integral of a function

[kol123]

f(x) = 1/x^2.5

what I did was

f(x) = (ln(x^2.5)) / (2.5 * (x)^1.5) + C

but then I need to use it to calculate area between functions and my answer is incorrect so I've checked if my integration was right, I tried online integration calculator and instead of doing it like me using :
y'(x) = 1/f(x)
y(x) = lnf(x) / f'(x) + C

they do it this way :
1 / x^2.5 = x^-2.5

and then integrate from this point. I know it's possible but my way suppose to be correct aswell using this :
y'(x) = 1/f(x)
y(x) = lnf(x) / f'(x) + C

but I still get wrong answer

 

[HallsofIvy]

f(x) = 1/x^2.5

what I did was

f(x) = (ln(x^2.5)) / (2.5 * (x)^1.5) + C

You have now defined f(x) as two different functions! Do you mean that the second is the integral of the first? If so, you are using the "chain rule" the wrong way. The chain rule is a rule for differentiating functions. It cannot be reversed to do integrals unless one of your functions is a constant.

For example if I want to integrate f(x)= cos(2x) I can argue that this is f(x)= cos(g(x)) and g(x)= 2x. Then I can argue that if the integral is some function a sin(2x) I must have (a sin(2x))'= a(2 cos(2x))= 2a cos(2x)= cos(2x) so that a= 1/2. That is, because, when I am differentiating I must multiply by the derivative of g(x), when integrating I must divide by it.

The problem is this: in terms of integrating, I would have to change from $\displaystyle \int f(g(x))dx$ to $\displaystyle \int f(g(x))g'(x)dx$ and I cannot just "stick" the g'(x) into the integral like that. I could, of course, both multiply and divide by g'(x): \(\displaystyle \int [(f(g(x))g'(x))/g'(x)]dx[itex] and then I would like to write it as \(\displaystyle \frac{1}{g'(x)}\int f(g(x))g'(x)dx= \frac{F(g(x))}{g'(x)}+ C\) where "F" is an integral of f.

That last step is the rub- I can write $\displaystyle \int f(g(x))dx= \int \frac{g'(x)}{g'(x)}f(g(x))dx$ but I cannot then write it as $\displaystyle \frac{1}{g'(x)}\int g'(x)f(g(x))dx$ because I cannot take a function of x outside the integral like that. If that g'(x) were a constant- that is if g(x)= ax- then I could: $\displaystyle \int f(2x)dx= \int \frac{2}{2}f(2x)dx= \frac{1}{2}\int 2 f(2x)dx= \frac{1}{2}F(2x)+ C$ where, again, F is an integral of f: F'= f so that $\displaystyle \left(\frac{1}{2}F(2x)+ C\right)'= \frac{1}{2}F'(2x)(2)= f(2x)$.

But if g'(x) is NOT a constant, I cannot do that. If I want to do $\displaystyle \int f(x^2) dx$, I could write that as $\displaystyle \int \frac{2x}{2x}f(x^2)dx$ but I cannot take that "1/2x" out of the integral. I cannot write it as "$\displaystyle \frac{1}{2x}\int 2x f(x^2)dx$.

Now, if the integral were $\displaystyle \int x f(x^2)dx$ then I could write it as $\displaystyle \int \frac{2x}{2}f(x^2)dx= \frac{1}{2}\int (2x)f(x^2)dx= F(x^2)+ C$ because the "x" was already in the integral and only needed to add the constant 2 which I could move in or out of the integral.

That is the whole point of "substitution". If I have $\displaystyle \int f(g(x))g'(x)dx$ where "g'(x)" is already I the integral (up to a multiplied constant) then I can say "let u= g(x) so that du= g'(x)dx" and then $\displaystyle \int f(g(x))g'(x)dx= \int f(u) du= F(u)+ C= F(g(x))+ C$.

But, that "g'(x)" has to already be in the integral, I cannot just move it in (unless it happens to be a constant).

but then I need to use it to calculate area between functions and my answer is incorrect so I've checked if my integration was right, I tried online integration calculator and instead of doing it like me using :
y'(x) = 1/f(x)
y(x) = lnf(x) / f'(x) + C

they do it this way :
1 / x^2.5 = x^-2.5

and then integrate from this point. I know it's possible but my way suppose to be correct aswell using this :
y'(x) = 1/f(x)
y(x) = lnf(x) / f'(x) + C

but I still get wrong answer

Yes, because you are effectively moving f'(x), a function of x, into and out of the integral- and you cannot do that. It is true that, for c a constant, $\displaystyle \int cf(x)dx= c\int f(x)dx$. But if g(x) is NOT a constant then $\displaystyle \int g(x)f(x)dx\ne g(x)\int f(x)dx$\)

 

[kol123]

What are you doing? You have now defined f(x) as two different functions! Do you mean that the second is the integral of the first? If so, you are using the "chain rule" the wrong way. The chain rule is a rule for differentiating functions. It cannot be reversed to do integrals unless one of your functions is a constant.

