Antiderivatives with initial conditions

Problem 3.526 · hard

1. \[ \frac{d}{d x} \left(2 x^{\frac{3}{2}} - \frac{2 x^{3}}{3}\right) = 3 \sqrt{x} - 2 x^{2} \]
   An antiderivative of f″; f′ is this plus a constant C₁.✓ Proved
2. \[ - \left. - \frac{2 x^{3}}{3} + 2 x^{\frac{3}{2}} \right|_{\substack{ x=1 }} - 4 = - \frac{16}{3} \]
   f′(1) = -4 fixes C₁.✓ Proved
3. \[ \frac{d}{d x} \left(\frac{4 x^{\frac{5}{2}}}{5} - \frac{x^{4}}{6} - \frac{16 x}{3}\right) = 2 x^{\frac{3}{2}} - \frac{2 x^{3}}{3} - \frac{16}{3} \]
   Antidifferentiate f′; add a constant C₀.✓ Proved
4. \[ 4 - \left. - \frac{x^{4}}{6} - \frac{16 x}{3} + \frac{4 x^{\frac{5}{2}}}{5} \right|_{\substack{ x=1 }} = \frac{87}{10} \]
   f(1) = 4 fixes C₀.✓ Proved
5. \[ - \frac{x^{4}}{6} - \frac{16 x}{3} + \frac{4 x^{\frac{5}{2}}}{5} + \frac{87}{10} = \frac{4 x^{\frac{5}{2}}}{5} - \frac{x^{4}}{6} - \frac{16 x}{3} + \frac{87}{10} \]
   The solution.✓ Proved