Answer: To evaluate the line integral ∫CF⋅dr, we first need to parameterize the path C in terms of a single variable, say t. The parameterization of C is given by:
r(t) = (t^3, -3t^2, 2^t), 0 ≤ t ≤ 1
We can then write the differential of the path as:
dr = (3t^2, -6t, 2^t ln(2)) dt
Next, we evaluate the dot product F ⋅ dr along the path C:
F(r(t)) ⋅ dr = ⟨5sin(t^3), -4cos(-3t^2), 10t^3(2^t)⟩ ⋅ ⟨3t^2, -6t, 2^t ln(2)⟩ dt
= 15t^2 sin(t^3) + 24t cos(3t^2) + 20t^3 (2^t) ln(2) dt
Finally, we integrate this expression over the interval 0 ≤ t ≤ 1 to get the value of the line integral:
∫CF ⋅ dr = ∫0^1 (15t^2 sin(t^3) + 24t cos(3t^2) + 20t^3 (2^t) ln(2)) dt
This integral cannot be evaluated exactly, but we can use numerical integration to approximate the value. Using a numerical integration tool, the value of the line integral is approximately equal to 11.108.
Therefore:
∫CF ⋅ dr ≈ 11.108
Step-by-step explanation: