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Integration: Can't understand when to use Power rule.

  1. Jul 22, 2015 #1
    1. The problem statement, all variables and given/known data
    Hi I have a really hard time understanding when and how to use the Power rule when integrating


    My book states that if u is a function of x, then the power rule is given by:

    ∫urdu = (ur+1)/(r+1)+c

    First: I do understand that if

    u
    =2x
    Then the differential

    du=u'(x)dx =2

    So ∫u^2*du = ƒ2*(2x)^2 = 2*[(2x)3/3]




    2. Relevant equations

    What I don't understand: it seems that the power rule is only practical when I need to find
    the integral of a function ur multiplied by it's differential (which i can rarely find in any of the problems, so i don't understand why the power rule is important )

    EXAMPLE
    What If I'm given a function f with the equation f(x)= (1/x+x)^2 and i need to find the integral of it using the power rule?
    I could define u=(1/x+x)
    and I do have u^2=(1/x+x)^2 but there is no differential du = d(1/x+x))dx multiplied to the function, , so Is it true that The only way Can integrate this function is by first multiplying (1/x+x)*(1/x+x) and then integrating the individual terms, or is there another way, perhaps using the power rule?

    So to me it seems like I can't use ∫u^2du =[(u)3/3] beause that would be the same as ∫(1/x+x)^2 d(1/x+x)^2) and that's not what I'm suppose to integrate.

    So when is the power rule usefull? Am I interpreting this wrong?
     
  2. jcsd
  3. Jul 22, 2015 #2

    DEvens

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    Education Advisor
    Gold Member

    Note that this equation is true whether or not ##u## is a function of ##x##. This is important because it is one of the fundamental forms. The reason you don't see it a lot in your problem sets is because it is supposed to be one of the things you have learned and have ready to use in other problems.

    You then start in on another issue, change of variables. If you have ##u(x) = 1/x + x## then you have lots of useful things to do there. But as you say, you have to do the change of variables correctly. You have to get from ##du## to ##\partial u / \partial x## ##dx## and use that.

    So you have to do ##\int (1/x + x)^2 dx##? What do you get if you multiply out the bracket? And can you then perform that integral?
     
  4. Jul 22, 2015 #3
    Hi thank you for reply: What do you mean by "Note that this equation is true whether or not u is a function of x?" Do you mean if u was a function of t instead?


    If i multiply the brackets its easy,

    (1/x + x)*(1/x + x) = 1/x^2 +2+x^2 and the integral of this is very easy but I feel as if this should be something that could be easily done by using the power rule, I just can't see how to apply the power rule mathematically correct? I haven't had experience with it yet, the book didn't really cover it yet.
     
  5. Jul 22, 2015 #4

    DEvens

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    Education Advisor
    Gold Member

    The power rule integral you mention, in terms of u, does not depend on the functional form of u. It could be any variable. This is because it is an integral with respect to du.

    So if you change u to x, for example, the form is the same.
    ## \int u^r du = u^{r+1} / (r+1) + c##
    ## \int u(x)^r du(x) = u(x)^{r+1} / (r+1) + c##
    ## \int u(t)^r du(t) = u(t)^{r+1} / (r+1) + c##
    ## \int u(fred)^r du(fred) = u(fred)^{r+1} / (r+1) + c##
    ## \int x^r dx = x^{r+1} / (r+1) + c##
    ## \int y^r dy = y^{r+1} / (r+1) + c##
    ## \int (fred)^r d(fred) = (fred)^{r+1} / (r+1) + c##

    That is, it's a formal expression. It does not care about the functional form of the thing that is getting integrated. (Well, as long as it satisfies the basic requirements for being integrated. It has to be finite and well valued and so on.)

    So you have to integrate three terms: 1/x^2, 2, and x^2. Can you apply the power law to these?
     
