Step 1: Problem Overview:
This problem involves calculating a complex line integral along a specified path. The standard method involves parameterizing the path to convert the complex integral into a real variable integral.
Step 2: Key Steps:
1. Path Parameterization: The path is \( y = x^2 \). Using \(x\) as the parameter, the path starts at \(z=0\) (i.e., \(x=0, y=0\)) and ends at \(z=1+i\) (i.e., \(x=1, y=1\)). Therefore, \(x\) ranges from 0 to 1.
2. Expressing everything in terms of \(x\):
- \( z(x) = x + iy = x + ix^2 \)
- \( dz = \frac{dz}{dx} dx = (1 + 2ix) dx \)
- The integrand: \( f(z) = x^2 - iy = x^2 - ix^2 = x^2(1-i) \)
3. Substitution and Integration Setup:
\[ \int_C f(z) dz = \int_{0}^{1} x^2(1-i) \cdot (1+2ix) dx \]
Step 3: Detailed Solution:
The integral is evaluated as follows:
\[ \int_{0}^{1} x^2(1-i)(1+2ix) dx = (1-i) \int_{0}^{1} x^2(1+2ix) dx \]
\[ = (1-i) \int_{0}^{1} (x^2 + 2ix^3) dx \]
Integrating term by term:
\[ = (1-i) \left[ \frac{x^3}{3} + 2i\frac{x^4}{4} \right]_{0}^{1} = (1-i) \left[ \frac{x^3}{3} + i\frac{x^4}{2} \right]_{0}^{1} \]
\[ = (1-i) \left( \left(\frac{1}{3} + i\frac{1}{2}\right) - (0) \right) \]
Multiplying the complex numbers:
\[ = (1-i) \left(\frac{1}{3} + \frac{i}{2}\right) = 1\left(\frac{1}{3} + \frac{i}{2}\right) - i\left(\frac{1}{3} + \frac{i}{2}\right) \]
\[ = \frac{1}{3} + \frac{i}{2} - \frac{i}{3} - \frac{i^2}{2} = \frac{1}{3} + \frac{1}{2} + i\left(\frac{1}{2} - \frac{1}{3}\right) \]
\[ = \frac{2+3}{6} + i\left(\frac{3-2}{6}\right) = \frac{5}{6} + \frac{1}{6}i \]
Step 4: Final Answer:
The integral's value is \( \frac{5}{6} + \frac{1}{6}i \), corresponding to option (B).
(Note: If the path were \(y=x\), the answer would be \( \frac{5}{6} - \frac{1}{6}i \), which is option A. A possible question typo.)