Chapter 16 - Vector Calculus - 16.2 Exercises - Page 1096: 7

$\dfrac{5}{2}$

$I=\int_{C_1}(x+2y) dx+x^2 dy+\int_{C_2}(x+2y) dx+x^2 dy=\int_{0}^1 [2t+2(t)](2 dt)+(2t)^2 \times (1 dt)+\int_{0}^1 [(2+t)+2(1-t) \times 1 dt +(2+t)^2(-1 dt)$ $=\int_{0}^1 8(t)+4(t^2) dt+\int_0^1 (4-t) dt-(4+4t+t^2) dt$ $I=4\cdot t^2+[\dfrac{4}{3}]_0^1-[\dfrac{t^3}{3}]_0^1-\dfrac{5t^2}{2}=\dfrac{16}{3}-\dfrac{17}{6}$ Hence, $I=\dfrac{5}{2}$