- #1
Scholar1
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Homework Statement
∫ from 2 to 6 of dx/(x-4)
Homework Equations
None
The Attempt at a Solution
u=x-4
du=dx
ln abs value x-4 abs value
ln2-ln2=0
What am I doing wrong?
Math_QED said:How does the graph of 1/(x-4) look like ? What do you know when x = 4?
You integrate from 2 to 6. 4 is in this interval. Your function is not continuous in x = 4.Scholar1 said:Vertical asymptote what does that have to do with this integration?
Scholar1 said:Homework Statement
∫ from 2 to 6 of dx/(x-4)
Homework Equations
None
The Attempt at a Solution
u=x-4
du=dx
ln abs value x-4 abs value
ln2-ln2=0
What am I doing wrong?
Scholar1 said:So since ln4-4=0 its divergent regardless of all the other stuff?
Ray Vickson said:I cannot tell who you are responding to, but if it is to my post #6, then you need to check if the limit of ##I(a,b)## exists.
There is also the issue of whether or not you mean the Cauchy Principal Value integral, which can exist even if a more general type does not. The Cauchy principal value (if it exists) would be ##\lim_{\epsilon \downarrow 0} I(4-\epsilon, 4 + \epsilon)## in the notation of post #6.
Scholar1 said:I apologize I was replying to you. I am only Calc 2 so I don't know what the Cauchy Principal Value is. I am confused how to check if the limit exist. Please show me.
He just said he doesn't know what it is, you really think he'd understand that equation? I'm in AP Calc BC which is the calc 2 equivalent and I have never seen anything similar to that equation. You should explain it to him from a calculus 2 standpoint.Ray Vickson said:I have just explained what the Cauchy Principal value is, by giving an actual formula.
PF rules do not allow me to complete your problem for you.
For your purposes, the integral doesn't exist for the reason Math_QED gave back in post #4. You can't just mechanically follow a procedure and get an answer. You have to know the limitations.Scholar1 said:So since ln4-4=0 its divergent regardless of all the other stuff?
nfcfox said:He just said he doesn't know what it is, you really think he'd understand that equation? I'm in AP Calc BC which is the calc 2 equivalent and I have never seen anything similar to that equation. You should explain it to him from a calculus 2 standpoint.
I am actually in AP Calculus BC I just said calc 2 because people are more familiar with that. It wouldn't serve me well to spend time learning something so specific when I am preparing for an exam in a limited spectrum course.Ray Vickson said:Yes, I would hope that he would understand that equation after working at it for a while---not instantly, maybe, but after spending some time on it. Alternatively, he could Google 'Cauchy principal value' and find numerous web pages dealing with the topic, although skipping the Mathworld and Wikipedia pages on it is probably a good idea for a beginning student.
Anyway, I am assuming that 'Scholar1' (as a Calc 2 student) is capable of evaluating both of the very elementary integrals in the definition of ##I(a,b)##, and that he can then put ##a = 4 - \epsilon, b = 4 + \epsilon## to see what happens. Am I really expecting too much?
By the way: one reason that he/she might be unfamiliar with the Cauchy Principal value is that it is usually found only in some applications in Physics, Engineering and the like. Some writers regard it as a bit of meaningless nonsense in general.
Scholar1 said:I am actually in AP Calculus BC I just said calc 2 because people are more familiar with that. It wouldn't serve me well to spend time learning something so specific when I am preparing for an exam in a limited spectrum course.
Why are you worried about this integral man? I'm worried about sequences and seriesScholar1 said:I am actually in AP Calculus BC I just said calc 2 because people are more familiar with that. It wouldn't serve me well to spend time learning something so specific when I am preparing for an exam in a limited spectrum course.
Ray Vickson said:You are dealing with an improper integral, so you need to define what it means. In your case, you need to evaluate
[tex] I(a,b) = \int_2^a \frac{dx}{x-4} + \int_b^6 \frac{dx}{x-4}, [/tex]
then see of the limit of ##I(a,b)## exists as ##a \uparrow 4## and ##b \downarrow 4##.
The integral of ∫dx/(x-4) from 2 to 6 is equal to ln| x-4 | evaluated from 2 to 6, which is equal to ln| 6-4 | - ln| 2-4 |, or ln2 - ln(-2).
To solve ∫dx/(x-4) from 2 to 6, you can use the formula for the integral of a function, which is ∫f(x)dx = F(x) + C, where F(x) is the antiderivative of f(x). In this case, the antiderivative of 1/(x-4) is ln| x-4 |. Then, evaluate the antiderivative at 6 and 2, and subtract the two values to get the final answer.
Yes, you can use a calculator to solve this integral. However, make sure that your calculator is in radian mode and that you use proper parentheses and brackets to indicate the limits of integration. Also, double check your calculator's answer with the steps outlined in question 2 to ensure accuracy.
The integral of a function represents the area under the curve of that function. In this case, solving ∫dx/(x-4) from 2 to 6 gives us the area under the curve of 1/(x-4) between the limits of 2 and 6. This can be useful in various applications, such as calculating work done by a varying force or finding the average value of a function.
Yes, there are various techniques that can be used to solve this integral, such as substitution, integration by parts, and partial fractions. However, for this particular integral, the most straightforward method is to use the formula for the integral of a function and evaluate the antiderivative at the given limits.