Calculus 2 probably easy questions

[gnaib]

Hello Super Friends, I have been out of school for about 8 years and I headed back in at the calc 2 level. Needless to say, I forgot a lot and since I'm taking it online, I have no one to discuss questions with and my tutor flat out told me that he can't give me any help on homework. Here goes nerds:

1) Solve the Inequality listed below for x

log₂x > log₃x
lnx/ln2 > lnx/ln3
ln3 > ln2

Now my x factors out so I'm wondering if I did something wrong or would my answer then become x > 0 since it is a log function.


2) a. Explain why

0 < x²tan^-1x < (pi)x²/4 for all 0 < x < 1 (NOTE: all the inequalities can also be equal to, unsure how to implement that symbol)

I understand this question but decided to add it in just in case

b) Use the properties of integrals to show that the value of the integral

x²tan^-1x dx

lies on the interval [0, pi/12]

This is my first homework assignment and I've read ahead and I now know how to solve it using integration by parts. Is there another way to solve this using the properties of integrals alone that I'm missing? I haven't covered trig sub, integration by parts, rational fractions or improper integrals at this time.

3) Explain why the limit

lim as x -> - infinity of sin^-1((3-x)/(x-1))

is not well defined.

i attempted to take the limit which resulted in an answer of -pi/2. Then i thought about the graph and how this inverse function needs restricted intervals so that it can actually be a function. I guess my question now is does arcsinx exist outside of the intveral [-pi/2, pi/2] at all? Can I just shift the interval of sinx from [-pi/2, pi/2] to something else and then get a similar inverse function but with different bounds? I hope this makes sense...

If x goes to negative infinity, it may not be well defined because the inverse function depends on it's related, uninverted function and therefore....

4) Use Mathematical induction to show that the integral of


xⁿe^-xdx = n! from 0 to infinity

I have no idea what I'm doing here and the class notes and book have done a terrible job of explaining this section. I'm not looking for the answer (well I am..) but moreso on how to go about even doing this question. I attempted to use integration by parts which resulted in another integral that was "one degree" lower in terms of x, going onto infinity.

 

[MarkFL]

For the first question, we have:

[MATH]\log_2(x)>\log_3(x)[/MATH]
Using the change of base theorem, I would then arrange as:

[MATH]\frac{\ln(x)}{\ln(2)}-\frac{\ln(x)}{\ln(3)}>0[/MATH]
Factor:

[MATH]\ln(x)\left(\frac{1}{\ln(2)}-\frac{1}{\ln(3)}\right)>0[/MATH]
This implies:

[MATH]\ln(x)>0[/MATH]
Or:

[MATH]1<x[/MATH]

 

[Dr.Peterson]

1) Solve the Inequality listed below for x

log₂x > log₃x
lnx/ln2 > lnx/ln3
ln3 > ln2

Now my x factors out so I'm wondering if I did something wrong or would my answer then become x > 0 since it is a log function.

Your error is in dividing an inequality by ln(x), without considering whether it is positive. If you had actually factored it out and then thought carefully, you would have been okay (see MarkFL's version).

2) a. Explain why

0 < x²tan^-1x < (pi)x²/4 for all 0 < x < 1 (NOTE: all the inequalities can also be equal to, unsure how to implement that symbol)

I understand this question but decided to add it in just in case

I'm not sure what you're asking here. If you want to verify that you understand it, you'll have to show your answer.

To display ≤, you can either learn LaTeX, e.g. [MATH]\le[/MATH], or paste in the symbol from some outside source (as I did), or use "<=".

b) Use the properties of integrals to show that the value of the integral

x²tan^-1x dx

lies on the interval [0, pi/12]

This is my first homework assignment and I've read ahead and I now know how to solve it using integration by parts. Is there another way to solve this using the properties of integrals alone that I'm missing? I haven't covered trig sub, integration by parts, rational fractions or improper integrals at this time.

Did you omit something here? An indefinite integral is not a number; was it supposed to be a definite integral?

3) Explain why the limit

lim as x -> - infinity of sin^-1((3-x)/(x-1))

is not well defined.

i attempted to take the limit which resulted in an answer of -pi/2. Then i thought about the graph and how this inverse function needs restricted intervals so that it can actually be a function. I guess my question now is does arcsinx exist outside of the intveral [-pi/2, pi/2] at all? Can I just shift the interval of sinx from [-pi/2, pi/2] to something else and then get a similar inverse function but with different bounds? I hope this makes sense...

