# Homework Help: Integration of a composite function

1. Dec 10, 2013

### Tensel

y=f(x),K=F(y),
and dF(y)/dy=y^2/y', (1)
then
dF(x)=y^2*dx;
so, F(x)=int(y^2*dx)=int((f(x)^2)*dx);
then we obtain,
F(x)=(f(x))^3/(3*f'(x))+C;
substitution of y=f(x) into F(x), we get, F(y)=y^3/(3*y')+C; (2)

using the result above(eq.(2)), the dF(y)/dy can be computed as follow:

dF(y)/dy=y^2/y' - (y^3*y'')/(3*(y')^3) (3)

Probelm: Eq.(3) is not equal to the eq.(1).

I don't know which step is not correct.

2. Dec 10, 2013

### ShayanJ

I have trouble figuring out which part of your calculation is correct!

3. Dec 10, 2013

### HallsofIvy

What reason do you have for introducing "K" when it is not used again?

I don't believe this is possible. Do you mean y' to be the dervative of y with respect to x?
The derivative of F with respect to y has nothing to do with y's dependence of x.

If you meant dF/dx, then $dF/dx= (dF/dy)(dy/dx)$. If you meant that $F(y)= y^3/3$, then $dF/dx= y^2$ times y', not divided by it.

and this doesn't follow from what you wrote before. I don't know where you got the $y^2$ but, again, if you meant that $F(y)= y^3/3$ then $dF= y^3 y'(x)dx$.

Last edited by a moderator: Dec 10, 2013
4. Dec 10, 2013

### Tensel

Yes, you are correct that K is not used, we can get rid of it.

Yes, y' means the derivative of y with respect ot x.

So, you meant that we should consider y' as constant when computing the derivative of F with respect to y? I need more information to understand this.

At the beginning, we have dF(y)/dy=y^2/y', then dF(y)/dy=y^2/(dy/dx),
eliminate the dy on the two side of the equation, we obtain dF(y)=y^2*dx,
substitution of y=f(x) into it, dF(f(x))=dF(x)=y^2*dx

5. Dec 10, 2013

### pwsnafu

Suppose that $F(y) = y^3 / 3$ and $y$ is a function of x. Then
$\frac{dF}{dy} = y^2$
$\frac{dF}{dx} = y^2 \frac{dy}{dx}$

How? Where did this come from?

You are using $\frac{1}{dy/dx} = \frac{dx}{dy}$ which requires y to be an invertible function.

6. Dec 11, 2013

### Tensel

dF(y)/dy=y^2/y'-----this equation is what I want to construct, it can be considered as a known condition, I want to get F(y).

Last edited: Dec 11, 2013
7. Dec 11, 2013

### Tensel

I find the problem.
given that, $\frac{dF(y)}{dy}=\frac{y^{2}}{y'}$
then,
$\frac{dF(x)}{dx}=\frac{dF(y)}{dy}.\frac{dy}{dx}=\frac{y^{2}}{y'}.y'=y^{2}$,
so,
$F(x)=\int y^{2}.dx=\int f^{2}(x).dx$

I made a mistake in the integration of F(x)
i.e. $F(x)≠\frac{f^{3}(x)}{3.f'(x)}+C$

But, what should be the answer for F(x) from which I can obtain F(y)? should the specific expression for f(x) be given?

Last edited: Dec 11, 2013