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Here's the definition of an integral from Wikipedia:

Given a function $f$ of a real variable $x$ and an interval $[a, b]$ of the real line, the definite integral

$\int_a^b f(x) \, dx$

is defined informally to be the area of the region in the $xy$-plane bounded by the graph of $f$, the $x$-axis, and the vertical lines $x = a$ and $x = b$, such that area above the $x$-axis adds to the total, and that below the $x$-axis subtracts from the total.

Why do they specify a closed interval? Wouldn't using $(a,b)$ make no difference as the contribution to the integral from the endpoints $a$ and $b$ is zero as points have no width?

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    We had this question again here: http://math.stackexchange.com/questions/461459/is-there-any-notable-difference-between-studying-the-riemann-integral-over-open/464512#4645122013-08-22

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In addition to the facts mentioned in the comments under the question, notice that Riemann sums use the endpoints. (If we were dealing with Lebesgue integrals rather than Riemann integrals, then no attention would be paid to the endpoints.)

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You can take it further and remove any set of measure 0 from the interval (or rather a subinterval on which the function's all-positive or all-negative). That preseres the semantics of the integral, but it wouldn't be clear how the procedure accomplishes it. The real reason it's a closed interval comes down to topology, being able to treat the function as a continuous function from the interval, as a directed path, to the curve. You'll see this if you look at how paths are used and defined abstractly. So, the difference is a homeomorphism from the space uses the whole space.