1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Disproof derivative of integral

  1. Jan 19, 2014 #1
    Disproof:
    (d/dx)(integral of f(t)dt from 0 to x^4)=integral of (d/dt)(f(t))dt from 0 to x^4

    I arrived here:

    4(x^3)f(x^4)=f(x^4)-f(0)

    I am stuck!
    Please give clues. I don't understand how I can disproof this.
     
  2. jcsd
  3. Jan 20, 2014 #2

    maajdl

    User Avatar
    Gold Member

    The original assumption is not valid for any function f, but only for a small club of functions such that

    f(x^4) = f(0)/(1-4x^3)

    where f(0) is an arbitrary constant.
     
  4. Jan 20, 2014 #3

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Show:
    $$\frac{d}{dx}\int_0^{x^4} f(t)\text{d}t \neq \int_0^{x^4}\frac{d}{dt}f(t)\text{d}t$$

    You got as far as

    $$4x^3f(u)+f(0)=f(u): u=x^4$$

    As maajdl points out, this means that you now have a condition that must be met that makes the expression true. But the expression is asserted to be true for any f(t) not just some specific subset.

    It is that assertion which is false.
     
  5. Jan 20, 2014 #4

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    I think, it's much easier to start from the definition
    [tex]F(u)=\int_0^{u} \mathrm{d} x f(x).[/tex]
    Then you can use the chain rule and the fundamental theorem of calculus to evaluate
    [tex]\frac{\mathrm{d}}{\mathrm{d} x} F(x^4)[/tex]
    to get the correct derivative asked for.
     
  6. Jan 20, 2014 #5

    maajdl

    User Avatar
    Gold Member

    Didn't he just do that?
     
  7. Jan 20, 2014 #6

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    No, obviously not :-)).
     
  8. Jan 20, 2014 #7

    maajdl

    User Avatar
    Gold Member

    He was asked to disproof that

    expr1 = expr2

    where, as he derived:

    expr1 == D[F[x^4],x] = 4 x^3 f(x^4)
    expr2 == Integrate[D[f[t],t],{t,0,x^4}] = f[x^4] - f[0]
     
  9. Jan 20, 2014 #8

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    That is not what the problem is asking for. It's easy to find an f(t) such that the above inequality becomes an equality.

    There's a huge difference between ##\frac{d}{dx}\int_0^{x^4} f(t)\,dt \neq \int_0^{x^4}\frac{d}{dt}f(t)\,dt\,\,\forall\,f(t) \in C^1## (the two expressions are unequal to one another for all differentiable functions f(t)) and ##\neg \left(\,\forall\,f(t) \in C^1: \frac{d}{dx}\int_0^{x^4} f(t)\,dt = \int_0^{x^4}\frac{d}{dt}f(t)\,dt\right)## (it is not the case that for all differentiable functions the two expressions are equal to one another). The former means the two expressions are never equal, no matter what f(t) is. The latter merely means that this equality is not an identity.

    One way to disprove the conjecture is to find some function f(t) for which the equality does not hold. alingy1, can you find a simple function f(t) for which ##4x^3f(x^4) \ne f(x^4)-f(0)## ?

    Yes, obviously he did. The left hand side of the equality in the original post is your ##\frac{d}{dx}F(x^4)##.
     
  10. Jan 20, 2014 #9

    pasmith

    User Avatar
    Homework Helper

    What happens for [itex]f: x \mapsto 1[/itex]?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Disproof derivative of integral
Loading...