设f(x)连续,且f(0)=1,令F(t)=f(x2+y2dxdy(t≥0),求F’’(0).

admin2018-01-23  20

问题 设f(x)连续,且f(0)=1,令F(t)=f(x2+y2dxdy(t≥0),求F’’(0).

选项

答案由F(t)=∫0dθ∫0trf(r2)dr=2π∫0trf(r2)dr=π∫2t2f(u)du, 得F’(t)=2πtf(t2),F’(0)=0, [*]=2πf(0)=2π.

解析
转载请注明原文地址:https://kaotiyun.com/show/UfX4777K
0

最新回复(0)