Annoying integral ~ Help please!

  • Thread starter Peppino
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  • #1
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I have to find the integral of [tex](4-x)x^{-3}[/tex]. My TI-89 says it should be [tex]\frac{x-2}{x^{2}}+C[/tex] but I can't seem to get it myself.


I rearranged it to get [tex](4x^{-1}-1)x^{-2}[/tex] and then I used U-Substitution. And set [tex]U = 4x^{-1}-1[/tex] so that [tex]dU = -4x^{-2}dx[/tex]Then I rewrote the integral as [tex]-\frac{1}{4}\int Udu[/tex] and evaluated it to get [tex]-\frac{1}{8}U^{2}+C[/tex]or [tex]-\frac{1}{8}(4x^{-1}-1)^{2}+C[/tex] which works out to [tex]-\frac{(x-4)^{2}}{8x^{2}}+C[/tex]which 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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Answers and Replies

  • #2
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1) If you hadn't dropped the [itex]\frac{1}{8}[/itex], 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 [itex]x^{-1}[/itex] through instead of [itex]x^{-3}[/itex]?
 
  • #3
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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
 
  • #4
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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?
 
  • #5
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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
 
  • #6
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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
 
  • #7
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I have to find the integral of [tex](4-x)x^{-3}[/tex]. My TI-89 says it should be [tex]\frac{x-2}{x^{2}}+C[/tex] but I can't seem to get it myself.

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

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.
 
  • #8
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Yes, I realized that. Any guess as to why my circuitous method didn't work?
 
  • #9
SammyS
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[itex]\displaystyle -\frac{(x-4)^{2}}{8x^{2}}-\frac{x-2}{x^2}=-\frac{1}{8}[/itex]

The answers differ only by a constant, so both have the same derivative.
 
  • #10
Curious3141
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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 [itex]C[/itex], remember that they're actually different in value (they differ by [itex]\frac{1}{8}[/itex] in this case). It would be more mathematically correct to write the constant in one case as [itex]C_1[/itex] and the other as [itex]C_2[/itex] so you don't get confused.
 
  • #11
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Oh, I see it. That's interesting. So when using different methods to find an integral, the value of C (or, whats included in C) would change?
 
  • #12
SammyS
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Oh, I see it. That's interesting. So when using different methods to find an integral, the value of C (or, whats included in C) would change?
C can be different.

Two correct answers can differ by a constant.
 
  • #13
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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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