If x goes to negative infinity, it may not be well defined because the inverse function depends on it's related, uninverted function and therefore....

Try graphing the function! What do you find for negative values of x?

The issue is not the range of the inverse sine (which you can't change -- it's been defined for you); it's the domain.

4) Use Mathematical induction to show that the integral of

xⁿe^-xdx = n! from 0 to infinity

I have no idea what I'm doing here and the class notes and book have done a terrible job of explaining this section. I'm not looking for the answer (well I am..) but moreso on how to go about even doing this question. I attempted to use integration by parts which resulted in another integral that was "one degree" lower in terms of x, going onto infinity.

Give it a try and show us what you did. You might be more right than you realize.

In the future, please don't ask more than one question at once, unless they are closely related.

 

[MarkFL]

For the second problem, if you've shown that:

[MATH]0\le x^2\arctan(x)\le \frac{\pi}{4}x^2[/MATH] for \(0\le x\le1\)

And the integral under consideration is:

[MATH]I=\int_0^1 x^2\arctan(x)\,dx[/MATH]
Then, we may state:

[MATH]\int_0^1 0\,dx\le \int_0^1 x^2\arctan(x)\,dx\le\int_0^1 \frac{\pi}{4}x^2\,dx[/MATH]
Or:

[MATH]0\le I\le\int_0^1 \frac{\pi}{4}x^2\,dx[/MATH]
Evaluating the definite integral on the far right will give the upper bound for the value of \(I\).

 

[gnaib]

Your error is in dividing an inequality by ln(x), without considering whether it is positive. If you had actually factored it out and then thought carefully, you would have been okay (see MarkFL's version).


I'm not sure what you're asking here. If you want to verify that you understand it, you'll have to show your answer.

To display ≤, you can either learn LaTeX, e.g. [MATH]\le[/MATH], or paste in the symbol from some outside source (as I did), or use "<=".


Did you omit something here? An indefinite integral is not a number; was it supposed to be a definite integral?


Try graphing the function! What do you find for negative values of x?

The issue is not the range of the inverse sine (which you can't change -- it's been defined for you); it's the domain.


Give it a try and show us what you did. You might be more right than you realize.

In the future, please don't ask more than one question at once, unless they are closely related.

Your error is in dividing an inequality by ln(x), without considering whether it is positive. If you had actually factored it out and then thought carefully, you would have been okay (see MarkFL's version).


I'm not sure what you're asking here. If you want to verify that you understand it, you'll have to show your answer.

To display ≤, you can either learn LaTeX, e.g. [MATH]\le[/MATH], or paste in the symbol from some outside source (as I did), or use "<=".


Did you omit something here? An indefinite integral is not a number; was it supposed to be a definite integral?


Try graphing the function! What do you find for negative values of x?

The issue is not the range of the inverse sine (which you can't change -- it's been defined for you); it's the domain.


Give it a try and show us what you did. You might be more right than you realize.

In the future, please don't ask more than one question at once, unless they are closely related.

I don't know how to quote you the way you quoted me since I'm new here :S. But here goes:

1) Ah, that makes sense and I would have never thought of bringing them both to one side. Thank you! Would you mind expanding upon dividing by lnx without considering if "it is was positive"? I understand now that what I did was wrong but I don't know why.

2) a) Don't worry about this part it was just for completion, my main question was b).

b) Yes I did forget something... the integral goes from x = 0 to 1.

3) I think I understand a little bit more about this question now, I was much more confused before and I couldn't wrap my head around it. The graph of arcsinx is very limited because it is the inverse of a sin function with it's interval restricted. As an example of what was perplexing me, the normal interval for sin(x) is [-pi/2, pi/2] - I understand this but can't you just shift this interval to the left or to the right while maintaining the ability to be inverted? Can't the interval instead be [pi/2, 3pi/2] (which still passes the horizontal line test and therefore is invertible (Now that I think about it, it doesn't change the fact that the domain of the inverse is [-1, 1].)

Basically what I'm thinking now is the domain of the inverse function is [-1,1] no matter if you can or cannot shift the intervals of sinx...and x cannot take on any value less than -1 so that it is why it's not well defined? Taking the limit of the function gave me an actual answer but I don't think it was correct for me to do that.

