Natural log/definite integral

In summary, the problem is confusing and the book provides only an equation for integrals in the form of du/u, which is not the form used in this problem. It appears that if I take u = x^6 + 6x and plug in the lower and upper limit, I will get a new limit that can be utilized in the correct answer. I already know what the answer is, but I want to figure out how to get to the answer: factor out du = 6(x^5 + 1)).
  • #1
16
0

Homework Statement



Evaluate.

Integral of (x^5 + 1) / (x^6 + 6x) on x from 2 to 3.

[tex]
\int_2^3 (x^5 + 1) / (x^6 +6x) dx
[/tex]

Homework Equations



Not sure! The book only seems to give an equation for integrals in the form of du/u, which is integral 1/u*du = ln |u| + C. It also says that if u = f(x) THEN du = f'(x)?? Is that supposed to be a logical change? All of the integrals of this type are actually du/u in the book. This problem is not and is confusing the heck out of me.

The Attempt at a Solution



It appears that if I take u = x^6 + 6x and plug in the lower and upper limit, I will get a new limit that can be utilized in the correct answer. I already know what the answer is, but I want to figure out how to get to the answer:

1/6 ln (747/76)

The upper and lower limit when plugged into u are there: ln(747/76). I don't understand where the 1/6 is coming from and I don't understand how/why those two limits appears as a fraction.

Thanks for any help!
 
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  • #2
If you have u=x^6+6x then what is du? (Factor it to see what you get)
 
  • #3
rock.freak667 said:
If you have u=x^6+6x then what is du? (Factor it to see what you get)

if u is x^6 + 6x then du = 6x^5 + 6 -> 6(x^5 + 1)...
 
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  • #4
mathor345 said:
if u is x^6 + 6x then du = 6x^5 + 6 -> 6(x^5 + 1)...
The dx is very important!

Actually, du = 6(x^5 + 1) dx

→ (1/6) du = (x^5 + 1) dx
 
  • #5
SammyS said:
The dx is very important!

Actually, du = 6(x^5 + 1) dx

→ (1/6) du = (x^5 + 1) dx

Okay... I see how the 1/6 comes into the answer. I don't see why the upper and lower limit appear inside of the ln (u) though? I've never been this confused over a single problem before.
 
  • #6
You have to undo the substitution. You can't just substitute 2 and 3 in for u; these are values of x.
 
  • #7
Mark44 said:
You have to undo the substitution. You can't just substitute 2 and 3 in for u; these are values of x.

I know, I meant that those limits are substituted for x in u (f(x)).

Ok, so apparently for a definite integral

[tex]\int_2^3\frac{x^5 + 1}{x^6+6x}dx[/tex]

must be converted into the form:

[tex]\int_2^3\frac{1}{u}du[/tex]

So,

[tex]u = x^6 + 6x[/tex] so,

[tex]du = 6x^5 + 6[/tex]

factor out [tex]du = 6(x^5 + 1)[/tex] --> [tex]\frac{1}{6}du[/tex]

Plugging in the limits, x, into u:

[tex]u(2) = 76[/tex] and [tex]u(3) = 747[/tex]

So,

[tex]\int_x^y\frac{1}{6}\frac{du}{u} = \frac{1}{6} ln \|u}|]\right_x^y[/tex]

(x and y are 76 and 747 respectively. LaTeX is acting funny)

-> ln |747| - ln |76| -> 1/6 ln (747/76)

So my questions are:

1) Can someone explain why the 1/6 can be brought out in front like that?
2) In the last formula du/u = ln |u| why are the terms SUBTRACTED as above so allowing for the quotient rule to be used?
 
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  • #8
mathor345 said:
I know, I meant that those limits are substituted for x in u (f(x)).

Ok, so apparently for a definite integral

[tex]\int_2^3\frac{x^5 + 1}{x^6+6x}dx[/tex]

must be converted into the form:
It's not that it "must be converted" the form below; that's what you get when you do the substution u = x6 + 6x, and du = 6(x5 + 1)dx.

