Let f(x) = x - [x], for every real number x, | vedclass

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(A) The function $f(x) = x - [x]$ is the fractional part function,denoted as $\{x\}$.For the interval $[-1, 1]$,we split the integral at $x = 0$:$\int

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(A) The function $f(x) = x - [x]$ is the fractional part function,denoted as $\{x\}$.For the interval $[-1, 1]$,we split the integral at $x = 0$:$\int_{-1}^{1} f(x) \, dx = \int_{-1}^{0} f(x) \, dx + \int_{0}^{1} f(x) \, dx$.For $-1 \le x For $0 \le x Now,calculate the integrals:$\int_{-1}^{0} (x + 1) \, dx = \left[ \frac{x^2}{2} + x \right]_{-1}^{0} = (0 + 0) - \left( \frac{1}{2} - 1 \right) = -(-1/2) = 1/2$.$\int_{0}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{1}{2} - 0 = 1/2$.Adding these results: $1/2 + 1/2 = 1$.

Let $f(x) = x - [x],$ for every real number $x,$ where $[x]$ is the integral part of $x.$ Then $\int_{-1}^{1} f(x) \, dx =$

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