Answer
$\int_C r \cdot dr =\dfrac{1}{2}[|r(b)|^2-|r(a)|^2]$
Work Step by Step
Suppose, we have $r(t)=x(t) i+y(t) j+z(t) k$
and $\int_C r \cdot dr =\int_a^b (r) [r'(t)] dt=\int_a^b (x(t) i+y(t) j+z(t) k) \cdot [x'(t) i+y'(t) j+z'(t) k] dt$
$= \int_a^b [x(t) x'(t) dt +y(t) y'(t) +z(t) z'(t)] dt$
$= (\dfrac{1}{2}) \int_a^b \dfrac{d}{dt} [x^2(t) +y^2(t) +z^2(t) ] dt$
$=(\dfrac{1}{2}) [x^2(t) +y^2(t) +z^2(t) ]_a^b $
Hence, it has been shown that
$\int_C r \cdot dr =\dfrac{1}{2}[|r(b)|^2-|r(a)|^2]$
