Answer
$\int_C v \cdot dr =v \cdot [r(b)-r(a)]$
Work Step by Step
Suppose $r(t)=x(t) i+y(t) j+z(t) k; v=v_1i+v_2j+v_3k$
and $\int_C v \cdot dr =\int_a^b v r'(t) dt=\int_a^b (v_1i+v_2j+v_3k) [x'(t) i+y'(t) j+z'(t) k] dt$
or, $=(v_1) \int_a^b x'(t) dt +(v_2) \int_a^b y'(t) dt+(v_3) \int_a^b z'(t) dt$
or, $=v_1[x(b)-x(a)] +v_2 [y(b)-y(a)]+v_3[z(b)-z(a)]$
Hence, the result has been proved.
$\int_C v \cdot dr =v \cdot [r(b)-r(a)]$
