Concept question - integrals requiring substitution to solve

  • #1

Homework Statement



Random example: 2[([x^2] + 3)^7](7x)

Homework Equations



?

The Attempt at a Solution



I know that somehow you substitute [x^2] + 3 with 'u', but I don't understand the process going forward for it, and my teacher and textbook has some rather convoluted stuff in it, so if someone could explain this concept and how this ends up integrating anything, that would be nice.
 

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  • #2
Ray Vickson
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Homework Statement



Random example: 2[([x^2] + 3)^7](7x)

Homework Equations



?

The Attempt at a Solution



I know that somehow you substitute [x^2] + 3 with 'u', but I don't understand the process going forward for it, and my teacher and textbook has some rather convoluted stuff in it, so if someone could explain this concept and how this ends up integrating anything, that would be nice.
If ##u = x^2 + 3##, then ##du/dx = 2 x##, so ##du = 2 x \, dx##. Therefore, ##\int x (x^2+3)^7 \, dx = \int u^7 \, du/2 = \frac{1}{2} \int u^7 \, du##.
 
  • #3
Okay, hold on, I'm missing something.

When I'm given something like ∫(2x2)(3x), do they mean ∫(f(x)g(x) or ∫f(x)g'(x)? If the former, how do you use the product rule for integration by parts bit? And why in my first given equation, do you have to use du at all and whythe first x in your ∫ x(x2 + 3)7dx?
 
  • #4
SteamKing
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Okay, hold on, I'm missing something.

When I'm given something like ∫(2x2)(3x), do they mean ∫(f(x)g(x) or ∫f(x)g'(x)? If the former, how do you use the product rule for integration by parts bit? And why in my first given equation, do you have to use du at all and whythe first x in your ∫ x(x2 + 3)7dx?
First of all, learn how to write correct integrals.

You would never be given something like ∫(2x2)(3x) without the dx to indicate which variable is being integrated.

As far as product rules go, while there is one for taking the derivative of f(x) ⋅ g(x), there is no explicit product rule for integration.

If you are given an integral like ∫(2x2)(3x) dx, you can combine terms to give ∫ 6x3 dx and use the power rule to find the integral. Any other integrals will have to be considered on a case by case basis to determine which technique to use.
 
  • #5
First of all, learn how to write correct integrals.

You would never be given something like ∫(2x2)(3x) without the dx to indicate which variable is being integrated.

As far as product rules go, while there is one for taking the derivative of f(x) ⋅ g(x), there is no explicit product rule for integration.

If you are given an integral like ∫(2x2)(3x) dx, you can combine terms to give ∫ 6x3 dx and use the power rule to find the integral. Any other integrals will have to be considered on a case by case basis to determine which technique to use.
Well, sheesh, sorry for forgetting the dx. For the purposes of my course, x or t are the only variables ever being integrated, so unless there's another variable in the equation given (and that's another question for another time on how THAT works...), the dx is usually something you just stick on the end because it's proper math than for any use for it, so it ends up being implied. But yeah, I have to remember to stick it at the end...

Say that integral was written right and had a more complicated thing, like 2x3+3x2+2 instead of 2x2 and something like that instead of 3x where multiplying out the whole thing takes a long time. And when I say product rule, I mean the integral equivalent. What's the actual term, other than integration by parts?

What's the rule for, say, my original example in the first post, when dealing with the 7x at the end?

Also, still looking for and answer to this bit, let's see if I can simplify things: given the format ∫f(x)g'(x)dx = f(x)g(x) - ∫f'(x)g(x), is any given integral that you have to solve for involving such a format actually in the form of the left side integral, or are you looking for ∫f(x)g(x), and if the latter, how are you supposed to solve that?
 
  • #6
Ray Vickson
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Well, sheesh, sorry for forgetting the dx. For the purposes of my course, x or t are the only variables ever being integrated, so unless there's another variable in the equation given (and that's another question for another time on how THAT works...), the dx is usually something you just stick on the end because it's proper math than for any use for it, so it ends up being implied. But yeah, I have to remember to stick it at the end...

Say that integral was written right and had a more complicated thing, like 2x3+3x2+2 instead of 2x2 and something like that instead of 3x where multiplying out the whole thing takes a long time. And when I say product rule, I mean the integral equivalent. What's the actual term, other than integration by parts?

What's the rule for, say, my original example in the first post, when dealing with the 7x at the end?