Notice that in the numerator above you don't quite have du, so you multiply by 6 and divide by 6 (i.e., multiply by 1) to get what you need. The 6 in the numerator goes to make up what is needed for du. The 6 in the denominator goes outside the integral as a factor of 1/6. I think you have been missing that throughout this thread.
mathor345 said:
[tex]\int_2^3\frac{1}{u}du[/tex]
Not quite. As explained above, there is a factor of 1/6, so the integral is
[tex]\frac{1}{6}\int_{x = 2}^3\frac{du}{u}[/tex]

mathor345 said:
So,

[tex]u = x^6 + 6x[/tex] so,

[tex]du = 6x^5 + 6[/tex]
Please read the responses more carefully. This is du = (6x5 + 6)dx, or du = 6(x5 + 1))dx. If you consistently leave off the dx you will be opening yourself up to grievous problems up ahead. That's guaranteed.
mathor345 said:
factor out [tex]du = 6(x^5 + 1)[/tex] --> [tex]\frac{1}{6}du[/tex]
?
du = 6(x5 + 1)dx ==> du/6 = (x5 + 1)dx
mathor345 said:
Plugging in the limits, x, into u:
No, no, not yet! You forgot to integrate!
mathor345 said:
[tex]u(2) = 76[/tex] and [tex]u(3) = 747[/tex]

So,

[tex]\int_x^y\frac{1}{6}\frac{du}{u} = \frac{1}{6} ln \|u}|]\right_x^y[/tex]

(x and y are 76 and 747 respectively. LaTeX is acting funny)

-> ln |747| - ln |76| -> 1/6 ln (747/76)

So my questions are:

1) Can someone explain why the 1/6 can be brought out in front like that?
2) In the last formula du/u = ln |u| why are the terms SUBTRACTED as above so allowing for the quotient rule to be used?

1) Explained above and elsewhere in this thread.
2) No, the formula isn't du/u = ln |u|; it's
[tex](1/6)\int \frac{du}{u} = (1/6) ln |u| + C[/tex]
In this case, after undoing the substitution, the right side above becomes
(1/6) ln|x6 + 6x|, which you evaluate at x = 3 and x = 2. (For a definite integral you don't need to include the constant C.)

This gives you (1/6){ ln(747) - ln(76)}. The two log terms can be combined using the properties of logs.
 
  • #9
Thanks for responses, especially Mark for the detailed response.

It's been a year since my last math class, so I was obviously missing some of the basics. I was completely ignoring the dx as if it had no meaning, and was a bit confused with moving constants in front of the integral sign. I understand what I was doing wrong. Thanks again.
 

1. What is a natural log?

A natural log, or ln, is a type of logarithm that uses the constant e (approximately equal to 2.71828) as its base. It is the inverse function of the exponential function, and is commonly used in mathematical and scientific calculations.

2. How is the natural log related to definite integrals?

The natural log and definite integrals are related through the fundamental theorem of calculus. This theorem states that the natural log of a function f(x) can be calculated by taking the definite integral of the function's derivative, f'(x), over a given interval. In other words, the natural log of a function is the area under its derivative curve.

3. What is the difference between natural log and logarithm with other bases?

The main difference between natural log and logarithms with other bases is the base number used in the calculation. Natural log uses the constant e as its base, while other logarithms can use any positive number as their base. This leads to differences in the calculations and properties of these logarithms.

4. How is the natural log used in real-world applications?

The natural log is used in a variety of real-world applications, including finance, biology, and physics. In finance, it is used to calculate compound interest and growth rates. In biology, it is used to model population growth. In physics, it is used to describe certain physical phenomena, such as radioactive decay.

5. What is the importance of the natural log in mathematics?

The natural log is important in mathematics because it is a fundamental function that is used in many areas of mathematics, including calculus, differential equations, and complex analysis. It is also closely related to the exponential function, which has numerous applications in science and engineering.

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