0.0.0.3 (1) の解答

$\Vector{f}(\Vector{\varphi}(t))= \Vector{f}(t,t^2)= \twovector{t^2}{t^2}$, $\Vector{\varphi}'(t)=\twovector{1}{2t}$ であるから、

$$\Vector{f}(\Vector{\varphi}(t))\cdot\Vector{\varphi}'(t) =t^{2}\cdot 1+t^2\cdot 2t=t^{2}+2t^3.$$

ゆえに

$$\int_{C_1}\Vector{f}\cdot\D\Vector{r} =\int_0^1\left(t^{2}+2t^3\r... ...c{t^4}{2}\right]_0^1 =\frac{1}{3}+\frac{1}{2}=\frac{2+3}{6}=\frac{5}{6}. \quad$$