Antiderivatives with initial conditions

Problem 3.477 · hard

1. \[ \frac{d}{d x} \left(\frac{2 x^{\frac{3}{2}}}{3} + x^{4}\right) = \sqrt{x} + 4 x^{3} \]
   An antiderivative of f″; f′ is this plus a constant C₁.✓ Proved
2. \[ - \left. x^{4} + \frac{2 x^{\frac{3}{2}}}{3} \right|_{\substack{ x=1 }} - 2 = - \frac{11}{3} \]
   f′(1) = -2 fixes C₁.✓ Proved
3. \[ \frac{d}{d x} \left(\frac{4 x^{\frac{5}{2}}}{15} + \frac{x^{5}}{5} - \frac{11 x}{3}\right) = \frac{2 x^{\frac{3}{2}}}{3} + x^{4} - \frac{11}{3} \]
   Antidifferentiate f′; add a constant C₀.✓ Proved
4. \[ - \left. \frac{x^{5}}{5} - \frac{11 x}{3} + \frac{4 x^{\frac{5}{2}}}{15} \right|_{\substack{ x=1 }} - 4 = - \frac{4}{5} \]
   f(1) = -4 fixes C₀.✓ Proved
5. \[ \frac{x^{5}}{5} - \frac{11 x}{3} + \frac{4 x^{\frac{5}{2}}}{15} - \frac{4}{5} = \frac{4 x^{\frac{5}{2}}}{15} + \frac{x^{5}}{5} - \frac{11 x}{3} - \frac{4}{5} \]
   The solution.✓ Proved