For example if I want to integrate f(x)= cos(2x) I can argue that this is f(x)= cos(g(x)) and g(x)= 2x. Then I can argue that if the integral is some function a sin(2x) I must have (a sin(2x))'= a(2 cos(2x))= 2a cos(2x)= cos(2x) so that a= 1/2. That is, because, when I am differentiating I must multiply by the derivative of g(x), when integrating I must divide by it.

The problem is this: in terms of integrating, I would have to change from $\displaystyle \int f(g(x))dx$ to $\displaystyle \int f(g(x))g'(x)dx$ and I cannot just "stick" the g'(x) into the integral like that. I could, of course, both multiply and divide by g'(x): \(\displaystyle \int [(f(g(x))g'(x))/g'(x)]dx[/itex] and then I would like to write it as \(\displaystyle \frac{1}{g'(x)}\int f(g(x))g'(x)dx= \frac{F(g(x))}{g'(x)}+ C[/itex] where "F" is an integral of f.

That last step is the rub- I can write $\displaystyle \int f(g(x))dx= \int \frac{g'(x)}{g'(x)}f(g(x))dx$ but I cannot then write it as $\displaystyle \frac{1}{g'(x)}\int g'(x)f(g(x))dx$ because I cannot take a function of x outside the integral like that. If that g'(x) were a constant- that is if g(x)= ax- then I could: \(\displaystyle \int f(2x)dx= \int \frac{2}{2}f(2x)dx= \frac{1}{2}\int 2 f(2x)dx= \frac{1}{2}F(2x)+ C where, again, F is an integral of f: F'= f so that \(\displaystyle \left(\frac{1}{2}F(2x)+ C\right)'= \frac{1}{2}F'(2x)(2)= f(2x)\).

But if g'(x) is NOT a constant, I cannot do that. If I want to do $\displaystyle \int f(x^2) dx$, I could write that as $\displaystyle \int \frac{2x}{2x}f(x^2)dx$ but I cannot take that "1/2x" out of the integral. I cannot write it as "$\displaystyle \frac{1}{2x}\int 2x f(x^2)dx$.

Now, if the integral were $\displaystyle \int x f(x^2)dx$ then I could write it as $\displaystyle \int \frac{2x}{2}f(x^2)dx= \frac{1}{2}\int (2x)f(x^2)dx= F(x^2)+ C$ because the "x" was already in the integral and only needed to add the constant 2 which I could move in or out of the integral.

That is the whole point of "substitution". If I have $\displaystyle \int f(g(x))g'(x)dx$ where "g'(x)" is already I the integral (up to a multiplied constant) then I can say "let u= g(x) so that du= g'(x)dx" and then $\displaystyle \int f(g(x))g'(x)dx= \int f(u) du= F(u)+ C= F(g(x))+ C$.

But, one more time, that "g'(x)" has to already be in the integral, I cannot just move it in (unless it happens to be a constant).


Yes, because you are effectively moving f'(x), a function of x, into and out of the integral- and you cannot do that. It is true that, for c a constant, $\displaystyle \int cf(x)dx= c\int f(x)dx$. But if g(x) is NOT a constant then $\displaystyle \int g(x)f(x)dx\ne g(x)\int f(x)dx$\)\)\)

\(\displaystyle \(\displaystyle \(\displaystyle
I've got confused by this :S

I was given by a friend a list of examples on how to do integration to some functions and one of the integration showed this :

y' = 1/f(x)

then y = ln(f(x) / f'(x) + C

it's not true then ?\)\)\)

 

[HallsofIvy]

I've got confused by this :S

I was given by a friend a list of examples on how to do integration to some functions and one of the integration showed this :

y' = 1/f(x)

then y = ln(f(x) / f'(x) + C

it's not true then ?

No, that's NOT true unless f'(x) happens to be a constant.

If you are asked to, say, differentiate, y= ln(f(x)), then it is true that y'= f'(x)/f(x) (chain rule) so that it is true that $\displaystyle \int f'(x)/f(x) dx= ln(x)+ C$. But if the "f'(x)" is not already in the integral, you cannot put it in.

It is true that $\displaystyle \int (1/f(x))dx= \int (f'(x)/f(x)) (1/f'(x))dx$
Now, if we could take that "1/f'(x)" outside the integral then we would have
$\displaystyle \int (1/f(x))dx= (1/f'(x))\int f'(x)/f(x) dx= (1/f(x))(ln(x)+ C)$

But that is NOT TRUE. If f'(x) is a function of x, not just a constant, we cannot take it outside the integral that way.