If Ik=∫−2kπ2kπ |sin⁡x|[sin⁡x]dx,∀k∈N, where [.] denotes the greatest integer function, then ∑k=110 Ik is equal to

A
-110
B
-440
C
-330
D
-220

Solution:

$\begin{array}{l} ∵ I_{k} = \int_{- 2 kπ}^{0} | \sin x | [\sin x] dx + \int_{0}^{2 kπ} | \sin x | [\sin x] dx \\ = - \int_{2 kπ}^{0} | \sin x | [- \sin x] dx + \int_{0}^{2 kπ} | \sin x | [\sin x] dx \\ = \int_{0}^{2 kπ} | \sin x | ([\sin x] + [- \sin x]) dx \\ = 2 k \int_{0}^{π} | \sin x | ([\sin x] + [- \sin x]) dx \\ = 2 k \int_{0}^{π} \sin x (0 - 1) dx \\ = - 2 k [- \cos x]_{0}^{π} \\ = - 4 k \\ ∴ \sum_{k = 1}^{10} I_{k} = - 4 \sum_{k = 1}^{10} k = - 4 \cdot \frac{10 \cdot 11}{2} = - 220 \\ ∴ - \sin x < 0, x \in (0, π) \end{array}$

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

Solution:

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