MHB Is the Calculation of the Integral Along the Curve $\gamma(t)=1+it+t^2$ Correct?

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The integral calculation along the curve defined by $\gamma(t) = 1 + it + t^2$ for $0 \leq t \leq 1$ is questioned for accuracy. The integral is expressed as $\int_0^1 (1 + it + t^2)(i + 2t) dt$, which simplifies to two separate integrals. The resulting evaluation yields $1 + 2i$, but there is uncertainty regarding its correctness. The original problem involves solving the integral $\int_{\gamma} z\,dz$ along the specified curve. Clarification is sought on the validity of the computed result.
Dustinsfl
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$\gamma(t)=1+it+t^2, \ 0\leq t\leq 1$

$\displaystyle\int_0^1 (1+it+t^2)(i+2t)dt=\int_0^1(2t^3+t)dt+i\int_0^1(1+3t^2)dt = 1 + 2i$

I was told that was wrong. What is wrong with it?
 
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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?

What was the original problem? To solve $\displaystyle \int_{\gamma} z\,dz$ where $\gamma(t) = 1+it+t^2$, $t\in [0,1]$??
 
Chris L T521 said:
What was the original problem? To solve $\displaystyle \int_{\gamma} z\,dz$ where $\gamma(t) = 1+it+t^2$, $t\in [0,1]$??
Yes
 

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