1. **State the problem:** We are given a vector field \(\mathbf{E} = -\nabla f(x,y,z)\) where \(f(x,y,z) = 2x^4 y + 3xyz\). We need to find the line integral \(\int_{n_1}^{n_2} \mathbf{E} \cdot \mathbf{t} \, dl\) from \(n_1 = (0,0,0)\) to \(n_2 = (2,1.5,3)\). 2. **Recall the hint and relevant theorem:** Since \(\mathbf{E} = -\nabla f\), \(\mathbf{E}\) is a conservative vector field. The line integral of a conservative field along a path depends only on the endpoints and equals the difference in the potential function values: $$\int_{n_1}^{n_2} \mathbf{E} \cdot d\mathbf{l} = f(n_1) - f(n_2)$$ 3. **Evaluate \(f\) at the points:** - At \(n_1 = (0,0,0)\): $$f(0,0,0) = 2 \cdot 0^4 \cdot 0 + 3 \cdot 0 \cdot 0 \cdot 0 = 0$$ - At \(n_2 = (2,1.5,3)\): $$f(2,1.5,3) = 2 \cdot (2)^4 \cdot 1.5 + 3 \cdot 2 \cdot 1.5 \cdot 3$$ Calculate each term: $$2^4 = 16$$ $$2 \cdot 16 \cdot 1.5 = 48$$ $$3 \cdot 2 \cdot 1.5 \cdot 3 = 27$$ Sum: $$48 + 27 = 75$$ 4. **Calculate the integral:** $$\int_{n_1}^{n_2} \mathbf{E} \cdot d\mathbf{l} = f(0,0,0) - f(2,1.5,3) = 0 - 75 = -75$$ 5. **Final answer:** The value of the integral is \(-75.00\) (two decimals, including sign). This result is scalar and in SI units as requested.