f^'(x)=cos(x),f(0)=-5

Lösung

f^{^{'}} (x) = cos (x), f (0) = - 5

f (x) = sin (x) - 5

Schritte zur Lösung

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

Wenn

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

dann

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

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

\int cos (x) d x = sin (x) + c_{1}

f (x) = sin (x) + c_{1}

Wende die Anfangsbedingung an:

f (x) = sin (x) - 5

f (x) = sin (x) - 5

s l o p e in t erce pt (- 10, 3), (- 8, - 8)

n = 0 \sum \infty (\frac{1}{5})^{n + 1}

\int (e^{x} + 2 x)^{2} d x

d er i v a t i v e f (x) = 5 - 3 x

\int_{- 1}^{1} (x^{2} - x) d x