The value of $\int_{- 20 π}^{20 π} [\sin x + \cos x] d x$ is: (Where $[.]$ denotes greatest integer function)

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$20 π$

$10 π$

$- 20 π$

$- 10 π$

answer is B.

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Detailed Solution

$[\sin x + \cos] = [\sqrt{2} \sin (x + \frac{π}{4})] = {\begin{cases} 1 0 \leq x \leq \frac{π}{2} \\ 0 \frac{π}{2} \leq x \leq \frac{3 π}{4} \\ - 1 \frac{3 π}{4} < x < π \\ - 2 π < x \leq \frac{3 π}{2} \\ - 1 \frac{3 π}{2} < x \leq \frac{7 π}{2} \\ 0 \frac{7 π}{4} < x \leq 2 π \end{cases} I = 20 [\int_{0}^{π / 2} 1 d x + \int_{π / 2}^{3 π / 4} 0 d x + \int_{3 π / 4}^{π} - 1 d x + \int_{π}^{3 π / 2} - 2 d x + \int_{3 π / 2}^{7 π / 4} - 1 d x + \int_{7 π / 4}^{2 π} 0 d x] = 20 \times [\frac{π}{2} - π + \frac{3 π}{4} - 2 \times \frac{3 π}{2} + 2 π - \frac{7 π}{4} + \frac{3 π}{2}] = 20 \times - π = - 20 π$