MHB Understanding Integral Substitution: Finding Equivalent Ranges for Functions

[TheFallen018]

Hi, I posted a question here a few days ago regarding some questions I've been doing on an online quiz. I seem to be getting stuck on the integral substitution questions. I've been slowly making progress, but some of these questions have been confusing me, and reading up on them is only giving me a vague idea of how to progress. I figured that asking someone might be more beneficial to help me find my footing.

Anyways, here's the question. It has to do with finding equivalent ranges for a function once x has been put into a u substitution. I'm not really sure where to start for this one. My textbook is somewhat unclear on this subject.

I would really be grateful for some help. Thanks again

View attachment 7951

 

[MarkFL]

Hi, I posted a question here a few days ago regarding some questions I've been doing on an online quiz. I seem to be getting stuck on the integral substitution questions. I've been slowly making progress, but some of these questions have been confusing me, and reading up on them is only giving me a vague idea of how to progress. I figured that asking someone might be more beneficial to help me find my footing.

Anyways, here's the question. It has to do with finding equivalent ranges for a function once x has been put into a u substitution. I'm not really sure where to start for this one. My textbook is somewhat unclear on this subject.

I would really be grateful for some help. Thanks again

I would try the substitution:

$$u=2x$$

Using this substitution, what do you get?

 

[TheFallen018]

I would try the substitution:

$$u=2x$$

Using this substitution, what do you get?

Oh, wow, for some reason I wasn't thinking that the constant would be affected by the square of u. That clears some things up. So basically that would work out to a=10, b=12, and c being 1/2. That works nicely.

I have another question about another problem, if you would be so kind as to have a look. I managed to do the first bit, which wasn't too hard, but when I integrate the problem, I end up getting a really ugly formula, and it doesn't seem to be right anyways. I'm sure with how most of these questions have been working out, that there's something that will make it integrate in a neat way. The subsequent questions also seem to be eluding me, but I'm guessed they will clear up as I figure out the previous ones.

View attachment 7952

Thanks again for your help. You guys are amazing

 

[MarkFL]

Okay, if we make the suggested substitution, this means:

$$u=3+e^{4x}\implies du=4e^{4x}\,dx$$

And so for part (a) we may write:

$$\int e^{4x}\left(3+e^{4x}\right)^3\,dx=\frac{1}{4}\int u^{3}\,du$$

Can you continue?

 

[TheFallen018]

Okay, if we make the suggested substitution, this means:

$$u=3+e^{4x}\implies du=4e^{4x}\,dx$$

And so for part (a) we may write:

$$\int e^{4x}\left(3+e^{4x}\right)^3\,dx=\frac{1}{4}\int u^{3}\,du$$

Can you continue?

One of the limitations of part a is I can't move things outside of the integral, due to the way the question is put. So the way it worked out was $$\left(\frac{u^3}{4(u-3)}\right)$$

The software accepted that as the right solution, so that's what I've been working with. Part b in particular was where I was having difficulty

 

[MarkFL]

One of the limitations of part a is I can't move things outside of the integral, due to the way the question is put. So the way it worked out was $$\left(\frac{u^3}{4(u-3)}\right)$$

The software accepted that as the right solution, so that's what I've been working with. Part b in particular was where I was having difficulty

Where did the $u-3$ in the denominator of the integrand come from?

It should just be:

$$\frac{u^3}{4}$$

If we pull the constant inside.

 

[TheFallen018]

Where did the $u-3$ in the denominator of the integrand come from?

It should just be:

$$\frac{u^3}{4}$$

If we pull the constant inside.

I think it came from the $$e^4x$$ at the front of the integral equation. The $$u-3$$ came about to represent it in terms of u.

 

[MarkFL]

I think it came from the $$e^4x$$ at the front of the integral equation. The $$u-3$$ came about to represent it in terms of u.

Let's step through the substitution a bit more slowly...

$$\int e^{4x}\left(3+e^{4x}\right)^3\,dx=\int \left(3+e^{4x}\right)^3\,\left(e^{4x}\,dx\right)=\frac{1}{4}\int \left(3+e^{4x}\right)^3\,\left(4e^{4x}\,dx\right)=\frac{1}{4}\int u^{3}\,du$$

 

[TheFallen018]

Let's step through the substitution a bit more slowly...

$$\int e^{4x}\left(3+e^{4x}\right)^3\,dx=\int \left(3+e^{4x}\right)^3\,\left(e^{4x}\,dx\right)=\frac{1}{4}\int \left(3+e^{4x}\right)^3\,\left(4e^{4x}\,dx\right)=\frac{1}{4}\int u^{3}\,du$$

Hmm, that's strange. I can see what you mean. It's strange that the software accepted it as a correct solution. I just changed it to reflect the above, and somehow it accepted it. So it seems the software is taking two different solutions as correct, even though they are both very different.

After taking a step back, and changing part a, part b followed along with no problem. Part c is now giving some issues. In my mind, the correct answer should be $$3+\left(\frac{1}{e^4}\right)$$. However, apparently I am missing something in the answer. Have I done something wrong there?

 

[MarkFL]

Hmm, that's strange. I can see what you mean. It's strange that the software accepted it as a correct solution. I just changed it to reflect the above, and somehow it accepted it. So it seems the software is taking two different solutions as correct, even though they are both very different.

After taking a step back, and changing part a, part b followed along with no problem. Part c is now giving some issues. In my mind, the correct answer should be $$3+\left(\frac{1}{e^4}\right)$$. However, apparently I am missing something in the answer. Have I done something wrong there?

Okay, proceeding to part (b), I get (leaving off the constant of integration):

$$\int e^{4x}\left(3+e^{4x}\right)^3\,dx=\frac{1}{4}\int u^{3}\,du=\frac{1}{16}u^4$$

For part (c) I get:

$$u(-1)=3+e^{-4}$$

This is equivalent to what you've stated.

 

[TheFallen018]

Okay, proceeding to part (b), I get (leaving off the constant of integration):

$$\int e^{4x}\left(3+e^{4x}\right)^3\,dx=\frac{1}{4}\int u^{3}\,du=\frac{1}{16}u^4$$

For part (c) I get:

$$u(-1)=3+e^{-4}$$

This is equivalent to what you've stated.

That's strange. I've tried both values, but it doesn't seem to accept them. I'm out of ideas at this point. I really appreciate the help you've provided thus far.

 

[MarkFL]

That's strange. I've tried both values, but it doesn't seem to accept them. I'm out of ideas at this point. I really appreciate the help you've provided thus far.

Is it possible it wants a decimal approximation there? Of course, then the issue is how many digits of accuracy? Gotta love these automated systems, accepting incorrect results, and denying correct ones. :)

 

[TheFallen018]

Is it possible it wants a decimal approximation there? Of course, then the issue is how many digits of accuracy? Gotta love these automated systems, accepting incorrect results, and denying correct ones. :)

Yeah, gave that a shot. It usually wants decimals rounded to 3 significant figures, but no such luck.