If $I_{k} = \int_{- 2k\pi}^{2k\pi}{|\sin{x|\lbrack\sin{x\rbrack\ dx,\ \forall k \in N,}}}$ where $\lbrack \bullet \rbrack$ denotes the greatest integer function, then $\sum_{k = 1}^{10}I_{k}$ is equal to

Question

Dr. Abubacker · Accepted Answer

(d)
$\because I_{k} = \int_{- 2k\pi}^{0}{|\sin{x|\lbrack\sin{xbrack\ dx + \int_{0}^{2k\pi}{|\sin{x|\lbrack\sin{xbrack\ dx}}}}}}$
$= - \int_{2k\pi}^{0}{|\sin{x|\lbrack - \sin{xbrack\ dx}}} + \int_{0}^{2k\pi}{|\sin{x|\lbrack\sin{xbrack\ dx}}}$
$= \int_{0}^{2k\pi}{|\sin{x|(\lbrack\sin{xbrack + \lbrack - \sin{xbrack)}dx}}}$
$= 2k\int_{0}^{\pi}{|\sin{x|(\lbrack\sin{xbrack + \lbrack - \sin{xbrack)}\ dx}}}$
$= 2k\int_{0}^{\pi}{\sin{x(0 - 1)dx}}$
$= - 2k\lbrack - {\cos{xbrack}}_{0}^{\pi}\ \ \ \ \ \ \ \ \ \ \ \ \begin{pmatrix}
\because\sin{x > 0,\ x \in (0,\ \pi)} \
	herefore\  - \sin{x < 0,\ x \in (0,\pi)}
\end{pmatrix}$
$= - 4k$
$	herefore\ \ \sum_{k = 1}^{10}{I_{k} = - 4}\sum_{k = 1}^{10}{k = - 4 \bullet \frac{10.11}{2} = - 220}$

Q. If $I_{k} = \int_{- 2k\pi}^{2k\pi}{|\sin{x|\lbrack\sin{x\rbrack\ dx,\ \forall k \in N,}}}$ where $\lbrack \bullet \rbrack$ denotes the greatest integer function, then $\sum_{k = 1}^{10}I_{k}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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