Question

${\int }_{\frac{\pi }{2}}^{\frac{3\pi }{2}}$ [2 sinx] dx, where [.] denotes G.I.F. $\le$ x

Answer 1

We have,

${\int }_{\frac{\pi }{2}}^{\frac{3\pi }{2}}\left[2\sin x\right]dx$where $[.]$ denotes the greatest integer function.

We know that,

$2to-2$

Then,

$\text{between}\text{limit}(2,1,0,-1,-2)$

$\downarrow \downarrow \downarrow \downarrow \downarrow$

$\text{Then,}\frac{\pi }{2},\frac{5\pi }{6},\pi ,\frac{7\pi }{6},\frac{3\pi }{2}$

Applying limits and solve that,

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

$={\int }_{\frac{\pi }{2}}^{\frac{5\pi }{2}}1dx+{\int }_{\frac{5\pi }{2}}^{\pi }0dx+{\int }_{\pi }^{\frac{7\pi }{6}}(-1)dx+{\int }_{\frac{7\pi }{6}}^{\frac{3\pi }{2}}(-2)dx$

$=(\frac{5\pi }{2}-\frac{\pi }{2})+(-\frac{7\pi }{6}+\pi )+(-2\times )(\frac{3\pi }{2}-\frac{7\pi }{6})$

$=\frac{4\pi }{2}-\frac{\pi }{6}-\frac{4\pi }{6}$

$=\frac{12\pi -\pi -4\pi }{6}$

$=\frac{7\pi }{6}$

Hence, this is the answer.