xfsinx在0到派的积分推导

鲍莘背1407若函数f(x)在[0,1]上连续,证 积分(0,兀)Xf(sinX)dX=兀/2积分(0,兀)f(sinX)dX -
桓筠隶13151223024 ______ ∫(0,pi)xf(sinx)dx=(令y=pi-x)∫(pi,0) (pi-y)f(siny)(-dy)= ∫(0,pi) (pi-x)f(sinx)dx=pi*∫(0,pi) f(sinx)dx-∫(0,pi)xf(sinx)dx 故∫(0,pi)xf(sinx)dx=pi/2*∫(0,pi) f(sinx)dx

鲍莘背1407证明:若函数f(x)在[0,1]上连续,则∫xf(sinx)dx=π/2∫f(sinx)dx (上限 π,下限 0) -
桓筠隶13151223024 ______ 令u=π-x,du=-dx,u:π--->0,则 ∫[0--->π] xf(sinx)dx=-∫[π--->0] (π-u)f(sin(π-u))du=∫[0--->π] (π-u)f(sinu)du=π∫[0--->π] f(sinu)du-∫[0--->π] uf(sinu)du 积分变量可随便换字母=π∫[0--->π] f(sinx)dx-∫[0--->π] xf(sinx)dx 将 -∫[0--->π] xf(sinx)dx 移到等式左边与左边合并,然后除去系数 ∫[0--->π] xf(sinx)dx=π/2∫[0--->π] f(sinx)dx

鲍莘背1407证明∫(上π,下0)xf(sinx)dx=π/2∫(上π,下0)f(sinx)dxf(x)在区间[0,1]连续 -
桓筠隶13151223024 ______[答案] ∫(上π,下π/2)xf(sinx)dx=(令t=x-π/2)=∫(上π/2,下0)(t+π/2)f(sint)dt=∫(上π/2,下0)tf(sint)dt+π/2∫(上π/2,下0)f(sint)dtπ/2∫(上π,下π/2)f(sinx)dx=(令t=x-π/2)=π/2∫(上π/2,下0)f(sint)dt看清楚了...

鲍莘背1407函数f(x)=xsinx在(0,π)上有最大值吗 -
桓筠隶13151223024 ______ 对其求导,在0到π/2上单调递增,所以在π/2上取到最大值f(x)max=π/2

鲍莘背1407根据公式(积分区间为0到π)∫xf(sinx)dx=(π/2)∫f(sinx)dx,为什么算不出来?(积分区间为0到π)∫xsin2xdx.按照分部积分计算出来记过是 - π/2但是如果用上面... -
桓筠隶13151223024 ______[答案] f(sinx)不能表示出 sin2x sin2x=2sinx*cosx 积分后,sin2x=2sinx*(1-(sinx)^2)^0.5 和-2sinx*(1-(sinx)^2)^0.5 积分区间[0,pi] 内,被分为两个区间,2sinx*(1-(sinx)^2)^0.5在[0,pi/2]-2sinx*(1-(sinx)^2)^0.5在[pi/2,p...

鲍莘背1407定积分证明题 ——请证明:【积分区间为0到π】∫xf(sinx)dx=(π/2)∫f(sinx)dx -
桓筠隶13151223024 ______[答案] 移到一边,积分限内: (x-π/2)f(sinx) 令x-π/2=p pf(Cosp),P积分限为-π/2至π/2,p为奇函数,f(Cosp)为偶函数,pf(Cosp)为奇函数,对称区间中积分为0.

鲍莘背1407∫(0,pai) x(sinx)^3dx -
桓筠隶13151223024 ______[答案] 用公式∫(0~π) xf(sinx) dx = (π/2)∫(0~π) f(sinx) dx会快很多. 设f(sinx) = sin³x 则∫(0~π) xsin³x dx = ∫(0~π) xf(sinx) dx = (π/2)∫(0~π) f(sinx) dx = (π/2)∫(0~π) sin³x dx = (π/2)∫(0~π) (cos²x - 1) dcosx = (π/2)[(1/3)cos³x - cosx] |(0~π) = (π/2)[(1/3)(-1) - (-1)] - (π/2)...

鲍莘背1407设f(x)=∫sint/π - tdt(0→x),求∫f(x)dx(0→π) -
桓筠隶13151223024 ______[答案] 记f'(x) = sinx/(π - x) ∫(0~π) f(x) dx = xf(x) - ∫(0~π) xf(x)' dx、 = πf(π) - ∫(0~π) x · sinx/(π - x) dx = π∫(0~π) sint/(π - t) dt - ∫(0~π) xsinx/(π - x) dx = π∫(0~π) sinx/(π - x) dx - ∫(0~π) xsinx/(π - x) dx = ∫(0~π) (πsinx - xsinx)/(π - x) dx = ∫(0~π) (π - x)sinx/(π - x) dx = ∫(0~π) ...

鲍莘背1407函数f(x)=sinx在[0,π/2]的平均值怎么求! -
桓筠隶13151223024 ______[答案] 用微积分...就是求它的ANTI DERIVATIVE...也就是-COS 派+COS 0 也就是2...然后除以这个X值的宽度派/2得到的就是4/派

鲍莘背1407f(x)是[1, - 1]上连续的偶函数,则∫(π, - π)xf(sinx)dx= -
桓筠隶13151223024 ______[答案] xf(sinx)是对称区间[π,-π]上的奇函数,故积分值为0.