The integral calculation along the curve defined by $\gamma(t)=1+it+t^2$ for $0 \leq t \leq 1$ is confirmed to be $\int_{\gamma} z\,dz = 1 + 2i$. The computation involves evaluating the integral $\int_0^1 (1+it+t^2)(i+2t)dt$, which simplifies to $\int_0^1(2t^3+t)dt + i\int_0^1(1+3t^2)dt$. The result is accurate, and the confusion regarding its correctness stems from misinterpretation of the integral's formulation.
PREREQUISITESStudents and professionals in mathematics, particularly those focusing on complex analysis, as well as anyone interested in understanding contour integrals and their evaluations.
dwsmith said:$\gamma(t)=1+it+t^2, \ 0\leq t\leq 1$
$\displaystyle\int_0^1 (1+it+t^2)(i+2t)dt=\int_0^1(2t^3+t)dt+i\int_0^1(1+3t^2)dt = 1 + 2i$
I was told that was wrong. What is wrong with it?
YesChris L T521 said:What was the original problem? To solve $\displaystyle \int_{\gamma} z\,dz$ where $\gamma(t) = 1+it+t^2$, $t\in [0,1]$??