Answer
$\dfrac{-1}{6}$
Work Step by Step
Here, we have
$\int_C f(x,y,z) ds=\int_{C_1} f(x,y,z) ds+\int_{C_2} f(x,y,z) ds+\int_{C_1} f(x,y,z) ds$
$\int_C f(x,y,z) ds=\int_{0}^{1} -t^2 dt+\int_{0}^{1} -t^{1/2}-1 dt+\int_{0}^{1} (t) dt$
This implies that
$[\dfrac{-t^3}{3}]_0^1+[\dfrac{2}{3}t^{3/2}-t]_0^1+[\dfrac{t^2}{2}]_0^1=-\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{-1}{6}$
