a. Parameterize C₁ by
[tex](x(t),y(t)) = (1-t)(-13,-7) + t(0,6) = (-13+13t,-7+13t)[/tex]
with 0 ≤ t ≤ 1. Then both dx = 13 dt and dy = 13 dt, so that the line integral along C₁ is
[tex]\displaystyle \int_C y^2\,dx + x\,dy = \int_0^1 (-7+13t)^2(13\, dt) + (-13+13t)(13\,dt)[/tex]
[tex]\displaystyle \int_C y^2\,dx + x\,dy = 13 \int_0^1 (169t^2 - 169t + 36) \, dt[/tex]
[tex]\displaystyle \int_C y^2\,dx + x\,dy = 13 \left(\frac{169}3-\frac{169}2+36\right) = \boxed{\frac{611}6}[/tex]
b. Parameterize C₂ by
[tex](x(t),y(t)) = (36-t^2,t)[/tex]
with -7 ≤ t ≤ 6. Then dx = -2t dt and dy = dt, so the line integral is
[tex]\displaystyle \int_C y^2\,dx + x\,dy = \int_{-7}^6 t^2(-2t\,dt) + (36-t^2)\,dt[/tex]
[tex]\displaystyle \int_C y^2\,dx + x\,dy = \int_{-7}^6 (36-t^2-2t^3) \, dt[/tex]
[tex]\displaystyle \int_C y^2\,dx + x\,dy = \left(36\cdot6-\frac{6^3}3-\frac{6^4}2\right) - \left(36\cdot(-7)-\frac{(-7)^3}3-\frac{(-7)^4}2\right) = \boxed{\frac{5005}6}[/tex]
