1. **State the problem:** We need to find the definite integral of the vector function $$\vec{E}(u) = 9u^2 - 7u\mathbf{i} + 5u\mathbf{j} - 4\mathbf{k}$$ from $$u = -1$$ to $$u = 4$$. 2. **Recall the integral of a vector function:** The integral of a vector function is the vector of the integrals of its components: $$\int \vec{E}(u) \, du = \int (9u^2) \, du \mathbf{i} + \int (-7u) \, du \mathbf{j} + \int (5u) \, du \mathbf{k} + \int (-4) \, du \mathbf{k}$$ 3. **Integrate each component:** - $$\int 9u^2 \, du = 9 \cdot \frac{u^3}{3} = 3u^3$$ - $$\int (-7u) \, du = -7 \cdot \frac{u^2}{2} = -\frac{7}{2}u^2$$ - $$\int 5u \, du = 5 \cdot \frac{u^2}{2} = \frac{5}{2}u^2$$ - $$\int (-4) \, du = -4u$$ 4. **Write the antiderivative vector function:** $$\vec{F}(u) = 3u^3 \mathbf{i} - \frac{7}{2}u^2 \mathbf{j} + \left( \frac{5}{2}u^2 - 4u \right) \mathbf{k} + \vec{C}$$ 5. **Evaluate the definite integral:** $$\int_{-1}^{4} \vec{E}(u) \, du = \vec{F}(4) - \vec{F}(-1)$$ Calculate each component: - For $$\mathbf{i}$$: $$3(4)^3 - 3(-1)^3 = 3(64) - 3(-1) = 192 + 3 = 195$$ - For $$\mathbf{j}$$: $$-\frac{7}{2}(4)^2 + \frac{7}{2}(-1)^2 = -\frac{7}{2}(16) + \frac{7}{2}(1) = -56 + 3.5 = -52.5$$ - For $$\mathbf{k}$$: $$\left( \frac{5}{2}(4)^2 - 4(4) \right) - \left( \frac{5}{2}(-1)^2 - 4(-1) \right) = \left( \frac{5}{2} \cdot 16 - 16 \right) - \left( \frac{5}{2} \cdot 1 + 4 \right) = (40 - 16) - (2.5 + 4) = 24 - 6.5 = 17.5$$ 6. **Final answer:** $$\int_{-1}^{4} \vec{E}(u) \, du = 195 \mathbf{i} - 52.5 \mathbf{j} + 17.5 \mathbf{k}$$