Converting to a Riemann Integral

[lastlydreaming]

If one were to convert the Integral of the Real (z) dz of a segment (1, -2i) to (-3, 4i) into a Reimann Integral would it be :

S(from 0 to 1) of (1-4t)(-4+6i) dt. Then you can take the (-4+6i) out , so you'd be left with

(-4+6i) S (0, 1)(1-4t) dt = (-4 +6i) (-1/8 * (-4+6i)^2 )] 0 to 1?

Thanks.

 

[pka]

The contour is a line segment \(\displaystyle \left\langle {\left( { - 4t + 1} \right),\left( {6t - 2} \right)} \right\rangle \quad 0 \le t \le 1.\)

Then
\(\displaystyle \L
\begin{array}{rcl}
\int_C {Re(z)dz} & = & \int_C {xdx} + i\int_C {xdy} \\
& = &\int\limits_0^1 {\left( { - 4t + 1} \right)\left( { - 4dt} \right)} + i\int\limits_0^1 {\left( { - 4t + 1} \right)\left( {6dt} \right)} \\
\end{array}\)