Let [t] denote the greatest integer less than or equal to t . Then...

Given, ${∫}_{- 3}^{101}\left(\left[sin\pix\right] + {e}^{\left[\cos 2\pix\right]}\right)dx$

Now let ${I}_{1} = {∫}_{- 3}^{101}\left[sin\pix\right]dx$ and ${I}_{2} = {∫}_{- 3}^{101}{e}^{\left[\cos 2\pix\right]}dx$

Now checking periodicity of \left[sin\pix\right]$we get,</p><p>$0 < x < 1 ⇒ \left[sin\pix\right] = 0$</p><p>$1 < x < 2 ⇒ \left[sin\pix\right] = - 1$</p><p>\left[sin\pix\right] \to$ periodic with period 2

So, ${I}_{1} = 52{∫}_{0}^{2}\left[sin\pix\right]dx$

$= 52\left[{∫}_{0}^{1}0 + {∫}_{1}^{2} - 1dx\right] = - 52$

Now checking periodicity of \left[\cos 2\pix\right]$we get,</p><p>$0 < x < \dfrac{1}{4} ⇒ \left[\cos 2\pix\right] = 0$</p><p>$\dfrac{1}{4} < x < \dfrac{1}{2} ⇒ \left[\cos 2\pix\right] = - 1$</p><p>$\dfrac{1}{2} < x < \dfrac{3}{4} ⇒ \left[\cos 2\pix\right] = - 1$</p><p>$\dfrac{3}{4} < x < 1 ⇒ \left[\cos 2\pix\right] = 0$</p><p>So, \left[\cos 2\pix\right]$ has periodicity of $1$

So, {I}_{2} = 104\left({∫}_{0}^{$\dfrac{1}{4}$}{e}^{0}dx + {∫}_{$\dfrac{1}{4}$}^{$\dfrac{3}{4}$}{e}^{- 1}dx + {∫}_{$\dfrac{3}{4}$}^{1}{e}^{0}dx\right)</p><p>⇒ {I}_{2} = 104\left(\dfrac{1}{4} + \dfrac{1}{e} \times \dfrac{1}{2} + \dfrac{1}{4}\right)$</p><p>$⇒ {I}_{2} = 52 + \dfrac{52}{e}$</p><p>So,${I}_{1} + {I}_{2} = \dfrac{52}{e}$or${∫}_{- 3}^{101}\left(\left[sin\pix\right] + {e}^{\left[\cos 2\pix\right]}\right)dx = \dfrac{52}{e}