1. **State the problem:** We need to evaluate the line integral $$\int_C 2x \, ds$$ where the curve $C$ consists of two parts: - $C_1$: the parabola $y = x^2$ from $(0,0)$ to $(1,1)$ - $C_2$: the vertical line segment from $(1,1)$ to $(1,2)$ 2. **Recall the formula for line integrals with respect to arc length:** $$\int_C f(x,y) \, ds = \int_a^b f(x(t), y(t)) \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$$ where $t$ parametrizes the curve. 3. **Evaluate the integral over $C_1$: parabola $y = x^2$ from $x=0$ to $x=1$.** - Parametrize by $x = t$, $y = t^2$, with $t$ from 0 to 1. - Compute derivatives: $$\frac{dx}{dt} = 1, \quad \frac{dy}{dt} = 2t$$ - Compute the differential arc length: $$ds = \sqrt{1^2 + (2t)^2} \, dt = \sqrt{1 + 4t^2} \, dt$$ - The integrand is $2x = 2t$. - So the integral over $C_1$ is: $$\int_0^1 2t \sqrt{1 + 4t^2} \, dt$$ 4. **Evaluate the integral over $C_2$: vertical line from $(1,1)$ to $(1,2)$.** - Parametrize by $x=1$, $y = t$, with $t$ from 1 to 2. - Compute derivatives: $$\frac{dx}{dt} = 0, \quad \frac{dy}{dt} = 1$$ - Differential arc length: $$ds = \sqrt{0^2 + 1^2} \, dt = 1 \, dt$$ - The integrand is $2x = 2 \times 1 = 2$. - So the integral over $C_2$ is: $$\int_1^2 2 \, dt = 2(t) \Big|_1^2 = 2(2 - 1) = 2$$ 5. **Calculate the integral over $C_1$ using substitution:** - Let $u = 1 + 4t^2$, then $du = 8t \, dt$. - So $t \, dt = \frac{du}{8}$. - Rewrite the integral: $$\int_0^1 2t \sqrt{1 + 4t^2} \, dt = \int_{u=1}^{u=5} 2t \sqrt{u} \, dt = \int_1^5 2t u^{1/2} \, dt$$ - Substitute $t \, dt = \frac{du}{8}$: $$= \int_1^5 2 \times u^{1/2} \times \frac{du}{8} = \int_1^5 \frac{u^{1/2}}{4} \, du = \frac{1}{4} \int_1^5 u^{1/2} \, du$$ - Integrate: $$\frac{1}{4} \times \frac{2}{3} u^{3/2} \Big|_1^5 = \frac{1}{6} (5^{3/2} - 1)$$ 6. **Sum the integrals over $C_1$ and $C_2$ to get the total integral:** $$\int_C 2x \, ds = \frac{1}{6} (5^{3/2} - 1) + 2$$ **Final answer:** $$\boxed{\int_C 2x \, ds = \frac{1}{6} (5^{3/2} - 1) + 2}$$