Also, still looking for and answer to this bit, let's see if I can simplify things: given the format ∫f(x)g'(x)dx = f(x)g(x) - ∫f'(x)g(x), is any given integral that you have to solve for involving such a format actually in the form of the left side integral, or are you looking for ∫f(x)g(x), and if the latter, how are you supposed to solve that?
Including the "dx" is not an unnecessary luxury; it is absolutely crucial. That is especially important when you are changing variables from ##x## to some new integration variable ##u##, where ##x = f(u)## or maybe ##u = g(x)##. When you do that, ##dx## changes to ##f'(u) \, du##, where ##f'## is the first derivative of ##f##. If you had not written the ##dx## you would have missed all that and gotten a totally wrong answer. That is why we emphasize it: not to be picky or petty, but because it really, really matters.

As for dealing with the ##7x## in your first example: I, personally, like to take out all the constants and use the fact that ##\int 2 (7x)(x^2+3)^7 \, dx = 14 \int x (x^2+3)^7\, dx##, so we can just deal with the integral ## \int x (x^2+3)^7\,dx## and multiply the answer by 14 at the end. Of course, in this case it might be better to keep the "2" inside the integral, since ##\int 2x (x^2+3)^7 \, dx = \int u^7 \, du## when we put ##u = x^2 + 3##.

Many of the basic questions you are asking are answered in most decent textbooks, and are certainly dealt with in detail in numerous on-line sources. There are lots of tutorials available that take you through all this step-by-step.
 
  • #7
LCKurtz
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Also, still looking for and answer to this bit, let's see if I can simplify things: given the format ∫f(x)g'(x)dx = f(x)g(x) - ∫f'(x)g(x), is any given integral that you have to solve for involving such a format actually in the form of the left side integral, or are you looking for ∫f(x)g(x), and if the latter, how are you supposed to solve that?
You may be confusing u substitutions with integration by parts. u substitutions are used to "reverse" the chain rule. Reversing the product rule is less direct, but the process is called integration by parts. In this case you integral appears in the form ##\int f(x)g(x)dx## and it is your job to see if it could be viewed as being in the form ##\int f(x)g'(x)dx##. For example, suppose you see the integral ##\int xe^{2x}dx##. That looks like the product of two functions, but you could view it as the product of a function ##x## and a derivative of the function ##\frac{e^{2x}}{2}##, that is$$\int x\left(\frac{e^{2x}}{2}\right)'~dx$$So even though the integrand doesn't appear on the face of it to be of the form f(x)g'(x), it can be viewed that way. Of course, it is awkward to write things that way, so the usual method would be to use integration by parts letting ##u=x,~dv = e^{2x}##. That way you differentiate the u and integrate the dv so the x multiplier goes away in the next step.

Only practice and experience will help you distinguish when you want a u substitution versus integration by parts.
 
  • #8
You may be confusing u substitutions with integration by parts. u substitutions are used to "reverse" the chain rule. Reversing the product rule is less direct, but the process is called integration by parts. In this case you integral appears in the form ##\int f(x)g(x)dx## and it is your job to see if it could be viewed as being in the form ##\int f(x)g'(x)dx##. For example, suppose you see the integral ##\int xe^{2x}dx##. That looks like the product of two functions, but you could view it as the product of a function ##x## and a derivative of the function ##\frac{e^{2x}}{2}##, that is$$\int x\left(\frac{e^{2x}}{2}\right)'~dx$$So even though the integrand doesn't appear on the face of it to be of the form f(x)g'(x), it can be viewed that way. Of course, it is awkward to write things that way, so the usual method would be to use integration by parts letting ##u=x,~dv = e^{2x}##. That way you differentiate the u and integrate the dv so the x multiplier goes away in the next step.

Only practice and experience will help you distinguish when you want a u substitution versus integration by parts.
Okay, I just got what u-substitution is for, and I could smack myself...

Anyways, here's a problem I have: ∫32x2(3-2x4)6dx

u-substitution is out, 'cause it doesn't look like the reverse of the chain rule. So back to integration by parts.

Say f'(x) = 32x2, would g(x) = 3-2x4 or (3-2x4)6?
 
  • #9
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Well, sheesh, sorry for forgetting the dx. For the purposes of my course, x or t are the only variables ever being integrated, so unless there's another variable in the equation given (and that's another question for another time on how THAT works...), the dx is usually something you just stick on the end because it's proper math than for any use for it, so it ends up being implied. But yeah, I have to remember to stick it at the end...
In addition to what Ray Vickson said about dx not being "an unnecessary luxury," omitting dx will cause you grief when you get to more complex integration techniques, such as integration by parts and (especially) trig substitution.
 
