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maths help integration

how do i know that i need to integrate f'(x)?
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integ.png
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@RDKGames
Original post by mervin101
integ.png


I got you
y=f(x),
Therefore you have to integrate your f'(x) to get f(x).
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Original post by JD2015AS
I got you
y=f(x),
Therefore you have to integrate your f'(x) to get f(x).


Thanks for replying!

is this with all cases? so when there is f'(x) and i need to express y in terms of x i need to integrate and make sure i find the constant which in this case is 0 (make x=0)

is there was just f(x) and i needed to find y in terms of x would i differentiate to get f'(x)?
Original post by mervin101
Thanks for replying!

is this with all cases? so when there is f'(x) and i need to express y in terms of x i need to integrate and make sure i find the constant which in this case is 0 (make x=0)

is there was just f(x) and i needed to find y in terms of x would i differentiate to get f'(x)?


When you see a question like this, you do have to integrate AND find the constant of Integration.

Did you integrate this? if so, what did you get?
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Original post by JD2015AS
When you see a question like this, you do have to integrate AND find the constant of Integration.

Did you integrate this? if so, what did you get?



y=6x-2x^2 -x^3 + c
c=0

I'm just a bit confused
when it says find an expression for y in terms of x what does this mean?
is it finding f(x)? because initially its y=f('x) ?
Yes it is finding f(x), the curve is: y=f(x)..

The integral is correct. And the constant seems to be so, too.
Original post by mervin101
...


You're given that y=f(x)y=f(x) and differentiation with respect to xx of both sides yields y′=f′(x)y'=f'(x)

We know that f′(x)=6−4x−3x2f'(x)=6-4x-3x^2 therefore to get the original function we need to perfect the inverse operation of differentiation, which is integration.

So y=∫y′.dx=∫f′(x).dxy=\int y' .dx = \int f'(x) .dx

Original post by mervin101

is this with all cases? so when there is f'(x) and i need to express y in terms of x i need to integrate and make sure i find the constant which in this case is 0 (make x=0)


Well, yes in all cases similar to this where you're given the same info and being asked to find the same thing... This is assuming that y=f(x)y=f(x) of course

is there was just f(x) and i needed to find y in terms of x would i differentiate to get f'(x)?


If y=f′(x)y=f'(x) then yeah sure you would differentiate f(x)f(x) to get yy in terms of xx but I fear you are confusing yourself with some terminology here...
Thanks RDK -_-

It says on the curve that y=f(x)
Original post by JD2015AS
Thanks RDK -_-

It says on the curve that y=f(x)


That information wouldn't be the same "in all cases" so he cannot generalise like that which is what I'm saying.
Original post by mervin101
y=6x-2x^2 -x^3 + c
c=0

I'm just a bit confused
when it says find an expression for y in terms of x what does this mean?
is it finding f(x)? because initially its y=f('x) ?


Means you have yy on its own on one side of the equation, and everything else with the variable xx on the other side. Since y=f(x)y=f(x) you can replace f(x)f(x) by yy
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f'(x) is another way of saying dy/dx.
and f(x) without the ' is another way of saying y.

So you know if you differentiate y you get dy/dx. Integrate dy/dx and you get y.

Similarly differentiating f(x) gives you f'(x). Integrating f'(x) and you get f(x).
Original post by RDKGames
That information wouldn't be the same "in all cases" so he cannot generalise like that which is what I'm saying.


some cases*
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Original post by RDKGames
You're given that y=f(x)y=f(x) and differentiation with respect to xx of both sides yields y′=f′(x)y'=f'(x)

We know that f′(x)=6−4x−3x2f'(x)=6-4x-3x^2 therefore to get the original function we need to perfect the inverse operation of differentiation, which is integration.

So y=∫y′.dx=∫f′(x).dxy=\int y' .dx = \int f'(x) .dx



Well, yes in all cases similar to this where you're given the same info and being asked to find the same thing... This is assuming that y=f(x)y=f(x) of course



If y=f′(x)y=f'(x) then yeah sure you would differentiate f(x)f(x) to get yy in terms of xx but I fear you are confusing yourself with some terminology here...


Thank you, I understand now
for the bottom part. Yeah I think I am a bit confused
given that f(x) = 3x^2 6x + 5
f'(x) would be 6x + 6
right?
Original post by mervin101
Thank you, I understand now
for the bottom part. Yeah I think I am a bit confused
given that f(x) = 3x^2 6x + 5
f'(x) would be 6x + 6
right?


Assuming the typo there is supposed to say f(x)=3x2+6x+5f(x)=3x^2+6x+5 then yes, indeed f′(x)=6x+6f'(x)=6x+6
Original post by mervin101
integ.png


ummm...... you may have misread this... f'(x) = 6−4x−3x26 - 4x - 3x^2

And you said to differentiate that -_-
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Original post by JD2015AS
ummm...... you may have misread this... f'(x) = 6−4x−3x26 - 4x - 3x^2

And you said to differentiate that -_-


i said integrate
i was using examples above ^
Original post by mervin101
i said integrate
i was using examples above ^


I meant to quote @RDKGames in as well, sorry about that!

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