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Question: The value of \(\int _ { \pi } ^ { 2 \pi } [ 2 \sin x ] d x\) where ![](https://cdn.pureessence.tech/...

The value of ∫π2π[2sin⁡x]dx\int _ { \pi } ^ { 2 \pi } [ 2 \sin x ] d x where represents the greatest integer function, is

A

−π- \pi

B

−2π- 2 \pi

C

−5π3- \frac { 5 \pi } { 3 }

D

5π3\frac { 5 \pi } { 3 }

Answer

−5π3- \frac { 5 \pi } { 3 }

Explanation

Solution

∫π2π[2sin⁡x]dx=∫ππ+(π/6)(−1)dx+∫π+(π/6)π+(π/2)(−2)dx\int _ { \pi } ^ { 2 \pi } [ 2 \sin x ] d x = \int _ { \pi } ^ { \pi + ( \pi / 6 ) } ( - 1 ) d x + \int _ { \pi + ( \pi / 6 ) } ^ { \pi + ( \pi / 2 ) } ( - 2 ) d x

+∫π+(π/2)π+(π/2)+(π/3)(−2)dx+∫π+(π/2)+(π/3)2π(−1)dx+ \int _ { \pi + ( \pi / 2 ) } ^ { \pi + ( \pi / 2 ) + ( \pi / 3 ) } ( - 2 ) d x + \int _ { \pi + ( \pi / 2 ) + ( \pi / 3 ) } ^ { 2 \pi } ( - 1 ) d x

=−π6−2[π2−π6]−2[π3]−1[π2−π3]= - \frac { \pi } { 6 } - 2 \left[ \frac { \pi } { 2 } - \frac { \pi } { 6 } \right] - 2 \left[ \frac { \pi } { 3 } \right] - 1 \left[ \frac { \pi } { 2 } - \frac { \pi } { 3 } \right]

=−π6−2π3−2π3−π6= - \frac { \pi } { 6 } - \frac { 2 \pi } { 3 } - \frac { 2 \pi } { 3 } - \frac { \pi } { 6 } =−π6−8π6−π6=−10π6=−5π3= - \frac { \pi } { 6 } - \frac { 8 \pi } { 6 } - \frac { \pi } { 6 } = - \frac { 10 \pi } { 6 } = - \frac { 5 \pi } { 3 } .