The value of $\int_{- 1}^{15}{sgn\ \left( \left\{ x \right\} \right)dx,}$ where $\{ \bullet \}$ denotes the fractional part function, is

Question

Dr. Abubacker · Accepted Answer

(b)
$\int_{- 1}^{15}{sgn\left( \left\{ x \right\} \right)dx = \int_{0}^{16}{sgn\left( \left\{ x - 1 \right\} \right)dx}}$ (by property)
$= \int_{0}^{16}{sgn\left( \left\{ x \right\} \right)dx}$
$= 16\int_{0}^{1}{sgn\left( \left\{ x \right\} \right)dx =}16\int_{0}^{1}{sgn(x)dx}$
$= 16\int_{0}^{1}{1 \bullet dx = 16}$

Q. The value of $\int_{- 1}^{15}{sgn\ \left( \left\{ x \right\} \right)dx,}$ where $\{ \bullet \}$ denotes the fractional part function, is

Explanation

Reviewed by Dr. Abubacker

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