mm-2349 === Subject: Please help me understand these integral substitution problems I am having trouble with using substitution for definite integrals. I have worked through the chapter examples but am stumped by these two problems. I have tried to solve them as follows. Please show me what steps I am missing. I will use the dollar sign as the integral. Problem 1 $ x((4 + x^2)^10) dx u is given as (4 + x^2). My next step is to find du. I think du = 4x + (x^3/3). I'm lost on what to do next. I know that I have to substitute. I know what to do with u, but I'm not sure what to do with a du in that form. I think I should set it equal to xdx and solve for dx but I'm not sure what I get when I do that. Problem 2 $ 4 / (1 + 2x)^3 dx u given as 1 + 2x dx = du / 4 Rewrite: $ (u^-3) 1/4 du (u^-2)/-2*1/4 du -(u^-2)/8 -((1 + 2x)^-2)/8 The answer to this one is -(1 + 2x)^-2, so it looks like my 1/4 du was part of the problem. === Subject: Re: Please help me understand these integral substitution problems >I am having trouble with using substitution for definite integrals. I >have worked through the chapter examples but am stumped by these two >problems. I have tried to solve them as follows. Please show me what >steps I am missing. I will use the dollar sign as the integral. >Problem 1 >$ x((4 + x^2)^10) dx > >u is given as (4 + x^2). My next step is to find du. I think du = 4x + (x^3/3). You integrated instead of differentiated. du = 2x dx. I'm lost on what to do next. I know that I have to >substitute. (4 + x^2)^10 = u^10 x dx = 1/2 du I know what to do with u, but I'm not sure what to do with a du in >that form. I think I should set it equal to xdx and solve for dx but >I'm not sure what I get when I do that. You get u^10 * 1/2 du. You should be able to carry on from here. Problem 2 >$ 4 / (1 + 2x)^3 dx u given as 1 + 2x dx = du / 4 No. du = 2 dx. Rewrite: $ (u^-3) 1/4 du Then u^(-3) 2 du. (u^-2)/-2*1/4 du Once you integrate, the du disappears. -(u^-2)/8 -((1 + 2x)^-2)/8 The answer to this one is -(1 + 2x)^-2, so it looks like my 1/4 du >was part of the problem. <> === Subject: Re: Please help me understand these integral substitution problems >You integrated instead of differentiated. du = 2x dx. This sentence got me on the right track. Somehow, right in the middle of doing problems, I started integrating instead of differentiating. Then, I got so engrossed in looking for my mistake that I couldn't find it. Today, I did a lot of problems and differentiated and got the correct answers, of course. === Subject: Re: Please help me understand these integral substitution problems > I will use the dollar sign as the integral. > Problem 1 > $ x((4 + x^2)^10) dx > integral x (4 + x^2)^10 dx let u = x^2; x = sqr u; dx = 1/(sqr u) * du integral sqr u * (4 + u)^10 * 1/(sqr u) * du = integral (4 + u)^10 du let v = 4 + u, etc as suggested below in one step, make substitution 4 + x^2 > +Au is given as (4 + x^2). My next step is to find du. I think du = 4x + (x^3/3). > du = 2x. You integrated u instead. > I'm lost on what to do next. I know that I have to > substitute. > integral x (4 + x^2)^10 dx u = 4 + x^2; x = sqr(u - 4); dx = 1/sqr(u - 4) * du integral sqr(u - 4) * u^10 * 1/sqr(u - r) * du = etc > Problem 2 > $ 4 / (1 + 2x)^3 dx > integral 4/(1 + 2x)^3 * dx > u given as 1 + 2x > u = 1 + 2x x = (u - 1)/2 best not to skip this inversion step as it's needed for the next line > dx = du / 4 > dx = 1/2 * du > Rewrite: > $ (u^-3) 1/4 du > You forgot the 4 integral 4/(1 + 2x)^3 * dx = integral 4/u^3 * 1/2 * du = etc > (u^-2)/-2*1/4 du > That looks like (u^-2)/(-2*1/4) You omitted the integral sign. > -(u^-2)/8 > You forgot the constant of integration. > -((1 + 2x)^-2)/8 The answer to this one is -(1 + 2x)^-2, so it looks like my 1/4 du > was part of the problem. > Yup. Fortunately your 1/4 mistake and omitting the 4 mistake didn't === Subject: Re: Please help me understand these integral substitution problems > I will use the dollar sign as the integral. >> Problem 1 >> $ x((4 + x^2)^10) dx >integral x (4 + x^2)^10 dx let u = x^2; x = sqr u; dx = 1/(sqr u) * du >integral sqr u * (4 + u)^10 * 1/(sqr u) * du > = integral (4 + u)^10 du let v = 4 + u, etc as suggested below in one step, make substitution >4 + x^2 > Not the best way to do this. Better to let: u = 4 + x^2, du = 2x dx or x dx = (1/2) du giving Int ( (1/2)u^10 du) directly with no sqrt involved. >> +Au is given as (4 + x^2). >> My next step is to find du. I think du = 4x + (x^3/3). >du = 2x. You integrated u instead. > I'm lost on what to do next. I know that I have to >> substitute. >integral x (4 + x^2)^10 dx >u = 4 + x^2; x = sqr(u - 4); dx = 1/sqr(u - 4) * du integral sqr(u - 4) * u^10 * 1/sqr(u - r) * du = etc > Again, see above. >> Problem 2 >> $ 4 / (1 + 2x)^3 dx >integral 4/(1 + 2x)^3 * dx > u given as 1 + 2x >u = 1 + 2x >x = (u - 1)/2 best not to skip this inversion step as it's needed for >the next line No, that step is unnecessary: du = 2 dx so dx = (1/2) du --Lynn === Subject: Re: