设f(x)在区间[0.1]上连续,函数F(x)是上限为x下限为0,tf(cost)的定积分,判断F(x)在[-π/2,π/2]的奇偶性
问题描述:
设f(x)在区间[0.1]上连续,函数F(x)是上限为x下限为0,tf(cost)的定积分,判断F(x)在[-π/2,π/2]的奇偶性
答
F(x) = ∫(0->x) tf(cost) dtF(-x) =∫(0->-x) tf(cost) dtlety= -tdy = -dtt=0,y=0t=-x,y=xF(-x) =∫(0->-x) tf(cost) dt=∫(0->x) (-y)f(cos(-y)) (-dy)=∫(0->x) yf(cosy) dy=F(x)F(x)偶函数