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

Question

1918 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-21T14:39:30

See the explanation section, below.

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

#f'(x) = 4x-3x^2-10x^4 +C# (for constant #C#)

# = C+4x-3x^2-10x^4# (for constant #C#)

It is found using the power rule for antiderivatives (and the constant multiple rule):
the antiderivative of #kx^n# is #kx^(n+1)/(n+1)#

The function #f# is the antiderivative of #f'(x)#, which is:

#f(x) = 2x^2-x^3-2x^5+Cx+D# (for constants #C# and #D#)

# =D+Cx+ 2x^2-x^3-2x^5# (for constants #C# and #D#)