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First of all sorry for asking such a dumb question

What is an indefinite integral?

Is it the difference between y=f(a) from f(b)=0? I mean, if f(b)=0 and we are to determine the integral at x=a of f(x) does that mean indefinite integral for this condition is int of f(x) from b to a?

Blue
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MSKB
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    The indefinite integral of a $f(x)$ is the a function whose derivative is $f(x)$ – Physor Mar 09 '21 at 20:02
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    An indefinite integral is an antiderivative. For example, if $f(x)=2x$, then the indefinite integral of $f(x)$ would be $F(x)=x^2+C$ for some real constant $C$. – morrowmh Mar 09 '21 at 20:02

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The indefinite integral of $f(x)$ is the integral without upper and lower bounds, and is defined to be the antiderivative of $f(x)$ so is equal to the general function whose derivative is $f(x)$ . We must put a constant of integration because then all functions with this form of derivative are accounted for.

Fry3141
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  • Can we explain indefinite integral as mentioned in the question? Just curious about this. – MSKB Mar 09 '21 at 20:09
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    Perhaps thinking of the indefinite integral as a family of functions such that the change of slope is determined by $f(x)$. I do not think it makes much sense to define an integral at a point (at $x=a$ as you say) ,as computing a definite integral is a summing of infinitely small areas between 2 $x$ values, so at a point this definition doesn't make sense. – Fry3141 Mar 09 '21 at 20:15
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    But yes, if you know f(a) = d and f(b) = 0 where say a<b, then you can compute the definite integral with upper bound b and lower bound a. – Fry3141 Mar 09 '21 at 20:17