Evaluate Integral (3x+y)dx + (2y-x)dy along the curve y=x^2+1 from (0,1) to (3,0).

To evaluate the line integral of the given differential form \( (3x + y)dx + (2y - x)dy \) along the curve \( y = x^2 + 1 \) from the point \( (0, 1) \) to \( (3, 0) \), we can proceed as follows:

### Step 1: Parameterize the curve
The curve is given by \( y = x^2 + 1 \), so we can directly use \( x \) as the parameter. Therefore, the parametric equations for the curve are:

\[
x = t, \quad y = t^2 + 1
\]
where \( t \) goes from 0 to 3 (since \( x \) starts at 0 and ends at 3).

Now, compute the differentials:
\[
dx = dt, \quad dy = \frac{d}{dt}(t^2 + 1) = 2t \, dt
\]

### Step 2: Substitute into the integral
Now, substitute \( y = t^2 + 1 \), \( dx = dt \), and \( dy = 2t \, dt \) into the line integral:

\[
\int_C (3x + y)dx + (2y - x)dy
\]

Substitute \( x = t \) and \( y = t^2 + 1 \):

\[
= \int_0^3 \left(3t + (t^2 + 1)\right)dt + \int_0^3 \left(2(t^2 + 1) - t\right)2t \, dt
\]

Simplifying each part:

\[
= \int_0^3 (3t + t^2 + 1)dt + \int_0^3 \left(2(t^2 + 1) - t\right)2t \, dt
\]

### Step 3: Evaluate each part
Let's compute the two integrals separately.

#### First Integral:
\[
\int_0^3 (3t + t^2 + 1) dt = \int_0^3 3t \, dt + \int_0^3 t^2 \, dt + \int_0^3 1 \, dt
\]

\[
= \left[\frac{3t^2}{2}\right]_0^3 + \left[\frac{t^3}{3}\right]_0^3 + \left[t\right]_0^3
\]

\[
= \frac{3(9)}{2} + \frac{27}{3} + 3 = 13.5 + 9 + 3 = 25.5
\]

#### Second Integral:
\[
\int_0^3 \left(2(t^2 + 1) - t\right)2t \, dt = \int_0^3 \left(2t^3 + 2t - 2t^2\right) dt
\]

\[
= \int_0^3 2t^3 \, dt + \int_0^3 2t \, dt - \int_0^3 2t^2 \, dt
\]

Evaluating each part:

\[
\int_0^3 2t^3 \, dt = \left[\frac{2t^4}{4}\right]_0^3 = \frac{2(81)}{4} = 40.5
\]

\[
\int_0^3 2t \, dt = \left[t^2\right]_0^3 = 9
\]

\[
\int_0^3 2t^2 \, dt = \left[\frac{2t^3}{3}\right]_0^3 = \frac{2(27)}{3} = 18
\]

So the second integral is:

\[
40.5 + 9 - 18 = 31.5
\]

### Step 4: Add the results
The total value of the line integral is:

\[
25.5 + 31.5 = 57
\]

Thus, the value of the line integral is \( \boxed{57} \).

Do you have any questions or would you like more details on a specific part?

Here are five related questions:
1. How would the integral change if the curve was different?
2. Can you explain how to parameterize curves for other line integrals?
3. What if the differential form was more complex? How would you handle it?
4. What role do the limits of integration play in line integrals?
5. Can this method be applied to vector fields?

**Tip:** When dealing with line integrals, always ensure the parameterization is consistent with the curve's direction and limits of integration.

Evaluate the line integral (3x+y)dx + (2y-x)dy along the curve y=x^2+1 from (0,1) to (3,0). This problem covers key calculus concepts such as parameterization and line integrals, with a step-by-step solution provided. Suitable for undergraduate students.

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