To evaluate the integral of $(x^2 + ixy)\, dz$ along the curve $x = t, y = t^3$ with the limits $t = 1$ to $t = 2$, we need to parametrize the curve and compute the corresponding integral in terms of $t$.

Step 1: Parametrize the curve

Given the curve is parametrized by:

• $x = t$
• $y = t^3$

The variable $z$ will follow the path in the complex plane as: $dz = \frac{d}{dt}(x + iy)\, dt = \frac{d}{dt}(t + it^3)\, dt = (1 + 3it^2)\, dt$

Step 2: Substitute into the integral

The integral becomes: $\int_{C} (x^2 + ixy)\, dz$ Substitute $x = t$ and $y = t^3$: $x^2 + ixy = t^2 + it(t^3) = t^2 + it^4$ Also, $dz = (1 + 3it^2)\, dt$, so the integral is: $\int_{1}^{2} (t^2 + it^4)(1 + 3it^2)\, dt$

Step 3: Expand the integrand

Now expand $(t^2 + it^4)(1 + 3it^2)$: $(t^2 + it^4)(1 + 3it^2) = t^2(1) + t^2(3it^2) + it^4(1) + it^4(3it^2)$ Simplifying each term: $= t^2 + 3it^4 + it^4 - 3t^6$ $= t^2 - 3t^6 + i(3t^4 + t^4) = t^2 - 3t^6 + i4t^4$

Step 4: Integrate

The integral becomes: $\int_{1}^{2} (t^2 - 3t^6 + i4t^4)\, dt$ Now, break this into two integrals: $\int_{1}^{2} (t^2 - 3t^6)\, dt + i\int_{1}^{2} 4t^4\, dt$

First integral: $\int_{1}^{2} (t^2 - 3t^6)\, dt$

$\int_{1}^{2} t^2\, dt = \left[\frac{t^3}{3}\right]_1^2 = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}$ $\int_{1}^{2} t^6\, dt = \left[\frac{t^7}{7}\right]_1^2 = \frac{128}{7} - \frac{1}{7} = \frac{127}{7}$ So, $\int_{1}^{2} (t^2 - 3t^6)\, dt = \frac{7}{3} - 3 \cdot \frac{127}{7} = \frac{7}{3} - \frac{381}{7}$

To subtract, find a common denominator: $\frac{7}{3} = \frac{49}{21}, \quad \frac{381}{7} = \frac{1143}{21}$ $\frac{49}{21} - \frac{1143}{21} = \frac{-1094}{21}$

Second integral: $i \int_{1}^{2} 4t^4\, dt$

$\int_{1}^{2} 4t^4\, dt = 4 \cdot \left[\frac{t^5}{5}\right]_1^2 = 4 \cdot \left(\frac{32}{5} - \frac{1}{5}\right) = 4 \cdot \frac{31}{5} = \frac{124}{5}$ Thus, the second part is: $i \cdot \frac{124}{5}$

Step 5: Combine the results

The full result is: $\frac{-1094}{21} + i \cdot \frac{124}{5}$

Final Answer:

$\int_C (x^2 + ixy)\, dz = \frac{-1094}{21} + i \cdot \frac{124}{5}$

Let me know if you need further clarification or details about any of the steps!