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

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Homework Help Overview

The discussion revolves around finding the primitive function for the derivative y'(x) = xcos(x²). Participants are exploring the relationship between the derivative and its integral in the context of calculus.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the implications of removing constants from the integral and consider substitution methods, such as letting u = x². There are questions about the correctness of the proposed primitive function and its derivation.

Discussion Status

Some participants have offered guidance on substitution techniques and the manipulation of constants within integrals. There is an ongoing exploration of different interpretations of the problem, with no explicit consensus reached on the final form of the primitive function.

Contextual Notes

Participants express uncertainty about the steps involved in the integration process and the correctness of their interpretations. There is a mention of feeling overwhelmed by the complexity of the discussion.

jakobs
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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?
 
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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.
 

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