计算d/dx∫(x,0)(x/(1+t^2)dt)
计算d/dx∫(x,0)(x/(1+t^2)dt)
求导!d/dx∫[0,x^2]根号(1+t^2)dt
求d/dx (∫[0,x](根号(1+t^2)dt)=?
d /dx ∫ 上x^3 下0 (√(1+t^2)) dt = 判断对错,
设f(x)连续 则d∫(0,2x)xf(t)dt/dx=?
d/dx定积分(0~x^2) (1+t^2)^(1/2)dt d/dx定积分(0~x^2)(x^1/2)cost^2dt
已知∫[x^2,0]xf(t)dt,求d^2y/dx^2
d/dx∫[x^2→0]xsin(t^2)dt
设函数为连续函数,则d/dx∫(x----0)f(2t)dt=?
d/dx∫(1,e^-x)f(t)dt=e^x,则f(x)=-x^(-2)
d/dx积分号(0~x^2)1/(1+t^2)dt=?
y= ∫[0,x](t-1)^3(t-2)dt,dy/dx(x=0)