Compute ∫ 𝑧̅⁡𝑑𝑧
𝐵
𝐴
along the semi-cubical parabolic path 𝐶: 𝑦 = 2𝑥3 from 𝐴(0,0) to 𝐵(1,2) by
representing the path parametrically.

To solve the line integral \(\int_C \bar{z} \, dz\) along the semi-cubical parabolic path \(C: y = 2x^3\) from \(A(0,0)\) to \(B(1,2)\), we will approach the problem by parametrizing the path, representing \(z\) and \(dz\) accordingly, and then computing the integral.

### Step 1: Parametrize the Path
Given that the path \(C\) is defined by the equation \(y = 2x^3\), we can parametrize it using \(x = t\). Consequently, the parameterization will be:
\[
x = t, \quad y = 2t^3.
\]
Therefore, the complex number \(z = x + iy\) becomes:
\[
z(t) = t + i(2t^3).
\]

The limits of the parameter \(t\) will range from \(0\) to \(1\), corresponding to points \(A(0,0)\) and \(B(1,2)\).

### Step 2: Compute \(dz\)
Differentiating \(z(t)\) with respect to \(t\) gives:
\[
dz = \frac{d}{dt}(t + i(2t^3)) \, dt = (1 + i6t^2) \, dt.
\]

### Step 3: Express \(\bar{z}\)
The complex conjugate of \(z(t)\) is:
\[
\bar{z}(t) = t - i(2t^3).
\]

### Step 4: Set Up the Integral
Now, the line integral can be written as:
\[
\int_C \bar{z} \, dz = \int_0^1 \bar{z}(t) \cdot dz(t).
\]
Substituting the expressions for \(\bar{z}(t)\) and \(dz(t)\), we get:
\[
\int_0^1 (t - i2t^3) \cdot (1 + i6t^2) \, dt.
\]

### Step 5: Expand the Product
Expanding the product inside the integral:
\[
(t - i2t^3)(1 + i6t^2) = t(1 + i6t^2) - i2t^3(1 + i6t^2).
\]
Simplifying further:
\[
= t + i6t^3 - i2t^3 - 12t^5.
\]
Thus, the integrand becomes:
\[
t + (i6t^3 - i2t^3) - 12t^5 = t + i4t^3 - 12t^5.
\]

### Step 6: Split the Integral
We now split the integral into its real and imaginary parts:
\[
\int_0^1 \left( t + i4t^3 - 12t^5 \right) dt = \int_0^1 t \, dt + i \int_0^1 4t^3 \, dt - \int_0^1 12t^5 \, dt.
\]

### Step 7: Compute the Integrals
1. \(\int_0^1 t \, dt = \left[ \frac{t^2}{2} \right]_0^1 = \frac{1}{2}\),
2. \(\int_0^1 4t^3 \, dt = 4 \left[ \frac{t^4}{4} \right]_0^1 = 1\),
3. \(\int_0^1 12t^5 \, dt = 12 \left[ \frac{t^6}{6} \right]_0^1 = 2\).

### Step 8: Combine the Results
Thus, the integral becomes:
\[
\frac{1}{2} + i(1) - 2 = \left( \frac{1}{2} - 2 \right) + i(1) = -\frac{3}{2} + i.
\]

### Final Answer:
The value of the line integral is:
\[
\boxed{-\frac{3}{2} + i}.
\]

Do you want more details, or do you have any questions?

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**Further questions:**
1. How do we interpret the result of a line integral in the complex plane?
2. What is the geometric significance of this integral?
3. How does parametrization simplify the calculation of line integrals?
4. Can this method be extended to any parametric curve in the complex plane?
5. What are the real-world applications of complex line integrals?

**Tip:** Always check if the path can be parametrized smoothly, as this simplifies the computation of line integrals.

Learn how to compute the line integral ∫ 𝑧̅⁡𝑑𝑧 along the semi-cubical parabolic path 𝐶: 𝑦 = 2𝑥^3 from point 𝐴(0,0) to point 𝐵(1,2) by parametrizing the path. This problem involves complex analysis, line integrals, and parametric equations, suitable for undergraduate students.

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