First fundamental theorem of calculus uses a function F ( x ) = ∫ a x f ( t ) d t for f a continuous function between [ a , b ] where x is between [ a , b ] F ( x ) is an antiderivative of function f.So what closed interval is considered when we take indefinite integral?

Question

First fundamental theorem of calculus uses a function  F  (  x  )  =         ∫    a    x    f      (    t    )      d  t  for   f a continuous function between   [  a  ,  b  ] where   x is between   [  a  ,  b  ]  F  (  x  )  is an antiderivative of function   f.So what closed interval is considered when we take indefinite integral?

odvucimo1pp17 · Accepted Answer

An indefinite integral does not need a closed interval to integrate over since &quot;indefinite integral&quot; is just a way to say &quot;antiderivative&quot;. An antiderivative of a function   f is just a function whose derivative is   f, with no requirement that you say how you found it.Often you can just guess an antiderivative based on what you know about derivatives. For example, an antiderivative for the function   f  (  x  )  =  3      x    2   is the function   F  (  x  )  =      x    3  . There&#039;s nothing about integration in that fact.The fundamental theorem of calculus says that even when you can&#039;t guess an antiderivative you can still find one by calculating lots of definite integrals (the hard way, as limits of sums). In particular, in your question, for each value of   x the definite integral of   f over the closed interval   [  a  ,  x  ] is a number you are calling   F  (  x  ). That function   F turns out to have derivative   f.

First fundamental theorem of calculus uses a function F ( x ) = <mtext> </m

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