If the integral ∫010sin2πxex−xdx=αe−1+βe−12+γ , where  α,β,γ are integers and x denotes the greatest integer less than or equal to x ,then the value of α+β+γ is equal to:

Integrand is periodic function with period 1I=10∫01[sin2πx]e[x]dx                           since  x-x=x=10∫01/2Odx+∫1/21-1exdx           In third and fourth quadrant sinx is niegative=10e-1-10e-1/2α=10,β=-10,γ=0⇒α+β+γ=0