已知∫[x^2,0]xf(t)dt,求d^2y/dx^2
已知∫[x^2,0]xf(t)dt,求d^2y/dx^2
设f(x)连续 则d∫(0,2x)xf(t)dt/dx=?
f(x)=∫(0,x^2) e^(-t^2)dt,求∫(0,1)xf(x)dx
定积分,f(x)=∫(1,x^2)e^-t^2dt,求 ∫(0,1)xf(x)dx
f(x)=∫(x^2,1)sint/t dt,求∫(1,0)xf(x)dx
设f(x)=∫(1,x^2) e^(-t)/t dt,求∫(0,1)xf(x)dt
变限积分计算已知f(x)=∫(上限x^2下限1)e^(-t^2)dt,计算∫(上限1下限0)xf(x)dx
以知f(x)=∫(sint/t)dt(从1到t^2)求∫xf(x)dx(从0到1)
设f(x)=∫(x^2到0) sint/t dt ,求 ∫(1到0 )xf(x) dx=
已知 x=e^t ,dy/dx=dy/xdt .分析变换具体步骤 d^2y/dx^2=(d^2y/dt^2-dy/dt)
求d/dx (∫[0,x](根号(1+t^2)dt)=?
求d/dx{∫cos(t^2)dt}