Question

设f'(e^x)=x,求∫x²f(x)dx

Answer 1

f'(e^x)=x, 设 y = e^x ，则 x = ln y, 则 f'(y)=ln y
所以f(y) = y * lny - 1 （y>0)
从0积到正无穷
因为x^3*lnx = (1/4*x^4*ln x - 1/16*x^4)的导数
设q(x) = x^3*lnx - x^2
Q(x) = (1/4*x^4*ln x - 1/16*x^4 - 1/3*x^3)
则Q'(x) = q(x)
∫x²f(x)dx = ∫x²（x * lnx - 1）dx = ∫(x^3*lnx - x^2)dx
= ∫Q'(x)dx = Q(+∞) - Q(0)
Q(x—>0) = lim ln x/(4*x^-4) = lim (1/x)/(4*(-4)*x^-5) = lim -x^4/16 = 0
∫x²f(x)dx = Q(+∞)
求Q(+∞)的极限