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

Answer 1

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#)

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Answer 2

William Abdalla

The antiderivative of ( f''(x) = 4 - 6x - 40x^3 ) with respect to ( x ) is ( f'(x) = 4x - 3x^2 - 10x^4 + C ), where ( C ) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.