Finding the Integral of xdx on Arc C

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SUMMARY

The discussion focuses on evaluating the integral of xdx along the arc C defined by the curve y=x² from the points (0,0) to (1,1). The correct parameterization is established as x=t and y=t², with t ranging from 0 to 1. The integral simplifies to ∫ from 0 to 1 of t² dt, resulting in a value of 1/3. The confusion regarding the nature of the integral is clarified, emphasizing that this is an ordinary integral rather than a line integral.

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Homework Statement


Let C be the arc of y=x2 from (0,0) to (1,1). Evaluate [tex]\int[/tex]xdx

Homework Equations


C1:
x=t
y=t2
-1 [tex]\leq[/tex] t [tex]\leq[/tex] 1 so -1 [tex]\leq[/tex] x [tex]\leq[/tex] 1

The Attempt at a Solution



dx is x' dt, right?

For some reason, I just can't figure this out. Any help?
 
Last edited:
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ummm [tex]\int x dx[/tex] is not even a line integral; it's just an ordinary integral...do you mean [tex]\int_{\mathcal{P}} \vec{x} \cdot \vec{ds}[/tex]?
 
No, that is what he has down...Except the C is where the p is on yours. I did not know how to do that. I figured it out, though.

since dx=xdt and x=t,

[tex]\int[/tex] from 0 to 1 of t2dt is 1/3.
 

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