Let $f(x) = x - \lbrack x\rbrack$, for every real $x$, where [$x$]is the greatest integer less than or equal to $x$. Then, $\int_{- 1}^{1}{f(x)dx}\ is$

Question

Dr. Abubacker · Accepted Answer

(a)
$\int_{- 1}^{1}{\left( x - \lbrack x\rbrack \right)dx = \int_{- 1}^{0}{(x + 1)dx + \int_{0}^{1}{(x - 0)dx}}}$
$= \left\lbrack \frac{x^{2}}{2} + x \right\rbrack_{- 1}^{0} + \left\lbrack \frac{x^{2}}{2} \right\rbrack_{0}^{1} = \frac{1}{2} + \frac{1}{2} = 1$

Q. Let $f(x) = x - \lbrack x\rbrack$, for every real $x$, where [$x$]is the greatest integer less than or equal to $x$. Then, $\int_{- 1}^{1}{f(x)dx}\ is$

Explanation

Reviewed by Dr. Abubacker

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