Answer
$\int_C r \cdot dr =\dfrac{1}{2}[|r(b)|^2-|r(a)|^2]$
Work Step by Step
Here, we have $\int_C r \cdot dr =\int_a^b r r'(t) dt$
Let us consider $r(t)=x(t) i+y(t) j+z(t) k$
Now, we have $\int_C r \cdot dr =\int_a^b (x(t) i+y(t) j+z(t) k) \cdot [x'(t) i+y'(t) j+z'(t) k] dt$
$= \int_a^b [x(t) x'(t) dt +y(t) y'(t) +z(t) z'(t)] dt$
$= \int_a^b \dfrac{1}{2} \dfrac{d}{dT} [x^2(t) +y^2(t) +z^2(t) ] dt$
$=(\dfrac{1}{2}) [x^2(t) +y^2(t) +z^2(t) ]_a^b $
$=\dfrac{1}{2}[|r(b)|^2-|r(a)|^2]$
