f^'(x)=2sin(6x),f(pi/6)=-5/2

解

f^{^{'}} (x) = 2 sin (6 x), f (\frac{π}{6}) = - \frac{5}{2}

f (x) = - \frac{1}{3} cos (6 x) - \frac{17}{6}

解答ステップ

f^{'} (x) = 2 sin (6 x)

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

であれば

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

f (x) = \int 2 sin (6 x) d x

\int 2 sin (6 x) d x = - \frac{1}{3} cos (6 x) + c_{1}

f (x) = - \frac{1}{3} cos (6 x) + c_{1}

初期条件を適用する:

f (x) = - \frac{1}{3} cos (6 x) - \frac{17}{6}

f (x) = - \frac{1}{3} cos (6 x) - \frac{17}{6}