1. **State the problem:** We need to evaluate the line integral $$\int \vec{A} \cdot d\vec{r}$$ where $$\vec{A} = (5xy - 6x^2) \hat{i} + (2y - 4x) \hat{j}$$ along the curve $$y = x^3$$ from point $$(1,1)$$ to $$(2,8)$$. 2. **Parameterize the curve:** Since $$y = x^3$$, we can use $$x$$ as the parameter: $$\vec{r}(x) = x \hat{i} + x^3 \hat{j}$$ with $$x$$ going from 1 to 2. 3. **Compute $$d\vec{r}$$:** $$d\vec{r} = \frac{d\vec{r}}{dx} dx = (1 \hat{i} + 3x^2 \hat{j}) dx$$. 4. **Substitute $$y = x^3$$ into $$\vec{A}$$:** $$\vec{A}(x) = (5x(x^3) - 6x^2) \hat{i} + (2(x^3) - 4x) \hat{j} = (5x^4 - 6x^2) \hat{i} + (2x^3 - 4x) \hat{j}$$. 5. **Compute the dot product $$\vec{A} \cdot d\vec{r}$$:** $$\vec{A} \cdot d\vec{r} = (5x^4 - 6x^2)(1) dx + (2x^3 - 4x)(3x^2) dx = (5x^4 - 6x^2) dx + (6x^5 - 12x^3) dx$$ 6. **Combine terms:** $$= (5x^4 - 6x^2 + 6x^5 - 12x^3) dx = (6x^5 + 5x^4 - 12x^3 - 6x^2) dx$$ 7. **Set up the integral:** $$\int_1^2 (6x^5 + 5x^4 - 12x^3 - 6x^2) dx$$ 8. **Integrate term-by-term:** $$\int 6x^5 dx = x^6$$ $$\int 5x^4 dx = x^5$$ $$\int -12x^3 dx = -3x^4$$ $$\int -6x^2 dx = -2x^3$$ So the antiderivative is: $$F(x) = x^6 + x^5 - 3x^4 - 2x^3$$ 9. **Evaluate at the bounds:** $$F(2) = 2^6 + 2^5 - 3(2^4) - 2(2^3) = 64 + 32 - 3(16) - 2(8) = 64 + 32 - 48 - 16 = 32$$ $$F(1) = 1 + 1 - 3 - 2 = -3$$ 10. **Calculate the definite integral:** $$\int_1^2 \vec{A} \cdot d\vec{r} = F(2) - F(1) = 32 - (-3) = 35$$ **Final answer:** $$\boxed{35}$$