f^'(x)=3x^3+4x^2+2x-9

Lösung

f^{^{'}} (x) = 3 x^{3} + 4 x^{2} + 2 x - 9

f (x) = \frac{3 x ^{4}}{4} + \frac{4 x ^{3}}{3} + x^{2} - 9 x + c_{1}

Schritte zur Lösung

f^{'} (x) = 3 x^{3} + 4 x^{2} + 2 x - 9

Wenn

f^{^{'}} (x) = g (x)

dann

f (x) = \int g (x) d x

f (x) = \int 3 x^{3} + 4 x^{2} + 2 x - 9 d x

\int 3 x^{3} + 4 x^{2} + 2 x - 9 d x = \frac{3 x ^{4}}{4} + \frac{4 x ^{3}}{3} + x^{2} - 9 x + c_{1}

f (x) = \frac{3 x ^{4}}{4} + \frac{4 x ^{3}}{3} + x^{2} - 9 x + c_{1}

d m o d x - 1 - 2 = m o d x - 3

\frac{y ^{^{''}} + a}{y ^{2}} = 0

f^{^{'}} (x) = - 2 cos (x)

y^{^{'}} - 2 x^{2} = 0

y^{^{'}} = 3, 0.4 - 0.24 y^{2} y (0) = 0