Please help me understand these integral substitution problems I will use the dollar sign as the integral. > Problem 1 > $ x((4 + x^2)^10) dx >integral x (4 + x^2)^10 dx >>let u = x^2; x = sqr u; dx = 1/(sqr u) * du >>integral sqr u * (4 + u)^10 * 1/(sqr u) * du >> = integral (4 + u)^10 du >>let v = 4 + u, etc as suggested below in one step, make substitution >>4 + x^2 > >Not the best way to do this. Better to let: > u = 4 + x^2, du = 2x dx or x dx = (1/2) du giving Int ( (1/2)u^10 du) directly with no sqrt involved. > +Au is given as (4 + x^2). My next step is to find du. I think du = 4x + (x^3/3). >du = 2x. You integrated u instead. > I'm lost on what to do next. I know that I have to > substitute. >integral x (4 + x^2)^10 dx >>u = 4 + x^2; x = sqr(u - 4); dx = 1/sqr(u - 4) * du >>integral sqr(u - 4) * u^10 * 1/sqr(u - r) * du = etc > >Again, see above. > Problem 2 > $ 4 / (1 + 2x)^3 dx >integral 4/(1 + 2x)^3 * dx > u given as 1 + 2x >u = 1 + 2x >>x = (u - 1)/2 best not to skip this inversion step as it's needed for >>the next line No, that step is unnecessary: du = 2 dx so dx = (1/2) du --Lynn We are still at the introductory point with substitution so these problems have a given for u. For problem 1, u = 4 + x^2. For problem 2, u = 1 + 2x. Also, in William's solution, we are not to the point of using v yet. I have a conceptual problem with du. Is it what is left over after u is determined, or is it the derivative of u? I'd like a working definition of what du is. I'm trying to understand the series of steps involved in solving the problem and I have a big problem with defining du. Again, though, I need to stick with u as defined for each problem. === Subject: Re: Please help me understand these integral substitution problems We are still at the introductory point with substitution so these > problems have a given for u. For problem 1, u = 4 + x^2. For problem 2, u = 1 + 2x. Also, in William's solution, we are not to the point of using v yet. I have a conceptual problem with du. > du is not actually a thing, it's an indication of integration with respect to. With u = x^2, du/dx = 2x. So it's written for memnomics du = 2x dx as tho you can use du and dx as things to multiply, etc. The actual theorem is integral f(u) du/dx dx = integral f(u) du integral f(u) du/dx integrated with respect to x = integral f(u) integrated with respace to u. So for these substitutions du and dx are manipulated formally as tho they are numbers. The manipulations and the du, dx notation are excelent memnomics but informal math. dy/dx means intuitively the ratio of an infinitesimal increase in y to an infinitesimal increase in x integral f(x) dx means intuitively adding all the areas of all the rectangle f(x) high by dx wide for each value of x. dx is of course infinitesimal, so the area f(x) dx is also infinitesimal. Picture it as the pencil point thin line from (0,x) to (f(x), x). Then the area of all these pencil point thin lines all added together fill in the area below the curve f(x). That is integration. The method I worked thru for you problems, tho not the most efficient for those particular problems in terms of memnomics, was used because mathematical more robust and in the long run gave clearer view to the actual mathematics. > Is it what is left over after u is determined, or is it the derivative > of u? I'd like a working definition of what du is. I'm trying to understand the series of steps involved in solving the > problem and I have a big problem with defining du. Again, though, I > need to stick with u as defined for each problem. > As for using a different notation for the substution, it's most simple. integral x(1 + x^2) dx Substitue v = 1 + x^2. It makes no difference what the notation. y = f(x) or y = g(x) both say y is a function of x. The first calls that function f, the second calls that function g. So instead of calling the function 1 + x^2 by u, I'm calling it v. It's still the same function even were I to call it w = 1 + x^2, or to be clearer by including the variable one may wright let w = w(x) = 1 + x^2 === Subject: Re: Please help me understand these integral substitution problems [...] [The problem referred to is int(x*((4 + x^2)^10)*dx)] > I'm trying to understand the series of steps involved in solving the > problem and I have a big problem with defining du. Again, though, I > need to stick with u as defined for each problem. Your text suggests u = 4 + x^2. Here's the trick: du/dx(4 + x^2) = 2*x. So, cheating a bit, du = 2*x*dx. In other words, treat the notation du/dx for the derivative as if it were an honest to goodness fraction. Now, solve du = 2*x*dx for x*dx to get x*dx = (1/2)*du then substitute to get int(x*((4 + x^2)^10)*dx) = int((1/2)*u^10*du). [You are not, at this time supposed to do it this way, but the guess and check method often works well: Guess int(x*((4 + x^2)^10)*dx) = (1/11)*(4 + x^2)^11. Check: d/dx((1/11)*(4 + x^2)^11) = (4 + x^2)*2*x. Oops. Correct: (1/2)*(1/11)*(4 + x^2)^11 should (and