f^'(x)=sqrt(x)(3+5x),f(1)=7

解

f^{^{'}} (x) = x (3 + 5 x), f (1) = 7

f (x) = 2 x^{\frac{3}{2}} + 2 x^{\frac{5}{2}} + 3

解答ステップ

f^{'} (x) = x (3 + 5 x)

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

であれば

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

f (x) = \int x (3 + 5 x) d x

\int x (3 + 5 x) d x = 2 x^{\frac{3}{2}} + 2 x^{\frac{5}{2}} + c_{1}

f (x) = 2 x^{\frac{3}{2}} + 2 x^{\frac{5}{2}} + c_{1}

初期条件を適用する:

f (x) = 2 x^{\frac{3}{2}} + 2 x^{\frac{5}{2}} + 3

f (x) = 2 x^{\frac{3}{2}} + 2 x^{\frac{5}{2}} + 3