to the conclusion that both the user's and calculator's answers differ only by a constant, which is valid in indefinite integration. The discussion highlights the importance of recognizing that different methods can yield equivalent results differing by a constant.

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Peppino
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I have to find the integral of (4-x)x^{-3}. My TI-89 says it should be \frac{x-2}{x^{2}}+C but I can't seem to get it myself.I rearranged it to get (4x^{-1}-1)x^{-2} and then I used U-Substitution. And set U = 4x^{-1}-1 so that dU = -4x^{-2}dxThen I rewrote the integral as -\frac{1}{4}\int Udu and evaluated it to get -\frac{1}{8}U^{2}+Cor -\frac{1}{8}(4x^{-1}-1)^{2}+C which works out to -\frac{(x-4)^{2}}{8x^{2}}+Cwhich is not what my calculator says it should be.

What am I doing wrong?

...and if I have to make another Latex equation...
 
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1) If you hadn't dropped the \frac{1}{8}, your answer is correct.

2) Your calculator also has a correct answer.

3) They're actually the same answer. How can that be?

4) You're making this integral WAY more difficult than it actually is. You don't need a substitution. Why did you just distribute x^{-1} through instead of x^{-3}?
 
Right... I meant to put the 1/8 in there. But when I graph the two expressions I get two completely different tables of values. Are you sure they are the same?

As for the unnecessary U-Substitution... that's just me being a moron I guess
 
What do you see when you look at the graphs? How are the graphs comparable?

What is an indefinite integral? Is it a function, or perhaps something ... broader? Whats that +C for?
 
The table of values are completely different... on my calculator's answer, x = 1 yields -1, while on mine x = 1 yields -1.125.

Is it not a function?

Because an infinite number of different values of C could lead to a function with a derivative of the original function
 
Doing the non-moronic way I get the same answer as my calculator. I'll stick with that for now but any comments will still be appreciated
 
Peppino said:
I have to find the integral of (4-x)x^{-3}. My TI-89 says it should be \frac{x-2}{x^{2}}+C but I can't seem to get it myself.

gopher_p said:
4) You're making this integral WAY more difficult than it actually is. You don't need a substitution. Why did you just distribute x^{-1} through instead of x^{-3}?

In other words, carry out the multiplication to arrive at this integral:
$$ \int 4x^{-3} - x^{-2}~dx$$

It should be pretty simple after that.
 
Yes, I realized that. Any guess as to why my circuitous method didn't work?
 
\displaystyle -\frac{(x-4)^{2}}{8x^{2}}-\frac{x-2}{x^2}=-\frac{1}{8}

The answers differ only by a constant, so both have the same derivative.
 
  • #10
Peppino said:
Doing the non-moronic way I get the same answer as my calculator. I'll stick with that for now but any comments will still be appreciated

To add to what SammyS said, remember that that so-called "arbitrary constant" of indefinite integration is really that - completely arbitrary. Any constant added to your answer will still give a completely valid answer. So when two expressions differ only be a constant value, they are both equally valid solutions.

Even though you wrote the "expected" solution and yours with the constant having the same symbol C, remember that they're actually different in value (they differ by \frac{1}{8} in this case). It would be more mathematically correct to write the constant in one case as C_1 and the other as C_2 so you don't get confused.
 
  • #11
Oh, I see it. That's interesting. So when using different methods to find an integral, the value of C (or, what's included in C) would change?
 
  • #12
Peppino said:
Oh, I see it. That's interesting. So when using different methods to find an integral, the value of C (or, what's included in C) would change?
C can be different.

Two correct answers can differ by a constant.
 
  • #13
For a different example that shows how two different techniques can produce what seem to be different antiderivatives, consider
$$ \int sin(x)cos(x)dx$$

If you let u = sin(x), du = cos(x)dx, you get an antiderivative of (1/2)u2 + C = sin2(x) + C1

OTOH, if you let u = cos(x), du = -sin(x)dx, and you get an antiderivative of -(1/2)u2(x) + C2 = -(1/2)cos2(x) + C2.

Even though the two answers appear to be significantly different, they differ by a constant (1/2). The key is the identity sin2(x) + cos2(x) = 1.
 

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