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Algebra, analysis and differential equations 1 course of the University of Oxford

Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford

2010

{c}

The <complex line integral> along the smooth parametrized curve is
$$
\int_\gamma f(z)dz:=\int_a^b f(\gamma(t))\gamma'(t)dt.
$$
For the reversed curve, the <chain rule> gives $\widetilde\gamma'(t)=-\gamma'(a+b-t)$. Substitution $s=a+b-t$ yields
$$
\int_{\widetilde\gamma}f(z)dz
=-\int_a^b f(\gamma(a+b-t))\gamma'(a+b-t)dt
=-\int_a^b f(\gamma(s))\gamma'(s)ds.
$$
Consequently \b[reversing orientation negates the complex line integral].

Solution

Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2010 / 5 / i / Solution

Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2010 / 5 / i / Solution

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