Finding the Primitive Function for y'(x)=xcos(x2) | Homework Help and Equations

  • Thread starter Thread starter jakobs
  • Start date Start date
  • Tags Tags
    Primitive
jakobs
Messages
15
Reaction score
0

Homework Statement


I have y'(x)= xcos(x2)
How do I get the primitive function for this?

Homework Equations





The Attempt at a Solution


I know that f'(x)=2xcos(x2) is f(x)=sin(x2)
How will removing that 2 in front of xcos affect the primitive function?
 
Physics news on Phys.org
y(x)=1/2f(x)
 
A more "formal" way to do this would be to let u= x^2 so that du= 2xdx or (1/2)du= xdx. Then \int x sin(x^2)dx becomes (1/2)\int sin(u)du. Alternatively, write \int x sin(x^2)dx as (1/2)(2)\int x sin(x^2)dx= (1/2)\int sin(x^2)(2x)dx

Note that you can only move constants into and out of the integral like that. If that first "x" were not there, you could not do the integral.
 
Last edited by a moderator:
conquest said:
y(x)=1/2f(x)

So, the answer should be f(x)=1/2 sin(x) ?
 
HallsofIvy said:
A more "formal" way to do this would be to let u= x^2 so that du= 2xdx or (1/2)du= xdx. Then \int x sin(x^2)dx becomes (1/2)\int sin(u)du[/tex]. Alternatively, write \int x sin(x^2)dx as (1/2)(2)\int x sin(x^2)dx= (1/2)\int sin(x^2)(2x)dx<br />
<br /> <br /> I&#039;m not completely understand everything you wrote there, and I feel a bit stupid now.
 
@jakobs
well you actually put f(x)=sin(x²). So I meant y(x)= 1/2 sin(x²). Then indeed y'(x)=xcos(x²).

@Halls of Ivy
Agreed this would be the 'follow the recepy' way of doing it, but with a problem this straightforward it really isn't necessary.
 
Thread 'Use greedy vertex coloring algorithm to prove the upper bound of χ'
Hi! I am struggling with the exercise I mentioned under "Homework statement". The exercise is about a specific "greedy vertex coloring algorithm". One definition (which matches what my book uses) can be found here: https://people.cs.uchicago.edu/~laci/HANDOUTS/greedycoloring.pdf Here is also a screenshot of the relevant parts of the linked PDF, i.e. the def. of the algorithm: Sadly I don't have much to show as far as a solution attempt goes, as I am stuck on how to proceed. I thought...
Back
Top