1. We are asked to evaluate the definite integral $$\int_0^4 (2e^x + 3 \sin x) \, dx$$. 2. The integral of a sum is the sum of the integrals, so we split it: $$\int_0^4 2e^x \, dx + \int_0^4 3 \sin x \, dx$$ 3. Use the formulas: - $$\int e^x \, dx = e^x + C$$ - $$\int \sin x \, dx = -\cos x + C$$ 4. Compute each integral: $$\int_0^4 2e^x \, dx = 2 \int_0^4 e^x \, dx = 2 [e^x]_0^4 = 2(e^4 - e^0) = 2(e^4 - 1)$$ $$\int_0^4 3 \sin x \, dx = 3 \int_0^4 \sin x \, dx = 3 [-\cos x]_0^4 = 3(-\cos 4 + \cos 0) = 3(1 - \cos 4)$$ 5. Add the results: $$2(e^4 - 1) + 3(1 - \cos 4) = 2e^4 - 2 + 3 - 3\cos 4 = 2e^4 + 1 - 3\cos 4$$ 6. Final answer: $$\boxed{2e^4 + 1 - 3\cos 4}$$