Question

f(x)在[0,π]上连续，∫0πf(x)sinxdx=∫0πf(x)cosxdx,为什么∫0πf(x)sinxdx=0,怎么求的

Answer 1

f(x)在[0,π]上连续，∫0πf(x)sinxdx=∫0πf(x)cosxdx,为什么∫0πf(x)sinxdx=0
直接代入方程
r(x~2+-2+22)ds
= T4 ds
=16
或将方程参数化然后计算
z2+32+A2=4
x+3+2=0
将＝-x-3代入＾2+y～2+2＝4中
==>x2+y~2+xy=2
(x+y/2)2+(V3y/2)~2=2
fa: y/2 v2c0st
f v3y/2 28int
ニ2
(a =v2cost-(v6/3)sint, da =fv2sint-(V6/3)cost
dt
f y=(2v6/3)sint dy= -2v6/3)cost dt
f x=-v2cost -(v6/3)sint, d: v2sint-( v6/3)cos
dt
0st≤2a
ds =VI(da)2 +(dy)42+(dz)] dt v4 dt =2 dt
(x42+3y~2+22)ds
f(0-2)(v2cost-(v6/3)sint]/2 + [(2v6/3)sint]2
1v2cost-(v6/3)sint1 2) *2 dt
(0→2r)4大2dt
=16x