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

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In summary, an integral is a mathematical concept used to represent the area under a curve on a graph and is important in calculus for solving problems involving rates of change, accumulation, and optimization. The "Annoying integral" is difficult because it does not have a closed-form solution and can be solved using advanced techniques such as substitution, integration by parts, and trigonometric identities. While it may seem frustrating, it is important in fields such as physics, engineering, and economics. While a computer can be used to approximate the value of the "Annoying integral," it is still important to understand the underlying concepts and techniques used to solve it manually.
  • #1
Peppino
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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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  • #2
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
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
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
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
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
Peppino said:
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.

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 [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
Yes, I realized that. Any guess as to why my circuitous method didn't work?
 
  • #9
[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
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 [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
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.
 

1. What is an integral?

An integral is a mathematical concept that represents the area under a curve on a graph. It is a fundamental tool in calculus and is used to solve problems involving rates of change, accumulation, and optimization.

2. Why is the "Annoying integral" difficult?

The "Annoying integral" is difficult because it does not have a closed-form solution, meaning it cannot be expressed in terms of elementary functions. This makes it challenging to solve using traditional methods and requires more advanced techniques.

3. How can I solve the "Annoying integral"?

There are a few different methods that can be used to solve the "Annoying integral." Some common approaches include using substitution, integration by parts, and trigonometric identities. It may also be helpful to break the integral into smaller, more manageable parts.

4. Why is the "Annoying integral" important?

While the "Annoying integral" may seem frustrating and irrelevant, it is actually an important concept in mathematics and science. It helps us understand the relationship between a function and its derivative, which is crucial in many fields such as physics, engineering, and economics.

5. Can I use a computer to solve the "Annoying integral"?

Yes, you can use a computer to approximate the value of the "Annoying integral." There are many numerical methods and software programs available that can provide a close approximation of the integral. However, it is still important to understand the underlying concepts and techniques used to solve integrals manually.

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