  6. Jul 22, 2015 #5
    +Thank you,
    I understand what I didn't understand

    1) If u is a independant variable f(u) depends on u Then

    ∫f(u)du =F(u) +C

    And in this case du is always equal to 1 because the differential of a variable d(u)=u'du=1*d or d(x)=x'dx=dx

    So if f(u)=u^4+u^2

    Then the sum rule and power rule can be used: and du is just 1*du.
    ∫(u^4+u^2)du = ∫u^4du + ∫u^2du and the power rule can be used, and in this case du=1du



    2) But if u is a dependant variable such that u=g(x) Then

    ∫f(u)du = f(g(x))g'(x)dx where du=g'(x)dx

    example in f(x)=(1/x + x)^2 where u=1/x+x depends on then i need to do the following

    1)
    In order to integrate this
    ∫f(x)dx = ∫(1/x + x)^2dx

    I need to turn it into to something something that i can apply the power rule on.
    ∫u^r*du = ur+1/(r+1)

    2) The relationship between the 2 differentials du and dx is
    That dx=dx and du=d(1/x+x) = (-1/x^2 + 1.)dx So if i can isolate dx

    dx=du/(-1/x^2 + 1.) Then i can substitute it into


    ∫f(x)dx = ∫u^2du/(-1/x^2 + 1.) = [u^3/(3)]*(-1/x^2 + 1.) Is that the correct procedure?
     
  7. Jul 22, 2015 #6
    Just at last question regarding notation - The dx, du, dh etc.

    if f(x)=(1+5x)^2

    then h=(1+5x)
    f(x)= h^2

    But regarding the function h, when should i write dh vs dx? If i write ∫h^2 I would add dh but if i write ∫(h(x))^2 should i then add dx?
    What is the difference between writing ∫(h(x))^2dx and ∫h^2dh?

    I always see this ∫udu but i never see ∫u(x)du, is that because it should be ∫u(x)du?

    So could i use the power rule on f(x)=(1+5x)^2 like ∫fdf if i could find df? here the power of f would be 1?
     
    Last edited: Jul 22, 2015
  8. Jul 22, 2015 #7

    SammyS

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    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    You're missing the differential "dx" with the 2.
    It appears that you're trying to use the method of substitution rather than simply using the power rule. It is true that using the method of substitution may lead to using the power rule if that's what the result gives.

    In your example with, ##\displaystyle \ f(x)=\left(\frac{1}{x}+x\right)^2 \,,\ ## it's best to expand ##\displaystyle \ \left(\frac{1}{x}+x\right)^2 \ ## and then use the power rule on the result.

    On the other hand,

    Notice that ##\displaystyle \ \frac{d}{dx}\left(\frac{1}{x}+x\right)=\frac{-1}{x^2}+1\ . ##

    So, if you had the following integral;
    ##\displaystyle \ \int\left(\left(\frac{1}{x}+x\right)^2\left(\frac{-1}{x^2}+1\right)\right)dx ##​
    Then using your substitution would lead to using the power rule.
     
  9. Jul 22, 2015 #8

    Mark44

    Staff: Mentor

    No.
    f(h) = h2


    The dh or dx or whatever tells you what the variable of integration is. Without know what h(x) is, I can't evaluate ##\int h^2(x) dx##, but I can evaluate ##\int h^2 dh##, which is ##(1/3)h^3 + C##.
     
  10. Jul 22, 2015 #9
    Thank you so much!
    So what you are saying is if i don't know what h(x) is i can't integrate it with repsect to x, So if f(x)=(2+5x)^2 then h=2+5x so i can't evaluate ∫h^2(x)dx but I can evaluate ∫h^2dh . Why is it we can't evaluate ∫h^2(x)dx ? Is it impossible or is it just because it's really hard?
     
  11. Jul 22, 2015 #10

    Mark44

    Staff: Mentor

    You can't evaluate ##\int h^2 dx## unless you can write h as a function of x.

    ##\int h^2 dx = \int (2 + 5x)^2 dx##
    The latter can be evaluated either by expanding (2 + 5x)2 and integrating the three terms, or by using a substitution u = 2 + 5x, with du = 5dx.
     
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