does) work.] -- === Subject: Re: Please help me understand these integral substitution problems <150720050132300253%plsperry@sc.rr.com [The problem referred to is int(x*((4 + x^2)^10)*dx)] Your text suggests u = 4 + x^2. Here's the trick: du/dx(4 + x^2) = 2*x. So, cheating a bit, du = 2*x*dx. In other words, > treat the notation du/dx for the derivative as if it were an honest > to goodness fraction. > Sloppy, sloppy!! du/dx * (4 + x^2) = 2x(4 + x^2) du/(dx(4 + x^2)) = 2x/(4 + x^2) du/d(x(4 + x^2)) = (du/dx) / (d(x(4 + x^2))/dx) = 2x/(4 + x^2 + 2x^3) du/dx = (d/dx)(4 + x^2) = 2x = d(4 + x^2)/dx > Now, solve du = 2*x*dx for x*dx to get x*dx = (1/2)*du then substitute > to get int(x*((4 + x^2)^10)*dx) = int((1/2)*u^10*du). [You are not, at this time supposed to do it this way, but the guess > and check method often works well: Guess int(x*((4 + x^2)^10)*dx) = (1/11)*(4 + x^2)^11. > Check: d/dx((1/11)*(4 + x^2)^11) = (4 + x^2)*2*x. Oops. > Correct: (1/2)*(1/11)*(4 + x^2)^11 should (and does) work.] -- > Paul Sperry > Columbia, SC (USA) > === Subject: Re: Please help me understand these integral substitution problems >We are still at the introductory point with substitution so these >problems have a given for u. > These problems usually don't have a given for u. Part of what you need to be learning is when a u substitution is appropriate and what a judicious choice for u is. >For problem 1, u = 4 + x^2. There are two things about the problem that would suggest a u substitution. The first is that 4 + x^2 is a quantity raised to the 10th power. You certainly wouldn't want to multiply it out although, in principle, that would work. You might think to yourself, gee, I wish that had just been x^10 instead of (4 + x^2)^10. That suggests letting u = 4 + x^2, giving u^10. Once you have picked u, you have no choice about du. You must take du = f'(x) dx for whatever f(x) you have chosen for u. In this case you get du = 2x dx. Fortunately, there is a spare x in the problem so that you can substitute for x dx and get an integral with just u's in it. The fact that there is not a 2x but just an x in the original problem doesn't matter because you can just divide both sides of du = 2x dx by 2 and substitute in (1/2)du for x dx. For problem 2, u = 1 + 2x. Also, in William's solution, we are not to the point of using v yet. I have a conceptual problem with du. Is it what is left over after u is determined, or is it the derivative >of u? I'd like a working definition of what du is. I'm trying to understand the series of steps involved in solving the >problem and I have a big problem with defining du. Again, though, I >need to stick with u as defined for each problem. > You don't define du. *You* choose u to make the problem appear simpler, and you compute du. The definition u is up to you and the calculation of du is *always* f'(x)dx where u = f(x) is your choice for u. The whole idea is to get all the x variables changed to u in such a way as to give a simple integral in u. --Lynn === Subject: Re: Cinema projector theory of reality >> I suppose God might be the projectionist - but what I want to know is, >> who >> are the audience?!! LOL! God is doing it either because 'he' wants to, or has to, or both. God is > doing it for a reason. Why are you answering your own post as if you were a different person? Please stop doing that. It is very confusing. === Subject: Re: Cinema projector theory of reality ... >> Playing the projector forwards follows the law of >> entropy, playing it backwards breaks the 2nd >> law of thermodynamics. Actually, no. The definition of entropy is at its most basic, from most > available work to least available work, running the film in reverse > still has the second law intact. Or else the film wouldn't be ABLE to run at all if the second law weren't intact? Just wondering. I thought it somewhat simplistic when he said a 4th dimension should be slices of 3D space, arranged one above the other. Why one above the other? Seems kind of axis-centric. === Subject: Re: Cinema projector theory of reality >> ... > Playing the projector forwards follows the law of > entropy, playing it backwards breaks the 2nd > law of thermodynamics. >> Actually, no. The definition of entropy is at its >> most basic, from most available work to least >> available work, running the film in reverse still has the >> second law intact. Or else the film wouldn't be ABLE to run at all > if the second law weren't intact? Just wondering. In a time reversed Universe, energy/work can be obtained by concentrating heat. Once all available heat has been concentrated, no further energy/work is available. Of course, in such a Universe you radiate ~6000 K photons and direct them to > I thought it somewhat simplistic when he said a 4th > dimension should be slices of 3D space, arranged > one above the other. Why one above the other? > Seems kind of axis-centric. Provincial thinking is something we all suffer from. Together we might be able to unlearn it. David A. Smith