I am currently reviewing my notes from Complex Analysis, and my professor gave us an exercise to verify that $$\int_0^5 (1 + it)^5 dt = \frac{(1+it)^3}{3} \bigg|_0^5$$
This didn't seem right to me (and it's very possible I just copied down the exercise incorrectly), and I tried to evaluate both sides and found that they weren't equal. My thought is that the integral should equal $$\frac{-i(1+it)^6}{6} \bigg|_0^5$$
Is there something I'm misunderstanding?
Assuming exponent $2$, the teacher's answer is wrong.
Solution 1:
Let $z=1+it$, so that $dz=i\,dt$.
Then
$$\frac1i\int_{1}^{1+i5}z^2dz=-i\left.\frac{z^3}3\right|_{1}^{1+i5}=\frac{-110+i75}3.$$
Solution 2:
$$\int_0^5(1+it)^2dt=\int_0^5(1+i2t-t^2)dt=\left.\left(t+it^2-\frac{t^3}3\right)\right|_0^5=\frac{-110+i75}3.$$