# Chapter 16 - Vector Calculus - 16.2 Line Integrals - 16.2 Exercises - Page 1126: 50

$\int_C r \cdot dr =\dfrac{1}{2}[|r(b)|^2-|r(a)|^2]$

#### 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 $\int_C r \cdot dr =\int_a^b (x(t) i+y(t) j+z(t) k) \cdot [x'(t) i+y'(t) j+z'(t) k] dt$ $= \int_a^b [x(t) x'(t) dt +y(t) y'(t) +z(t) z'(t)] dt$ $= \int_a^b \dfrac{1}{2} \dfrac{d}{dT} [x^2(t) +y^2(t) +z^2(t) ] dt$ $=(\dfrac{1}{2}) [x^2(t) +y^2(t) +z^2(t) ]_a^b$ $=\dfrac{1}{2}[|r(b)|^2-|r(a)|^2]$

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