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what is difference between:

$$\int \sqrt {(a^2-x^2)}dx,$$


$$\int_{x_1}^{x_2} \sqrt {(a^2-x^2)}dx$$

with mathematical solution

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I don't see why this question has quite so many downvotes. – Clive Newstead Jan 25 '13 at 0:46
@CliveNewstead At a glance, I mistook this question for "Please solve my Calculus homework." Perhaps others have done the same. – Austin Mohr Jan 25 '13 at 0:48
@Clive Newstead this is exactly the question that i always ask myself when i use stackexchange – Neo Jan 25 '13 at 8:58
@Austin Mohr funny part is that no one really solve it, – Neo Jan 25 '13 at 9:01
I don't see why this question has quite so many features contrary to what constitutes a useful, adapted to the site, question: what the OP tried, what they know, which related questions they can solve, why they think that what they tried failed, and so on. Coming from somebody with already 20 questions asked, this could be mistook for open flouting. – Did Jan 25 '13 at 13:40
up vote 3 down vote accepted

One is a definite integral while one is indefinite. If $F(x)=\int\sqrt{a^2-x^2}dx$, then $\int\limits_{x_1}^{x_2}\sqrt{a^2-x^2}dx=F(x_2)-F(x_1)$

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all answers are right but i like this one – Neo Jan 25 '13 at 0:07
@Neo: So why not accept it? – Clive Newstead Jan 25 '13 at 0:48
feedback: $\int \sqrt {(a^2-x^2)} dx=\frac {1}{2}(x\sqrt {(a^2-x^2)}+a^2\sin^{-1}\frac {x}{a}$ – Neo Jan 25 '13 at 9:07

First is a function $F(x)$, the second is a number, specifically $F(x_2)-F(x_1)$.

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The first exercise requires to compute an antiderivative (a function). The second one asks to compute an integral (a number).

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The first is an indefinite integral; the second is a definite integral. The indefinite integral $\int f(x) dx$ is a function $F$ whose derivative is $f$. (This is only determined up to an additive constant; if $F$ is an antiderivative of $f$, then so is $F+c$.) The definite integral $\int_{a}^{b} f(x) dx$ is the difference $F(b)-F(a)$.

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