### Author Topic: 1.6 Q5  (Read 365 times)

#### Nathan

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##### 1.6 Q5
« on: October 06, 2020, 06:08:35 PM »
Question: $\int_y Re(z) dz$ where $y$ is the line segment from 1 to $i$.

I can't get the same answer as the one in the textbook.

Answer in textbook: $\frac{1}{2}(i-1)$

$y(t)=$
$= 1 + (i - 1)t$
$= (1 - t) + it$

$y'(t) = i - 1$

$\int_y Re(z) dz=$
=$\int_1^i (1-t)(i-1)dt$
$=\int_1^i i - 1 - ti + t dt$
$=ti - t - \frac{t^2i}{2} + \frac{t^2}{2}|^i_1$
$=-1 -i + \frac{i}{2} - \frac{1}{2} - (i - 1-\frac{i}{2} + \frac{1}{2})$
$=-i - 1$

What is wrong with my answer?

#### smarques

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##### Re: 1.6 Q5
« Reply #1 on: October 06, 2020, 06:29:36 PM »
Hi Nathan.

The mistake seems to be in your bounds of integration, not the process itself. With the way you parameterized the line, the integral should be from 0 to 1, not 0 to i.