Wegintegral/Vektorfeld/Eindimensional/Weg/3/Aufgabe/Lösung

Es ist

${}\gamma '(t)=3t^{2}+1\,.$

Damit ist das Wegintegral gleich

${}{\begin{aligned}\int _{0}^{1}f(\gamma (t))\gamma '(t)dt&=\int _{0}^{1}{\left({\left(t^{3}+t-5\right)}^{2}-4{\left(t^{3}+t-5\right)}+1\right)}{\left(3t^{2}+1\right)}dt\\&=\int _{0}^{1}{\left(t^{6}+t^{2}+25+2t^{4}-10t^{3}-10t-4t^{3}-4t+20+1\right)}{\left(3t^{2}+1\right)}dt\\&=\int _{0}^{1}{\left(t^{6}+2t^{4}-14t^{3}+t^{2}-14t+46\right)}{\left(3t^{2}+1\right)}dt\\&=\int _{0}^{1}3t^{8}+7t^{6}+-42t^{5}+5t^{4}-56t^{3}+139t^{2}-14t+46dt\\&=\left({\frac {1}{3}}t^{9}+t^{7}-7t^{6}+t^{5}-14t^{4}+{\frac {139}{3}}t^{3}-7t^{2}+46t\right)|_{0}^{1}\\&={\frac {1}{3}}+1-7+1-14+{\frac {139}{3}}-7+46\\&={\frac {140}{3}}+20\\&={\frac {200}{3}}.\end{aligned}}$

Zur gelösten Aufgabe