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