So as y depends on x we consider x-values in points huh

Exactly. Since \( y \) is given as a function of \( x \) (\( y = x^3 \)), we consider the \( x \)-values of the points to determine the limits of integration.

Here is a detailed breakdown:

1. **Curve Equation**: \( y = x^3 \).
2. **Starting Point**: \((1, 1)\)
   - At this point, \( x = 1 \) and \( y = 1^3 = 1 \).
3. **Ending Point**: \((2, 8)\)
   - At this point, \( x = 2 \) and \( y = 2^3 = 8 \).

Since \( y \) is expressed in terms of \( x \) and our points of interest (\( (1, 1) \) to \( (2, 8) \)) are given by these coordinates, the integral is taken with respect to \( x \) from 1 to 2.

### Integral Calculation:

1. **Given Vector Field**:
   \[
   \mathbf{F} = (5xy - 6x^2)\mathbf{i} + (2y - 4z)\mathbf{j}
   \]
   Since \( z = 0 \) in the \( xy \)-plane, the vector field simplifies to:
   \[
   \mathbf{F} = (5xy - 6x^2)\mathbf{i} + 2y\mathbf{j}
   \]

2. **Curve \( c \)**:
   \[
   y = x^3 \implies dy = 3x^2 dx
   \]

3. **Differential Line Element**:
   \[
   d\mathbf{r} = dx \mathbf{i} + dy \mathbf{j}
   \]
   Substituting \( dy = 3x^2 dx \):
   \[
   d\mathbf{r} = dx \mathbf{i} + 3x^2 dx \mathbf{j}
   \]

4. **Dot Product \( \mathbf{F} \cdot d\mathbf{r} \)**:
   \[
   \mathbf{F} \cdot d\mathbf{r} = (5xy - 6x^2)dx + 2y (3x^2 dx)
   \]
   Substitute \( y = x^3 \):
   \[
   \mathbf{F} \cdot d\mathbf{r} = (5x(x^3) - 6x^2)dx + 2(x^3)(3x^2 dx)
   \]
   Simplify:
   \[
   \mathbf{F} \cdot d\mathbf{r} = (5x^4 - 6x^2)dx + 6x^5 dx = (5x^4 - 6x^2 + 6x^5) dx
   \]

5. **Integral**:
   \[
   \int_{(1,1)}^{(2,8)} \mathbf{F} \cdot d\mathbf{r} = \int_{1}^{2} (5x^4 - 6x^2 + 6x^5) dx
   \]

   Now, integrate term-by-term:
   \[
   \int_{1}^{2} (5x^4 - 6x^2 + 6x^5) dx = \left[ \frac{5x^5}{5} - \frac{6x^3}{3} + \frac{6x^6}{6} \right]_{1}^{2}
   \]
   Simplify:
   \[
   = \left[ x^5 - 2x^3 + x^6 \right]_{1}^{2}
   \]

6. **Evaluate at Limits**:
   \[
   \left[ 2^5 - 2 \cdot 2^3 + 2^6 \right] - \left[ 1^5 - 2 \cdot 1^3 + 1^6 \right]
   = \left[ 32 - 16 + 64 \right] - \left[ 1 - 2 + 1 \right]
   = 80 - 0
   = 80
   \]

Thus, the value of the line integral is \(\int_c \mathbf{F} \cdot d\mathbf{r} = 80\).

Do you need further details or have any questions?

Here are some questions you might want to ask:
1. How is the vector field \(\mathbf{F}\) expressed in terms of \( x \) and \( y \)?
2. How do we convert the differential line element \( d\mathbf{r} \) in terms of \( dx \) and \( dy \)?
3. What are the steps to integrate the given functions over the specified limits?
4. How do we verify the result of the integral calculation?
5. Can you explain the concept of a line integral in the context of vector fields?

**Tip:** When performing line integrals, ensure that the limits of integration correspond to the parameter used to describe the path. In this case, \( x \) varies from 1 to 2 along the curve \( y = x^3 \).

Learn how to calculate the line integral of a vector field over the curve y = x^3. This detailed solution includes parameterizing the curve, computing the dot product of the vector field with the differential line element, and integrating over specified limits. Suitable for advanced undergraduate students and anyone studying vector calculus.

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