Toán nguyên hàm của hàm số f(x)=1 /3*x^3 -2*x^2 +x-2019 giúp mink voi 05/08/2021 By Allison nguyên hàm của hàm số f(x)=1 /3*x^3 -2*x^2 +x-2019 giúp mink voi
`f(x)=1/3x^3-2x^2+x-2019` `*` $\int f(x) dx = \int(\frac{1}{3}x^3-2x^2+x-2019) dx$ $= \int \frac{1}{3}x^3dx – \int 2x^2dx+ \int xdx – \int 2019$ $=\frac{1}{3} .\frac{x^4}{4}- 2. \frac{x^3}{3}+\frac{x^2}{2}-2019x +C$ $=\frac{x^4}{12}-\frac{2x^3}{3}+\frac{x^2}{2}-2019x+C$ Trả lời
u'(x) = f(x) = 1 /3*x^3 -2*x^2 +x-2019
=> u(x) = 1/12*x^4 – 2/3*x^3 + 1/2*x^2 – 2019x + C
`f(x)=1/3x^3-2x^2+x-2019`
`*`
$\int f(x) dx = \int(\frac{1}{3}x^3-2x^2+x-2019) dx$
$= \int \frac{1}{3}x^3dx – \int 2x^2dx+ \int xdx – \int 2019$
$=\frac{1}{3} .\frac{x^4}{4}- 2. \frac{x^3}{3}+\frac{x^2}{2}-2019x +C$
$=\frac{x^4}{12}-\frac{2x^3}{3}+\frac{x^2}{2}-2019x+C$