What is the antiderivative of f''(x)=2+x^3+x^6?

Jul 15, 2015

It is $f ' \left(x\right) = 2 x + {x}^{4} / 4 + {x}^{7} / 7 + C$ (where $c$ is an arbitrary constant).

Explanation:

The antiderivative of $f ' ' \left(x\right) = 2 + {x}^{3} + {x}^{6}$ Is $f ' \left(x\right)$ and is found by reversing the power rule for derivatives.

The anti derivative of ${x}^{n}$ is ${x}^{n + 1} / \left(n + 1\right)$, and the antiderivative of a constant, $k$ is $k x$. (In both cases we need to allow for a difference of a conastant $C$ when we state the answer.

So from $f ' ' \left(x\right) = 2 + {x}^{3} + {x}^{6}$, we get:

$f ' \left(x\right) = 2 x + {x}^{3 + 1} / \left(3 + 1\right) + {x}^{6 + 1} / \left(6 + 1\right) + C$

$= 2 x + {x}^{4} / 4 + {x}^{7} / 7 + C$

And by repeating the process, we can continue to:
the antiderivative of $f ' \left(x\right)$ is
$f \left(x\right) = {x}^{2} + {x}^{5} / 20 + {x}^{8} / 56 + C x + D$ (where $C$ and $D$ are constants).