Definite Integration Question 34
Question 34
- The value of $\int_{0}^{2 \pi}[2 \sin x] d x$, where [.] represents the greatest integral functions, is (a) $-\frac{5 \pi}{3}$ (b) $-\pi$ (c) $\frac{5 \pi}{3}$ (d) $-2 \pi$
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Solution:
- It is a question of greatest integer function. We have, subdivide the interval $\pi$ to $2 \pi$ as under keeping in view that we have to evaluate $[2 \sin x]$
We know that, $\sin \frac{\pi}{6}=\frac{1}{2}$
$$ \begin{array}{lll} \therefore & \sin \pi+\frac{\pi}{6} & =\sin \frac{7 \pi}{6}=-\frac{1}{2} \ \Rightarrow & & \sin \frac{11 \pi}{6}=\sin 2 \pi-\frac{\pi}{6}=-\sin \frac{\pi}{6}=-\frac{1}{2} \ \Rightarrow & \sin \frac{9 \pi}{6}=\sin \frac{3 \pi}{6}=-1 \end{array} $$
Hence, we divide the interval $\pi$ to $2 \pi$ as
$$ \begin{aligned} & \pi, \frac{7 \pi}{6}, \frac{7 \pi}{6}, \frac{11 \pi}{6}, \frac{11 \pi}{6}, 2 \pi \ & \sin x=0,-\frac{1}{2},-1,-\frac{1}{2},-\frac{1}{2}, 0 \ & \Rightarrow \quad 2 \sin x=(0,-1),(-2,-1),(-1,0) \ & \Rightarrow \quad[2 \sin x]=-1 \ &= \int_{\pi}^{7 \pi / 6}[2 \sin x] d x+\int_{7 \pi / 6}^{11 \pi / 6}[2 \sin x] d x \ &+\int_{11 \pi / 6}^{2 \pi}[2 \sin x] d x \ &= \int_{\pi}^{7 \pi / 6}(-1) d x+\int_{7 \pi / 6}^{11 \pi / 6}(-2) d x+\int_{11 \pi / 6}^{2 \pi}(-1) d x \ &=-\frac{\pi}{6}-2 \frac{4 \pi}{6}-\frac{\pi}{6}=-\frac{10 \pi}{6}=-\frac{5 \pi}{3} \end{aligned} $$
