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July 6th, 2017, 10:36 AM   #1
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Find the definite integral

Find the definite integral $$\int_{0}^{20}|X-[X+.5]|\,dx,$$ (where X is the greatest integer for [X+.5])

Attempt: I found out that if X> or = to X+.5, then we would have X-[X+.5] as our function. And if X<X+.5, then our function would be -(X-[X+.5]), then I thought about adding up two integrals $$\int_{0}^{X+.5}-(X-[X+.5]),dx,$$+$$\int_{X+.5}^{20}(X-[X+.5])\,dx,$$ but was unable to work through this integral.

Last edited by skipjack; July 7th, 2017 at 02:44 AM.
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July 6th, 2017, 11:48 AM   #2
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Quote:
Originally Posted by poopeyey2 View Post
(where X is the greatest integer for [X+.5])
what does this mean?

do you mean that

$[X+.5]$ is The Floor of $X+.5$, usually written $\lfloor X+.5 \rfloor$ ?
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July 6th, 2017, 11:55 AM   #3
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Quote:
Originally Posted by romsek View Post
what does this mean?

do you mean that

$[X+.5]$ is The Floor of $X+.5$, usually written $\lfloor X+.5 \rfloor$ ?
If so if you plot $|X - \lfloor X+.5 \rfloor|$

you'll see it's 20 isosceles triangles each $0.5$ tall, and $1$ wide

The area of each triangle is just $\dfrac 1 2 (1)(0.5) = \dfrac 1 4$

So 20 of these sum to an area of $5$
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July 6th, 2017, 03:34 PM   #4
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Yes, you wrote it correctly.

Last edited by skipjack; July 7th, 2017 at 02:43 AM.
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July 7th, 2017, 03:12 AM   #5
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Your attempt (if X $\small\geqslant$ X + .5, etc.) was obviously incorrect.
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