Find the line integral of 𝑓(𝑥, 𝑦) = 𝑦𝑒 # !along the curve r(t) = 4t i – 3t j, -1 ≤ t ≤ 2

To solve this problem, we need to follow these steps:

1. First, we need to parameterize the function f(x, y) in terms of t using the given curve r(t) = 4t i – 3t j. From r(t), we can see that x = 4t and y = -3t. So, f(x, y) becomes f(t) = -3t * e.

2. Next, we need to find the derivative of the vector function r(t). The derivative dr/dt = 4 i - 3 j.

3. Then, we need to find the magnitude of the derivative, which is ||dr/dt|| = sqrt((4)^2 + (-3)^2) = 5.

4. Now, we can compute the line integral from t = -1 to t = 2. The line integral ∫C f(r(t)) ||dr/dt|| dt becomes ∫ from -1 to 2 of (-3t * e) * 5 dt.

5. Finally, we need to solve this integral. The integral of -15t * e dt from -1 to 2 is [-15/2 * e * t^2] from -1 to 2 = -15/2 * e * 4 - (-15/2 * e * 1) = -30e + 7.5e = -22.5e.

So, the line integral of f(x, y) = y * e along the curve r(t) = 4t i – 3t j from t = -1 to t = 2 is -22.5e.

To compute the line integral of a vector field along a curve, we need to parameterize the curve and the function.

1. First, let's parameterize the curve r(t) = 4t i – 3t j, -1 ≤ t ≤ 2. This gives us x = 4t and y = -3t.

2. Next, we substitute these parameterizations into the function f(x, y) = ye^x. This gives us f(t) = (-3t)e^(4t).

3. Now, we compute the line integral ∫f(t) ds from t = -1 to t = 2.

4. But first, we need to find ds. ds is the magnitude of the derivative of the position vector r(t). So, we first find dr/dt = 4 i - 3 j. The magnitude of this is sqrt((4)^2 + (-3)^2) = 5. So, ds = 5 dt.

5. Now, we compute the line integral ∫f(t) ds from t = -1 to t = 2. This is ∫((-3t)e^(4t)) * 5 dt from t = -1 to t = 2.

6. This integral can be computed using integration by parts, with u = -3t, dv = e^(4t) dt. Then du = -3 dt, v = 1/4 e^(4t).

7. The formula for integration by parts is ∫udv = uv - ∫vdu. Applying this gives (-3t)(1/4)e^(4t) - ∫(1/4)e^(4t) * -3 dt from t = -1 to t = 2.

8. Simplifying this gives (-3/4)t e^(4t) + 3/4 ∫e^(4t) dt from t = -1 to t = 2.

9. The integral ∫e^(4t) dt = 1/4 e^(4t). So, the line integral is (-3/4)t e^(4t) + 3/16 e^(4t) from t = -1 to t = 2.

10. Finally, we substitute the limits of integration into this expression to get the final answer.

Please note that the integral computation might require some knowledge on integration techniques such as integration by parts.