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I'm having a hard time understanding how to set t boundaries in line integrals...

The question is: find the line integral of $f(x,y,z)$ over the straight line segment from $(1,2,3)$ to $(0,-1,1)$. I figured out that $\bar{r}(t)=(1-t) \hat{i} + (2-3t) \hat{j} +(3-2t) \hat{k} $, and $| v | = \sqrt{14} $.

But I don't know how I should set up boundaries for t. My textbook says $0\leq t \leq 1$, but why??? How do I calculate this??? I thank in advance!

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2 Answers 2

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The boundaries should be 0 and 1 since those values of $t$ correspond to the correct coordinates for your given begin- and endpoints.

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  • $\begingroup$ Thank you so much! Can I ask just one more question? I think I almost get this now. If $\int_{c} x ds$, where C is the parabolic curve $x=t$ and $y=t^2$ from (0,0) to (2,4), then would t be $0\leq t \leq 2$ or $-2\leq t\leq 2$? $\endgroup$
    – user125342
    Commented Jul 11, 2014 at 3:43
  • $\begingroup$ @user125342 try to plug in those values in the equations defining $x$ and $y$. You will see what your answer is. $\endgroup$
    – Danu
    Commented Jul 11, 2014 at 11:50
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To say what Danu said in other words: $r(0)=(1,2,3); r(1)=(0,1,-1)$,

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