Question

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

Answer 1

It is `f'(x) = 2x+x^4/4+x^7/7 +C` (where `c` is an arbitrary constant).

Explanation:

The antiderivative of `f''(x)=2+x^3+x^6` Is `f'(x)` and is found by reversing the power rule for derivatives.

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

So from `f''(x)=2+x^3+x^6`, we get:

`f'(x)=2x+x^(3+1)/(3+1)+x^(6+1)/(6+1) +C`

`= 2x+x^4/4+x^7/7 +C`

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