Disproof derivative of integral

[alingy1]

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.

 

[maajdl]

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.

 

[Simon Bridge]

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.

 

[vanhees71]

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

 

[maajdl]

Didn't he just do that?

 

[vanhees71]

No, obviously not :-)).

 

[maajdl]

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]

 

[D H]

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

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 $4x^3f(x^4) \ne f(x^4)-f(0)$ ?

No, obviously not :-)).

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

 

[pasmith]

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.

What happens for $f: x \mapsto 1$?