Certainly! In the solution provided, the integral's interval limits for $x$, from 1 to 2, come from the x-coordinates of the given points through which the path passes, specifically from $(1,2)$ to $(2,5)$.

Here's a breakdown of how these limits were chosen:

1. Path Definition: The path $y = mx^2 + c$ is defined to pass through points $(1, 2)$ and $(2, 5)$. By solving for $m$ and $c$ (through a system of equations), they determined that the path is given by $y = x^2 + 1$.

2. Position Vector for the Path: The path is parameterized with a position vector: $\vec{r}(x) = x \hat{i} + (x^2 + 1) \hat{j}.$ Here, $x$ is the variable of integration, and it varies along the x-coordinates of the endpoints of the path.

3. Integral Limits: Since $x$ starts at 1 and ends at 2 as per the given points $(1,2)$ and $(2,5)$, the integral limits are set from $x = 1$ to $x = 2$. Therefore, $x$ is integrated over this interval to find the work done along the entire path from $(1,2)$ to $(2,5)$.

To summarize, the integral limits are derived from the $x$-coordinates of the starting and ending points on the defined path.

Would you like more details on the solution steps, or have any questions about specific parts of the explanation?

Related Questions

1. How are the values of $m$ and $c$ calculated for the given points?
2. Why does the vector $\vec{F}(x, y(x))$ depend on both $x$ and $y(x)$?
3. What role does the differential $d\vec{r} = (\hat{i} + 2x \hat{j}) dx$ play in the integration?
4. How is the vector field $\vec{F}(x, y)$ calculated along the path?
5. Why is the constant $\alpha$ factored out in the work calculation?

Tip

In line integrals, choosing the correct limits depends on the parameterization of the path. Ensure the interval corresponds to the values over which the parameter varies along the path.