- #1

Byeonggon Lee

- 14

- 2

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\sum\limits_{k=1}^{n}\int_{k-1}^{k}[x]dx = \frac{n(n-1)}{2}

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I'm so confused because as I know, definite integral is possible only when the target function is continuous in closed interval.

In this case, function ##[x]## should be continuous in interval ##[k-1,k]##

but actually [x] is not continuous in ##[k,k-1]##

##[x]=k-1## when ##k-1\leq x<k##, ##[x]=k## when ##x=k##

So I thought that this expression is false,

but in my book's answer, it is true:

"""

Because##[x]=k-1## when ##k-1\leq x<k##, ##[x]=k## when ##x=k##

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\sum\limits_{k=1}^{n}\int_{k-1}^{k}[x]dx

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=\int_{0}^{1}[x]dx+\int_{1}^{2}[x]dx+\int_{2}^{3}[x]dx+...+\int_{n-1}^{n}[x]dx

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=\int_{0}^{1}0dx+\int_{1}^{2}1dx+\int_{2}^{3}2dx+...+\int_{n-1}^{n}(n-1)dx

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=0+\bigg[x\bigg]_{1}^{2}+\bigg[2x\bigg]_{2}^{3}+...+\bigg[(n-1)x\bigg]_{n-1}^{n}

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=0+1+2+...+(n-1)

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=\frac{n(n-1)}{2}

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"""

The book also says that I need to think interval as ##k\leq x<k+1##

and I don't understand how is it possible to omit equal sign at ##x<k+1##