Question

Solve ${\int }_{\pi /2}^{3\pi /2}\left[2\sin x\right]dx$

Answer 1

Drawing the graph of $2\sin x$ between $(\frac{\pi }{2},\frac{3\pi }{2})$

At Point A: $2\sin x=2$

At Point B: $2\sin x=0$

At Point C: $2\sin x=-2$

At Point P: $2\sin x=1\Rightarrow \sin x=\frac{1}{2}$ Thus, $x=\pi -\frac{\pi }{6}=\frac{5\pi }{6}$

At Point Q: $2\sin x=-1\Rightarrow \sin x=\frac{-1}{2}$ Thus, $x=\pi +\frac{\pi }{6}=\frac{7\pi }{6}$

Now, Between A and P: $\frac{\pi }{2}\le x\le \frac{5\pi }{6}$

Here $1\le \sin x\le 2$ and $\left[\sin x\right]=1$

Between P and B: $\frac{5\pi }{6}\le x\le \pi$

Here $0\le \sin x\le 0$ and $\left[\sin x\right]=0$

Between B and Q: $\pi \le x\le \frac{7\pi }{6}$

Here $-1\le \sin x\le 0$ and $\left[\sin x\right]=-1$

Between Q and C: $\frac{7\pi }{6}\le x\le \frac{3\pi }{2}$

Here $-2\le \sin x\le -1$ and $\left[\sin x\right]=-2$

Thus ${\int }_{\frac{\pi }{2}}^{\frac{3\pi }{2}}\left[\sin x\right]dx={\int }_{\frac{\pi }{2}}^{\frac{5\pi }{6}}\left[\sin x\right]dx+{\int }_{\frac{5\pi }{6}}^{\pi }\left[\sin x\right]dx+{\int }_{\pi }^{\frac{7\pi }{6}}\left[\sin x\right]dx+{\int }_{\frac{7\pi }{6}}^{\frac{3\pi }{2}}\left[\sin x\right]dx$

${\int }_{\frac{\pi }{2}}^{\frac{3\pi }{2}}\left[\sin x\right]dx=1+0-1-2$

${\int }_{\frac{\pi }{2}}^{\frac{3\pi }{2}}\left[\sin x\right]dx=-2$