Question

高等数学，这个题怎么做？

Answer 1

令：x=π-t
∫(0,π) xsin^mxdx
=∫(π,0) (π-t)sin^m(π-t)d(π-t)
=∫(0,π) πsin^mtdt -∫(0,π) tsin^mtdt
= π∫(0,π) sin^mtdt -∫(0,π) xsin^mxdx
【此项左移】
= π/2∫(0,π) sin^mtdt
= π/2∫(0,π/2) sin^mtdt + π/2∫(π/2，π) sin^mtdt
令：x=π-t
= π/2∫(0,π/2) sin^mtdt + π/2∫(0，π/2) sin^mxdx
= π∫(0，π/2) sin^mxdx
① = π[(m-1)!!/m!!] m为奇数；
② = π²/2 [(m-1)!!/m!!] m为偶数。

Answer 2

使用一次洛必达法则,再使用导数的定义
lim(h→0) [f(x+2h)－2f(x+h)＋f(x)]/h^2
＝lim(h→0) [2f'(x+2h)－2f'(x+h)]/(2h)
＝lim(h→0) [f'(x+2h)－f'(x+h)]/h
＝lim(h→0) ｛2×[f'(x+2h)－f'(x)]/(2h)－[f'(x+h)－f'(x)]/h｝
＝2×lim(h→0)[f'(x+2h)－f'(x)]/(2h)－lim(h→0)[f'(x+h)－f'(x)]/h
＝2×f''(x)－f''(x)
＝f''(x)