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This may sound like a silly question, but was just wondering if someone can clear this up for me

Consider the line segment joining the points a,b. Therefore we have f(t)=a+t(b-a) where t is contained between 0 and 1

Now if we were to integrate f(z) over this contour. I know the formula, we have to replace z with the path and multiply by derivative of the path and integrate over 0 and 1. In my textbook they stated the derivative of the path is 1

How is that the case ? surely the derivative is f'(t)=b-a

Thanks

Shalid
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  • Correct, $\vec{f},'(t)=\vec{b}-\vec{a}$. – murray May 31 '16 at 20:21
  • Are you confusing the function $f: \mathbb{C} \to \mathbb{C}$ to be integrated over the (linear) path with the path $\gamma: [0, 1] \to \mathbb{C}$? – murray May 31 '16 at 20:27

1 Answers1

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It depends on how you paramaterize the path. As you set it up

$f(t) = a + (b-a)t, t\in [0,1]\\ f'(t) = b-a$

However, you could say:

$f(t) = t , t\in [a,b]\\ f'(t) = 1$

Doug M
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