The problem involves calculating the line integral of a vector field over a curve $C$ given by the parametric equations:

• $x = t^2$
• $y = 3t$
• $0 \leq t \leq 1$

The integral to compute is:

$\int_C (2x \, dx + (x - y) \, dy)$

We will solve this step by step.

Step 1: Express $dx$ and $dy$ in terms of $dt$

Given the parametric equations:

• $x = t^2$
• $y = 3t$

The differentials $dx$ and $dy$ are:

$dx = \frac{d}{dt}(t^2) \, dt = 2t \, dt$ $dy = \frac{d}{dt}(3t) \, dt = 3 \, dt$

Step 2: Substitute into the integral

Now substitute $x = t^2$, $y = 3t$, $dx = 2t \, dt$, and $dy = 3 \, dt$ into the original integral:

$\int_0^1 \left( 2(t^2) \cdot 2t \, dt + \left( t^2 - 3t \right) \cdot 3 \, dt \right)$

This simplifies to:

$\int_0^1 \left( 4t^3 + 3(t^2 - 3t) \right) \, dt$

Simplify further:

$\int_0^1 \left( 4t^3 + 3t^2 - 9t \right) \, dt$

Step 3: Integrate term by term

Now, perform the integration:

$\int_0^1 4t^3 \, dt = \left[ t^4 \right]_0^1 = 1^4 - 0^4 = 1$ $\int_0^1 3t^2 \, dt = \left[ t^3 \right]_0^1 = 1^3 - 0^3 = 1$ $\int_0^1 -9t \, dt = \left[ -\frac{9}{2}t^2 \right]_0^1 = -\frac{9}{2}(1^2 - 0^2) = -\frac{9}{2}$

Step 4: Combine the results

Now, summing up all the terms:

$1 + 1 - \frac{9}{2} = 2 - \frac{9}{2} = \frac{4}{2} - \frac{9}{2} = -\frac{5}{2} = -2.5$

Conclusion:

The value of the line integral is $-2.5$, so the correct answer is:

$\boxed{-2.5}$

Let me know if you need further details or have any other questions!

Here are some related questions you might find interesting:

1. How would the result change if the curve was parametrized differently?
2. What would the result be if the limits of integration for $t$ were different?
3. How can we compute a line integral for a curve in three dimensions?
4. What would the integral look like if we had an additional term in the vector field, like $5x^2y$?
5. Can you compute a similar integral for a different type of vector field?

Tip: Always carefully check the differential expressions when transforming from parametric equations to the line integral!