=(1/3)â«(0â1) f(x) d(x-1)^3
=(1/3)f(x)(x-1)^3 (0â1) -(1/3)â«(0â1)(x-1)^3df(x).
(1/3)f(x)(x-1)^3 (0â1)
=(1/3)f(1)(1-1)^3-(1/3)f(0)(x-1)^3
=0-(1/3)(x-1)^3â«(0â0)e^(-y^2+2y)dy
=0-0
=0
f(x)=â«(0âx)e^(-y^2+2y)dy
f'(x)=e^(-x^2+2x)
â´df(x)=e^(-x^2+2x)dx
â´-(1/3)â«(0â1)(x-1)^3df(x).
=-(1/3)â«(0â1)(x-1)^3e^(-x^2+2x)dx
=-(e/3)â«(0â1) (x-1)^3e^[-(x-1)^2]dx
=-(e/6)â«(0â1)(x-1)^2e^[-(x-1)^2d(x-1)^2
=(e/6)â«(0â1)(x-1)^2 d e^[-(x-1)^2 =(e/6)(x-1)^2e^[-(x-1)^2] (0â1)+(e/6)e^[-(x-1)^2](0â1)
=(e/6)e^[-(x-1)^2](x^2-2x+2) (0â1)
=(e-2)/6.
设f(x)=积分(0->x)e^(-y^2+2y)dy,求积分(0->1)[(x-1)^2]f(x)dx_百度...
f(x)=∫(0→x)e^(-y^2+2y)dy f'(x)=e^(-x^2+2x)∴df(x)=e^(-x^2+2x)dx ∴-(1\/3)∫(0→1)(x-1)^3df(x).=-(1\/3)∫(0→1)(x-1)^3e^(-x^2+2x)dx =-(e\/3)∫(0→1) (x-1)^3e^[-(x-1)^2]dx =-(e\/6)∫(0→1)(x-1)^2e^[-(x-1)^2d...
设f(x)=积分(0->x)e^(-y^2+2y)dy,求积分(0->1)[(x-1)^2]f(x)dx_百度...
f(x)=∫(0→x)e^(-y^2+2y)dy f'(x)=e^(-x^2+2x)∴df(x)=e^(-x^2+2x)dx ∴-(1\/3)∫(0→1)(x-1)^3df(x).=-(1\/3)∫(0→1)(x-1)^3e^(-x^2+2x)dx =-(e\/3)∫(0→1) (x-1)^3e^[-(x-1)^2]dx =-(e\/6)∫(0→1)(x-1)^2e^[-(x-1)^2d...
设f(x)= ∫0-x e^(-y+2y)dy 求∫0-1 [(1-x)^2]f(x)dx
是f(x)= ∫0-x e^(-y^2+2y)dy吧 交换积分次序即可 ∫[0,1] [(1-x)^2]f(x)dx =∫[0,1] [(1-x)^2]∫[0,x] e^(-y^2+2y)dydx =∫[0,1]e^(-y^2+2y)∫[y,1] (1-x)^2dxdy 然后自己先算算吧
设F(x,y)=∫(上限xy,下限0) e^(-n^2)du,证明: dF(X,Y)=e^(-x^2y^2...
2016-12-08 设F(x)=∫(上限x,下限0)tf(x^2-t^2)dt,... 19 2011-10-04 设∫(下限0-上限x)e^t^2 dt 则f'(x)=? 1 2018-05-30 f(x,y)=e的x的平方的定积分(下限是0,上限是xy),... 2012-04-23 设y=f(x)是由方程x-积分(上限为y+x,下限为1)e^... 2017-04-09 设f(u)...
求F(y)=∫(0→1)ln(x^2+y^2)dx的导数,y>0
导数的求法如下：导数是函数的局部性质 一个函数在某一点的导数描述了这个函数在这一点附近的变化率。如果函数的自变量和取值都是实数的话，函数在某一点的导数就是该函数所代表的曲线在这一点上的切线斜率。导数的本质是通过极限的概念对函数进行局部的线性逼近。不是所有的函数都有导数，一个函数也不...
函数y=f(x)由方程e^x-y-x^2+2y=2确定,求y'求dy\/dx
de^(x-y)-dx²+d2y=d2 e^(x-y)d(x-y)-2xdx+2dy=0 e^(x-y)*dx-e^(x-y)*dy-2xdx+2dy=0 所以 y'=dy\/dx=[e^(x-y)-2x]\/[e^(x-y)-2]
设∫0到2xf(t\/2)dt=e^(2x-1),求微分方程y''-3y'+2y=f(x)的通解
可以考虑换元法，答案如图所示
...0>1) dy∫(0>2y) f(x,y)dx +∫(1>3) dy∫(0>3-y) f(x,y)dx 写下...
1. 确定积分区域 对本题而言，即{(x,y):0<y<1,0<x<2y}∪{(x,y):1<y<3,x>0,x+y<3} 2. （如果想不清楚的话）在坐标系下画出区域 3.交换时，先确定x的范围，再确定对于给定的x,y的取值范围 结果：∫(0>2)dx∫(x\/2>1)f(x,y)dy+∫(0>2)dx∫(1>3-x)f(x,y)dy ...
∫(0到∏)e^[-(y+1)t]sintdt怎么做这道题啊?求高手。
= 1 + e^[- (y + 1)π] + 0 - (y + 1)²N[1 + (y + 1)²] * N = 1 + e^[- (y + 1)π]==> N = ∫(0→π) e^[- (y + 1)t] sint dt = {1 + e^[- (y + 1)π]} \/ {(y² + 2y + 2)e^[- (y + 1)π]} 已赞过 已踩过< 你对这个回答的...
f(0)=0,f'(o)=1, [xy(x+y)-f(x)y]dx+[f'(x)+x^2y]dy]=0为全微分方程...
所以D[xy(x+y)-f(x)y]\/Dy=D[f'(x)+x^2y]\/Dx, D表偏导 x^2+2xy-f(x)=f''(x)+2xy f''(x)+f(x)=x^2 (*)先找齐次解，f''(x)+f(x)=0,特征根方程 r^2+1=0,r=±i f(x)=Acos x+Bsinx 特解，看着就像二次函数，x^2-2满足方程(*)，没凑出就直接f=ax^2...
