On the real line R,we define two functions f | vedclass

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(C) Let ${x} = x - [x]$. Then $f(x) = \min \{\{x\}, 1 - \{x\}\}$ and $g(x) = \max \{\{x\}, 1 - \{x\}\}$.Note that $g(x) - f(x) = |\{x\} - (1 - \{x\})|

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$200$

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(C) Let ${x} = x - [x]$. Then $f(x) = \min \{\{x\}, 1 - \{x\}\}$ and $g(x) = \max \{\{x\}, 1 - \{x\}\}$.Note that $g(x) - f(x) = |\{x\} - (1 - \{x\})| = |2\{x\} - 1|$.The function $h(x) = g(x) - f(x) = |2\{x\} - 1|$ is periodic with period $1$.We calculate the integral over one period $[0, 1]$:$\int_0^1 |2\{x\} - 1| \, dx = \int_0^{1/2} (1 - 2x) \, dx + \int_{1/2}^1 (2x - 1) \, dx$$= [x - x^2]_0^{1/2} + [x^2 - x]_{1/2}^1$$= (1/2 - 1/4) + ((1 - 1) - (1/4 - 1/2)) = 1/4 + 1/4 = 1/2$.Since the function is periodic with period $1$,$\int_0^n h(x) \, dx = n \int_0^1 h(x) \, dx = n 	imes \frac{1}{2}$.Given $\int_0^n (g(x) - f(x)) \, dx = 100$,we have $\frac{n}{2} = 100$,which implies $n = 200$.

On the real line $R$,we define two functions $f$ and $g$ as follows:
$f(x) = \min \{x - [x], 1 - x + [x]\}$
$g(x) = \max \{x - [x], 1 - x + [x]\}$
where $[x]$ denotes the greatest integer function. The positive integer $n$ for which $\int_0^n (g(x) - f(x)) \, dx = 100$ is:

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On the real line $R$,we define two functions $f$ and $g$ as follows:$f(x) = \min \{x - [x], 1 - x + [x]\}$$g(x) = \max \{x - [x], 1 - x + [x]\}$where $[x]$ denotes the greatest integer function. The positive integer $n$ for which $\int_0^n (g(x) - f(x)) \, dx = 100$ is: