Analysis: Integration

[Shackleford]

I'm not sure about #4. I know probably #5 and definitely #6 are wrong. For some reason I'm stumped on #6 especially.

http://i111.photobucket.com/albums/n149/camarolt4z28/4-1.png [Broken]

http://i111.photobucket.com/albums/n149/camarolt4z28/5.jpg [Broken]

Question for #6: http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled.png

http://i111.photobucket.com/albums/n149/camarolt4z28/6a.jpg [Broken]
http://i111.photobucket.com/albums/n149/camarolt4z28/6b.jpg [Broken]

 

[Mark44]

Speaking for myself, I would really appreciate seeing the problem directly in the post, instead of having to go to another site to see what the problem is and what you have done.

 

[Shackleford]

Speaking for myself, I would really appreciate seeing the problem directly in the post, instead of having to go to another site to see what the problem is and what you have done.

All you're doing is clicking a link to a picture of my work. That's the fastest, easiest, and clearest way for me to express my work.

 

[Mark44]

OTOH, I have to have two windows open - one to see what you did and another to make comments about what you did. And I can't insert a comment at a particular point if there's something wrong. Instead, I have to give some context, like "in the line where blah, blah, you have xyz and it should be abc."

It may be fast, easy, and clear for you, but it's a hassle for me that I don't need. When you're asking for help from someone, the key is to make it as easy as possible for them, not necessarily for yourself.

 

[Shackleford]

OTOH, I have to have two windows open - one to see what you did and another to make comments about what you did. And I can't insert a comment at a particular point if there's something wrong. Instead, I have to give some context, like "in the line where blah, blah, you have xyz and it should be abc."

It may be fast, easy, and clear for you, but it's a hassle for me that I don't need. When you're asking for help from someone, the key is to make it as easy as possible for them, not necessarily for yourself.

I understand your point, but since I don't know latex, in problems like these it's more clear to simply take a picture. Fortunately, there aren't too many lines to worry about.

 

[grzz]

Latex is not so difficult. One can try and learn how to use it. The 'preview' facility is a great help.

 

[Mark44]

Don't you think it's about time you learned? LaTeX isn't difficult.
https://www.physicsforums.com/showthread.php?t=546968

 

[deluks917]

Number 4 is not correct. How do you know the point c is the same for both functions? You need to use MVT for integrals but to show a "new" function is 0.

Number 5 looks ok, idk why you need to say F is uniformly cont.

I don't see what problem 6 is?

 

[Harrisonized]

#4

Suppose ∫_a^bf(x)dx = ∫_a^bg(x)dx
Then ∫_a^b[f(x)-g(x)]dx = 0

Now you can use continuity to complete the proof. I don't think the mean value theorem can help you in this problem. The mean value theorem is used to prove that a point can be a certain slope, which has little to do with your problem.

#5

What's the lower bound of integration in the last line? Hint: c∈[a,b]

#6

What is the question?

 

[Shackleford]

Number 4 is not correct. How do you know the point c is the same for both functions? You need to use MVT for integrals but to show a "new" function is 0.

Number 5 looks ok, idk why you need to say F is uniformly cont.

For #4, I can set h(x) = f(x) - g(x) and take it from there, right?

For #5, I probably don't need to say it's uniformly continuous.

Oops. Sorry, I forgot to include the question.

 

[Shackleford]

#4

Suppose ∫_a^bf(x)dx = ∫_a^bg(x)dx
Then ∫_a^b[f(x)-g(x)]dx = 0

Now you can use continuity to complete the proof. I don't think the mean value theorem can help you in this problem. The mean value theorem is used to prove that a point can be a certain slope, which has little to do with your problem.

#5

What's the lower bound of integration in the last line?

#6

What is the question?

For #5, I forgot the b at the bottom. I just swapped x and b.

 

[LCKurtz]

OTOH, I have to have two windows open - one to see what you did and another to make comments about what you did. And I can't insert a comment at a particular point if there's something wrong. Instead, I have to give some context, like "in the line where blah, blah, you have xyz and it should be abc."

It may be fast, easy, and clear for you, but it's a hassle for me that I don't need. When you're asking for help from someone, the key is to make it as easy as possible for them, not necessarily for yourself.

I heartily agree with these sentiments. It is also a pain to follow the replies. That is why I didn't respond to this thread when I first saw it. Well, that and the fact that looked like some kind of take-home assignment.

 

[Shackleford]

I reworked #4 and cleaned up #5. I'm going take another shot at #6.

 

[Shackleford]

Any ideas on #6? I'm sure it's an easy problem. I just keep running into dead ends.

 

[flyingpig]

Any ideas on #6? I'm sure it's an easy problem. I just keep running into dead ends.

You are going to have to get rid of that integral sign, and no, don't evaluate the integral would be my hint.

 

[Shackleford]

You are going to have to get rid of that integral sign, and no, don't evaluate the integral would be my hint.

Okay. I think I'm over-complicating this or approaching from the wrong side of the FTOC. Haha.

And if f is continuous at a point c in the interval

F'(c) = f(c)

How about?

$$\int_a^x f(t)dt = \int_a^b f(t)dt -\int_x^b f(t)dt$$

$$\int_a^x f(t)dt = F(x) = \int_x^b f(t)dt$$

$$F(x) = \alpha - F(x)$$

$$F'(x) = 0 - F'(x)$$

$$f(x) = -f(x)$$

$$\Rightarrow f(x) = 0$$