# What is the antiderivative of f''(x)=4-6x-40x^3?

Oct 21, 2015

See the explanation section, below.

#### Explanation:

The antiderivative of $f ' ' \left(x\right) = 4 - 6 x - 40 {x}^{3}$ is

$f ' \left(x\right) = 4 x - 3 {x}^{2} - 10 {x}^{4} + C$ (for constant $C$)

$= C + 4 x - 3 {x}^{2} - 10 {x}^{4}$ (for constant $C$)

It is found using the power rule for antiderivatives (and the constant multiple rule):
the antiderivative of $k {x}^{n}$ is $k {x}^{n + 1} / \left(n + 1\right)$

The function $f$ is the antiderivative of $f ' \left(x\right)$, which is:

$f \left(x\right) = 2 {x}^{2} - {x}^{3} - 2 {x}^{5} + C x + D$ (for constants $C$ and $D$)

$= D + C x + 2 {x}^{2} - {x}^{3} - 2 {x}^{5}$ (for constants $C$ and $D$)