4) The only thing I could think of doing was integrating by parts by taking u = xⁿ and dv = e^-x

I did the same type of integration 2-3 times to picture the infinite integrals that would come about and thought maybe somehow I could factor or something would stick out to me visualizing it. I didn't really know how to proceed further.

I will limit my questions to one per post in the future...also thanks for the help I really appreciate it . Math needs to be discussed and talked about for me to really understand it and I'm unable to find anyone around me who can. I'm very rusty.

 

[gnaib]

For the second problem, if you've shown that:

[MATH]0\le x^2\arctan(x)\le \frac{\pi}{4}[/MATH] for \(0\le x\le1\)

And the integral under consideration is:

[MATH]I=\int_0^1 x^2\arctan(x)\,dx[/MATH]
Then, we may state:

[MATH]\int_0^1 0\,dx\le \int_0^1 x^2\arctan(x)\,dx\le\int_0^1 \frac{\pi}{4}x^2\,dx[/MATH]
Or:

[MATH]0\le I\le\int_0^1 \frac{\pi}{4}x^2\,dx[/MATH]
Evaluating the definite integral on the far right will give the upper bound for the value of \(I\).

Oh my lord that is so simple and makes sense. I didn't think of just setting up an inequality like that. Appreciate it!

 

[Dr.Peterson]

If you either hit the Reply button in the message you are answering, or select a bit and hit the Reply button that pops up, you get a copy in what you are writing. (The latter method doesn't always copy properly.)

1) Ah, that makes sense and I would have never thought of bringing them both to one side. Thank you! Would you mind expanding upon dividing by lnx without considering if "it is was positive"? I understand now that what I did was wrong but I don't know why.

As I said, you divided both sides by ln(x), but called it "factoring out" (also called canceling). But if you multiply or divide an inequality by a negative number, it changes the direction. So if you remove ln(x) from ln(x)/ln(2) > ln(x)/ln(3), then when ln(x) < 0, the result is not 1/ln(2) > 1/ln(3), from which you got ln(3) > ln(2), but 1/ln(2) < 1/ln(3), which of course is false. So the inequality is true if ln(x) > 0, but not if ln(x) < 0.

The other method, bringing everything to one side and factoring, is standard. You need to learn that, especially for inequalities, but also for any equation in which what you want to "cancel" might be 0. For example, given x^2 = 2x, you can't blindly divide by x and leave x = 2; rather, change it to x^2 - 2x = 0 and factor: x(x - 2) = 0. Now you can see that x = 0 or x = 2.

2) a) Don't worry about this part it was just for completion, my main question was b).

b) Yes I did forget something... the integral goes from x = 0 to 1.

If I'd looked closely at the relationship between the two parts, I would have guessed the correction, as MarkFL did.

3) I think I understand a little bit more about this question now, I was much more confused before and I couldn't wrap my head around it. The graph of arcsinx is very limited because it is the inverse of a sin function with it's interval restricted. As an example of what was perplexing me, the normal interval for sin(x) is [-pi/2, pi/2] - I understand this but can't you just shift this interval to the left or to the right while maintaining the ability to be inverted? Can't the interval instead be [pi/2, 3pi/2] (which still passes the horizontal line test and therefore is invertible (Now that I think about it, it doesn't change the fact that the domain of the inverse is [-1, 1].)

Basically what I'm thinking now is the domain of the inverse function is [-1,1] no matter if you can or cannot shift the intervals of sinx...and x cannot take on any value less than -1 so that it is why it's not well defined? Taking the limit of the function gave me an actual answer but I don't think it was correct for me to do that.

You could use a different restricted domain for sin, and therefore a different range for arcsin; but we don't! That would not be the function the problem is about! When you are solving a problem, you must go by what it means as written, not by an alternate possible definition.

I think you are seeing the issue, though you didn't quite make it explicit. The fact is that for x < 1, (3-x)/(x-1) < -1. That puts it outside the domain of the arcsin, so that the given function is not defined for x < -1. And if a function is not defined for negative inputs, it can't have a limit at negative infinity.

4) The only thing I could think of doing was integrating by parts by taking u = xⁿ and dv = e^-x

I did the same type of integration 2-3 times to picture the infinite integrals that would come about and thought maybe somehow I could factor or something would stick out to me visualizing it. I didn't really know how to proceed further.

