设f(x)连续,Y=∫0~X tf(x^2-t^2)dt 则dy/dx=?

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设f(x)连续,Y=∫0~X tf(x^2-t^2)dt 则dy/dx=?

设f(x)连续,Y=∫0~X tf(x^2-t^2)dt 则dy/dx=?
设f(x)连续,Y=∫0~X tf(x^2-t^2)dt 则dy/dx=?

设f(x)连续,Y=∫0~X tf(x^2-t^2)dt 则dy/dx=?
y = ∫[0,x] t f(x² - t²) dt
令u = x² - t²,du = -2t dt
当t = 0,u = x²;当t = x,u = 0
y = ∫[x²,0] t f(u) * du/(-2t)
= 1/2 ∫[0,x²] f(u) du
dy/dx = 1/2 [2x * f(x²) - 0]
= x f(x²)
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楼上的方法也不错.
令z² = x² - t²,2z dz = -2t dt => dt = -z/t dz
当t = 0,z = x;当t = x,z = 0
∫[0,x] t f(x² - t²) dt
= ∫[x,0] t f(z²) * (-z/t) dz
= ∫[x,0] -z f(z²) dz
= ∫[0,x] z f(z²) dz
dy/dx = x f(x²) - 0
= x f(x²)
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两个方法也行,但楼上的替换怎么又会涉及因变量y呢?

不妨设x>0。x<0时类似做法。做变量替换x^2-t^2=y^2,则t=0时对应y=x,t=x时对应y=0,t=根号(x^2-y^2),dt=-ydy/根号(x^2-y^2)。原积分化为积分(从x到0)根号(x^2-y^2)f(y^2)(-ydy)/根号(x^2-y^2)=积分(从0到x)yf(y^2)dy,因此dY/dx=xf(x^2)。

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