Antiderivatives with initial conditions

Problem 3.578 · hard

1. \[ \frac{d}{d x} \left(4 x + 3 e^{x}\right) = 3 e^{x} + 4 \]
   An antiderivative of f″; f′ is this plus a constant C₁.✓ Proved
2. \[ - \left. 4 x + 3 e^{x} \right|_{\substack{ x=1 }} - 4 = - 3 e - 8 \]
   f′(1) = -4 fixes C₁.✓ Proved
3. \[ \frac{d}{d x} \left(2 x^{2} + x \left(- 3 e - 8\right) + 3 e^{x}\right) = 4 x + 3 e^{x} - 3 e - 8 \]
   Antidifferentiate f′; add a constant C₀.✓ Proved
4. \[ - \left. 2 x^{2} + x \left(- 3 e - 8\right) + 3 e^{x} \right|_{\substack{ x=1 }} - 5 = 1 \]
   f(1) = -5 fixes C₀.✓ Proved
5. \[ 2 x^{2} + x \left(- 3 e - 8\right) + 3 e^{x} + 1 = 2 x^{2} - 3 e x - 8 x + 3 e^{x} + 1 \]
   The solution.✓ Proved