The integration by parts is correct; but I don't see that you've said anything about using induction. Do you know how to start a proof by mathematical induction? Write out what you need to do (the induction hypothesis, the base step, the inductive step), and you should be able to see the details easily, including where integration by parts fits in. It's the setup that is important.

 

[gnaib]

I am many years out of school and absolutely out of practice. I have also forgotten many of the little bits and pieces that helped string together my understanding of calculus (used to be my strongest subject ). I am very unfamiliar with inequalities and I think I just need to practice and re familiarize. I didn't know the inequality changed direction if you divided by a negative but relating it back to a regular equation put it into perspective for me...the inequality sign threw me off.

3) Makes total sense now .

4) I do not know how to start a proof nor what the process of induction is. I have a very technical math book from my university that is incredibly hard to follow and understand. Particularly the section on mathematical induction and proofs. I suppose I need to study more to fully understand what the question is asking me.

Thanks again!

 

[JeffM]

Inequalties change direction if you multiply or divide by a negative number.

Why?

Consider the example of 3 and 5. If I multiply each by minus 1, I get - 5 and - 3. And - 5 < - 3.

[MATH]5 > 3 \text { but } -\ 5 < -\ 3.[/MATH]
Try some other examples. You get the same result. So we make the following general rule

[MATH]a < 0 \text { and } b > c \implies ab < ac. [/MATH]

 

[gnaib]

Makes sense. So essentially because the function is ln(x), if x is a fraction then it will be a negative number so I didn't that that aspect into account?

 

[Dr.Peterson]

4) I do not know how to start a proof nor what the process of induction is. I have a very technical math book from my university that is incredibly hard to follow and understand. Particularly the section on mathematical induction and proofs. I suppose I need to study more to fully understand what the question is asking me.

I was wondering if that might be the issue. You need to find a source that works better for you; what you have may either be assuming you only need review, or aimed at people with a different style of thinking than you, or just poorly written.

I'd suggest you just search for "mathematical induction" and read a few explanations of it, then go through a few examples. Each different presentation you read is a new chance to see what you missed in the previous one! (Don't keep struggling with one of them too much, if it goes over your head; just move on to the next.)

But here's the basic idea for your example:

The problem, clarified slightly, is this:

Use mathematical induction to show that ∫₀^∞ xⁿe^-xdx = n! for all non-negative integers n.​

Again, formatted more nicely,

Use mathematical induction to show that [MATH]\int_0^\infty x^n e^{-x} dx = n![/MATH] for all non-negative integers [MATH]n[/MATH].​

First, you have to show that it's true for the smallest value of n, namely 0. That is,

1. Show that [MATH]\int_0^\infty x^0 e^{-x} dx = 0![/MATH]​

Then suppose that it's true for some particular value of n, say n = k. That is,

2. Suppose that [MATH]\int_0^\infty x^k e^{-x} dx = k![/MATH]. Then ...​

Prove that it is also true for the next value of n:

... show that [MATH]\int_0^\infty x^{k+1} e^{-x} dx = (k+1)![/MATH].​

This last bit is where the integration by parts will show up.

 

[gnaib]

Ok that makes sense to me and I definitely didn't know how to do or come up with that until now.

Do you know of any sources that you would recommend to brush up or even learn calculus material? I've been searching the internet for good resources and to date the only thing I've come up with is Prof Leonards lectures on youtube (which by the way have been life saving and without him I would know nothing).

 

[Dr.Peterson]

I usually recommend Paul's notes, http://tutorial.math.lamar.edu/. That's been around a long time, and keeps getting better. (It doesn't appear to cover induction, though.)

 

[Otis]

Hello gnaib. There's a lot of good information in the forum's submission guidelines (aka: READ BEFORE POSTING). For example, you'll find three tips about posting math symbols in the forum.

If you haven't done any math in eight years, you might find it tough to study calculus alongside reviews of algebra, trig and pre-calculus topics. (For example, induction is a precalculus topic.) I'm guessing that you don't have 16 hours a day to work through material from four courses, so, if you start to run out of time trying to refresh your memory on some point, perhaps it would be better to just ask here (following the guidelines, of course). We could then explain or provide a link to an outside lesson.

 

[JeffM]

Makes sense. So essentially because the function is ln(x), if x is a fraction then it will be a negative number so I didn't that that aspect into account?

Exactly