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Find a function f such that ∇f = 3x²ⁱ - 3y²ʲ. f = ]+K Use f to evaluate: ∫_C (3x²ⁱ - 3y²ʲ).

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Final answer:

The function f such that ∇f = 3x²ᵃ - 3y²ᵇ is f(x, y) = x³ - y³ + K. The line integral of the gradient of f over a curve C is equivalent to evaluating f at the endpoints of C.

Step-by-step explanation:

The student is asking for a function f(x, y) whose gradient is ∇f = 3x²ᵃ - 3y²ᵇ. To find this function, we integrate the partial derivatives. By integrating 3x² with respect to x, we obtain x³, and by integrating -3y² with respect to y, we get -y³. Therefore, a potential function for f is f(x, y) = x³ - y³ + K, where K is the constant of integration.

Using the potential function, we can evaluate the line integral ∫_C (3x²ᵃ - 3y²ᵇ) over a curve C, which by the gradient theorem (also known as the fundamental theorem for line integrals), reduces to evaluating the function f at the endpoints of C.

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