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What is the antiderivative of # (x-3)^2#?

Answer 1

Evelyn Agard

It can be written as: #1/3(x-3)^3+C# or as #1/3x^x-3x^2+9x +C#.
(The #C#'s are different.)

Expansion :

#(x-3)^2 = x^2-6x+9#

Antidifferentiate term by term to get:

#1/3x^3-1/2(6x^2)+9x+C# which simplifies to:

#1/3x^x-3x^2+9x +C#.

Substitution Let #u=(x-3)#, so #(du)/dx = 1# and we have #u^2 (du)/dx#

and the antiderivative (by reversing the chain rule) of #u^2 (du)/dx#

is #1/3 u^3 +C#.

Replacing #u# by #x-3# gets us:

#1/3 (x-3)^3 +C#.

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Answer 2

Scarlett Allison

The antiderivative of ( (x - 3)^2 ) is ( \frac{1}{3}(x - 3)^3 + 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.