Let $$[t]$$ denote the greatest integer less than or equal to $$t$$. Then the value of the integral $$\displaystyle\int_{-3}^{101} ([\sin(\pi x)] + e^{[\cos(2\pi x)]}) dx$$ is equal to

We wish to evaluate the integral $$\displaystyle\int_{-3}^{101}\bigl([\sin(\pi x)] + e^{[\cos(2\pi x)]}\bigr)\,dx$$ where $$[t]$$ denotes the greatest integer function.

First, consider the integral of $$[\sin(\pi x)]$$. Since $$\sin(\pi x)$$ has period 2, we examine one full period on $$[0,2]$$. On $$(0,1)$$ we have $$\sin(\pi x)\in(0,1]$$, so $$[\sin(\pi x)]=0$$ (the single point $$x=\tfrac12$$ where $$\sin(\pi x)=1$$ does not affect the integral). At the endpoints $$x=0$$ and $$x=1$$ the value is zero, so the floor remains 0 there as well. On $$(1,2)$$ we have $$\sin(\pi x)\in[-1,0)$$, giving $$[\sin(\pi x)]=-1$$. Thus the contribution over one period is

$$\int_{0}^{2}[\sin(\pi x)]\,dx = 0\cdot1 + (-1)\cdot1 = -1.$$

The interval from $$x=-3$$ to $$x=101$$ has length $$104$$, which is $$52$$ periods of length 2. Hence

$$\int_{-3}^{101}[\sin(\pi x)]\,dx = 52\times(-1) = -52.$$

Next, consider $$e^{[\cos(2\pi x)]}$$. The function $$\cos(2\pi x)$$ has period 1, so we study $$[0,1]$$. For $$0

$$\int_{0}^{1}e^{[\cos(2\pi x)]}\,dx = e^0\cdot\tfrac14 + e^{-1}\cdot\tfrac12 + e^0\cdot\tfrac14 = \tfrac12 + \tfrac1{2e}.$$

Again using the fact that $$[-3,101]$$ spans $$104$$ periods of length 1, we get

$$\int_{-3}^{101}e^{[\cos(2\pi x)]}\,dx = 104\Bigl(\tfrac12 + \tfrac1{2e}\Bigr) = 52 + \tfrac{52}{e}.$$

Combining both parts gives

$$\int_{-3}^{101}\bigl([\sin(\pi x)] + e^{[\cos(2\pi x)]}\bigr)\,dx = -52 + \Bigl(52 + \tfrac{52}{e}\Bigr) = \tfrac{52}{e}.$$

The answer is Option B: $$\dfrac{52}{e}$$.