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

1 Answer
Jul 15, 2015

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).