# Thread: Check solution of integral

1. ## Check solution of integral

I'm preparing for an exam and would like someone to check this if possible

$\displaystyle \int_{C}(z^{2}+z)dz$ where $\displaystyle C$ is any path from $\displaystyle -i$ to $\displaystyle 2+i$

I figured out that $\displaystyle (z^{2}+z)$ was not analytic using cauchy riemann equations

So i parametrized $\displaystyle -i$ to $\displaystyle 2+i$ using the formula
$\displaystyle z(t) = a + (b-a)t$ where $\displaystyle 0<=t<=1$

Filling into the above formula i got $\displaystyle z(t) = 2t + i(t-1)$
and $\displaystyle \frac{dz}{dt} = 2+i$

So i said $\displaystyle \int_{C}(z^{2}+z)dz = \int^{1}_{0}[(2t + i(t-1))^{2} + (2t +i(t-1))](2+i)dt$

i proceed to work this out an got an answer of $\displaystyle -\frac{5}{3} - \frac{4}{3}i$

The answer isn't really that important to me, i just would like to know if the line above my answer is correct

Thanks,
Piglet

2. Shouldn't the parametrisation be 2t + i(2t - 1) instead, because yours never attains the point 2+i for t in [0,1]. Otherwise you have the right idea with regards to computing the integral.

3. Originally Posted by Magus01
Shouldn't the parametrisation be 2t + i(2t - 1) instead, because yours never attains the point 2+i for t in [0,1]. Otherwise you have the right idea with regards to computing the integral.
Yeah your spot on there, silly mistake

note to self: practice simple multiplication

4. suppose z=t+（t+1）i and t blong to [-1,1]
then your question can be transfered into
$\displaystyle \int_{C}(z^{2}+z)dz = \int^{-1}_{1}[-(t+1)+(2t^{2}+3t+1)i]dt=\int^{-1}_{1}(2t^{2}+3t+1)idt-2=(4/3+2)i-2$

i trust you can solve the problem above!

5. my result is (4/3+2)i-2