Simple question about antiderivative?So this has been confusing me a lot. Let f ( x...

Gretchen Schwartz

Gretchen Schwartz

Answered

2022-07-08

Simple question about antiderivative?
So this has been confusing me a lot. Let f ( x ) = x 2 and let F ( x ) = 1 x f ( x ) d x . Then F ( 1 ) = 0, obviously, but the antiderivative of f (which is the same as 1 x f ( x ) d x ) is F ( x ) = x 3 / 3, so F ( 1 ) = 1 / 3, not 0?
What caused the confusion is that I forgot that a function has infinitely many anti derivatives; they are defined in this case as F ( x ) = x 3 / 3 + C, and in this example F ( x ) = x 3 / 3 1 / 3, x 3 / 3 is just a special antiderivative where C = 0

Answer & Explanation

treccinair

treccinair

Expert

2022-07-09Added 18 answers

Explanation:
It does not matter what C is equal to.
The antiderivative of f is
F ( x ) = 0 x f ( t ) d t = 0 x t 2 d t = ( t 3 3 + C ) | 0 x = ( x 3 3 + C ) ( 0 3 3 + C ) = x 3 3
F ( 1 ) = 0 1 f ( t ) d t = 1 / 3
therightwomanwf

therightwomanwf

Expert

2022-07-10Added 4 answers

Step 1
The integral 1 x x 2 d x is equal to
x 3 3 | 1 x .
This is 1 3 x 3 1 3 . Note that 1 3 x 3 1 3 is 0 at x = 1, as you pointed out it should be.
Step 2
Remark: When we are finding definite integrals with limits of integration that involve variables, it is best to use a different "dummy" variable of integration. Things may be clearer if you write
F ( x ) = 1 x t 2 d t .
While in principle the notation you used is not wrong, it is, for good reason, frowned on.
Note also that talking about "the" antiderivative of a function f(t) is not quite right. The function has infinitely many (very closely related) antiderivatives.

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