Let [t] denote the greatest integer less than or equal to t. Then the value of the integral ∫−3101([sin(πx)]+e[cos(2πx)])dx is equal to
Solutions · 1
Official solution
[sinπx] has period 2 and e[cos2πx] has period 1; the combined integrand has period 2, and 101−(−3)=104=52×2:
\[I = 52\int_0^2\left([\sin\pi x]+e^{[\cos2\pi x]}\right)dx\]
Breaking into subintervals where each bracket is constant (using [sinπx]=−1 on (1,2) except a set of measure zero where it's 0, and [cos2πx]=−1 near x=1/4,3/4,…, =0 elsewhere):
\[I = 52\left[\int_1^2(-1)\,dx + \int_{1/4}^{3/4}e^{-1}dx+\int_{5/4}^{7/4}e^{-1}dx + \int_0^{1/4}e^0dx+\int_{3/4}^{5/4}e^0dx+\int_{7/4}^2e^0dx\right] = \frac{52}e\]