  • #10
In addition to what Ray Vickson said about dx not being "an unnecessary luxury," omitting dx will cause you grief when you get to more complex integration techniques, such as integration by parts and (especially) trig substitution.
Ok, ok, I get it! I'll probably see more proof of it when I figure out how to do the problem in my last post, which I'm hoping to get help with!
 
  • #11
Sorry if I'm a bit grumpy and stuff; I have to finish all this stuff for tonight, and my final is on Wednesday, not to mention I have to study for other stuff as well, so I'm pretty stressed.
 
  • #12
Okay, so far I've got ∫-512x6(3-2x4)5dx = ((32x3/3)(3-2x4)6) - ∫32x2(3-2x4)6dx

Is this right so far? I dearly hope there are enough helpers here for both me and the other people freaking out over exams...

Also, now what? Substituting just ends up with everything canceling themselves out.
 
  • #13
Samy_A
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Anyways, here's a problem I have: ∫32x2(3-2x4)6dx

u-substitution is out, 'cause it doesn't look like the reverse of the chain rule. So back to integration by parts.

Say f'(x) = 32x2, would g(x) = 3-2x4 or (3-2x4)6?
Okay, so far I've got ∫-512x6(3-2x4)5dx = ((32x3/3)(3-2x4)6) - ∫32x2(3-2x4)6dx

Is this right so far? I dearly hope there are enough helpers here for both me and the other people freaking out over exams...

Also, now what? Substituting just ends up with everything canceling themselves out.
That doesn't seem to get any simpler, does it?
Maybe the easiest way to solve ∫32x2(3-2x4)6dx is to use the binomial theorem to expand ##(3-2x^4)^6##. After that your integral is just the integral of a polynomial.
 
  • #14
That doesn't seem to get any simpler, does it?
Maybe the easiest way to solve ∫32x2(3-2x4)6dx is to use the binomial theorem to expand ##(3-2x)^6##. After that your integral is just the integral of a polynomial.
Well yeah, but I don't think my teacher would appreciate that on the exam, 'cause it doesn't teach what to do when that exponent is ridiculously big, like in the hundreds.
 
  • #15
LCKurtz
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My guess would be that if that problem showed up on an exam, it wouldn't be exactly that problem. I it would be ##\int 32x^3(3-2x^4)^6~dx##. In other words, it would be cooked up so that substitution worked. And you would be expected to recognize that substitution was going to work. Or it might be instead, ##\int 32x^2(3-2x^4)^2~dx##, in which case you should recognize that substitution isn't going to work but this time is wouldn't be undue work on an exam to just multiply it out. On exams, teachers usually want the problems to be workable in a reasonably short time. And, as I have said before, with time and practice you will recognize whether an integration is one of these types or integration by parts.
 
  • #16
My guess would be that if that problem showed up on an exam, it wouldn't be exactly that problem. I it would be ##\int 32x^3(3-2x^4)^6~dx##. In other words, it would be cooked up so that substitution worked. And you would be expected to recognize that substitution was going to work. Or it might be instead, ##\int 32x^2(3-2x^4)^2~dx##, in which case you should recognize that substitution isn't going to work but this time is wouldn't be undue work on an exam to just multiply it out. On exams, teachers usually want the problems to be workable in a reasonably short time. And, as I have said before, with time and practice you will recognize whether an integration is one of these types or integration by parts.
So for now I should just forget about how to do this properly with rules and expand it out and cross my fingers that my teacher doesn't give it on the exam?
 
  • #17
So for now I should just forget about how to do this properly with rules and expand it out and cross my fingers that my teacher doesn't give it on the exam?
Never mind. I mis-saw somehow or mistyped or miscopied onto a paper for transfer... It was your first example of what would be given... Boy am I sorry now... *hangs head in shame* This is how stressed I am...
 
  • #18
LCKurtz
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So for now I should just forget about how to do this properly with rules and expand it out and cross my fingers that my teacher doesn't give it on the exam?
I would have said don't waste your time on it. You know expanding it out would work, so move on.

Never mind. I mis-saw somehow or mistyped or miscopied onto a paper for transfer... It was your first example of what would be given... Boy am I sorry now... *hangs head in shame* This is how stressed I am...
Don't worry about it. We all make typos, and with integrals that can easily make them difficult or even impossible.

Relax and just be sure to get a good night's sleep before your exam. That's the most important thing.
 
  • #19
I would have said don't waste your time on it. You know expanding it out would work, so move on.



Don't worry about it. We all make typos, and with integrals that can easily make them difficult or even impossible.

Relax and just be sure to get a good night's sleep before your exam. That's the most important thing.
Thanks.
 

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