Parameterize the path from (0, 0) to (8, 1) by
• x(t) = 8t and y(t) = t
and the path from (8, 1) to (9, 0) by
• x(t) = 8 + t and y(t) = 1 - t
both with 0 ≤ t ≤ 1.
Over the first path, we have
![\displaystyle \int_(C_1) (x+8y) \, dx + x^2 \, dy = \int_0^1 (8t + 8\cdot t) \cdot (8 \, dt) + (8t)^2 \cdot dt = \int_0^1 (128t + 64t^2) \, dt = \frac{256}3](https://img.qammunity.org/2022/formulas/sat/high-school/vshkn8jcp568uqs46mu932umtlamylxzmo.png)
and over the second path,
![\displaystyle \int_(C_2) (x+8y) \, dx + x^2 \, dy = \int_0^1 ((8 + t) + 8(1-t)) \cdot dt + (8+t)^2 \cdot (-dt) \\\\ = \int_0^1 (-48 - 23t - t^2) \, dt = -\frac{359}6](https://img.qammunity.org/2022/formulas/sat/high-school/8jpfarir4loto2efo6k3vsv0s5131jh67z.png)
and so the total line integral is
![\displaystyle \int_C (x+8y) \, dx + x^2 \, dy = \frac{256}3 - \frac{359}6 = \boxed{\frac{51}2}](https://img.qammunity.org/2022/formulas/sat/high-school/xjh0i5g8d8apc7tvqoxyax9oya5b9zgavl.png)